NCERT Exemplar Class 10 Maths Solutions Chapter 6 Triangles

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NCERT Exemplar Class 10 Maths Solutions Chapter 6 Triangles

Edited By Ravindra Pindel | Updated on Sep 05, 2022 03:42 PM IST | #CBSE Class 10th

NCERT exemplar Class 10 Maths solutions chapter 6 provides the understanding for triangles and the similarity of triangles. It is the first chapter under the unit: Geometry of Class 10. The solutions available under NCERT exemplar Class 10 Maths chapter 6 solutions have been prepared and compiled by our subject matter experts of mathematics with cumulative experience of more than 10 years and equips the student to study NCERT Class 10 Maths effortlessly. These Class 10 Maths NCERT exemplar chapter 6 solutions provide step-by-step detailed solutions to enhance the learning of concepts based on Triangles. The CBSE Syllabus for Class 10 is the building block for these NCERT exemplar Class 10 Maths solutions chapter 6.

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This Story also Contains

NCERT Exemplar Class 10 Maths Solutions Chapter 6: Exercise-6.1
NCERT Exemplar Class 10 Maths Solutions Chapter 6: Exercise-6.2
NCERT Exemplar Class 10 Maths Solutions Chapter 6: Exercise-6.3
NCERT Exemplar Class 10 Maths Solutions Chapter 6: Exercise-6.4
NCERT Exemplar Class 10 Maths Chapter 6 Solutions Important Topics:
NCERT Class 10 Exemplar Solutions for Other Subjects:
NCERT Class 10 Maths Exemplar Solutions for Other Chapters:
Features of NCERT Exemplar Class 10 Maths Solutions Chapter 6:

NCERT Exemplar Class 10 Maths Solutions Chapter 6: Exercise-6.1

Question:1

In Fig., $\angle BAC=90^{o}$ and $AD\perp BC$ . Then,

(a) $BD.CD=BC^{2}$
(b) $AB.AC=BC^{2}$
(c) $BD.CD=AD^{2}$
(d) $AB.AC=AD^{2}$

Answer:

Given :- $\angle BAC=90^{o}$
First of all we have find $\angle DBA$ and $\angle DAC$ in triangles $ABD$ and $ADC$ respectively
In $\Delta ABD$ and $\angle ADC$
$\angle ADB=90^{o}$
$\angle ADC=90^{o}$
In $\Delta ABC$
$\angle BAC=90^{o}$ (given)
Now let us find angle CAD from triangle ADC
Here $\angle ADC+\angle DCA+\angle CAD=180^{o}$
(sum of interior angles of triangle = 180^o)
$90^{o}+\angle C+\angle CAD=180^{o}$
$\angle CAD=180^{o}-90^{o}-\angle C$
$\angle CAD=90^{o}-\angle C \; \; \; \; \; \; \; \; \; \; \; \; \; ....(1)$
Now let us find angle DBA using triangle ABC
Here $\angle ABC+\angle BCA+\angle CAB=180^{o}$
{sum of interior angles of triangle = 180^o}
$\angle ABC+\angle C+90^{o}=180^{o}$
$\angle ABC=90^{o}-\angle C$
$\angle ABC=\angle DBA=90^{o}-\angle C \; \; \; \; \; \; \; ...(2)$
In $\Delta ABD$ and $\Delta ADC$ we get
$\angle ADB=\angle ADC$ {90^o each}
$AD=AD$ {common side}
$\angle CAD=\angle DBA$ {from equation (1) and (2) it is clear that both is $90^{o}-\angle C$ }
$\therefore \Delta ABD\sim \Delta ADC$ (by ASA similarity)
$\therefore \frac{BD}{AD}=\frac{AD}{CD}$
$BD.CD=AD.AD$ {by cross multiplication}
$BD.CD=AD^{2}$
Hence option (C) is correct.

Question:2

The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Then, the length of the side of the rhombus is
(A) 9 cm
(B) 10 cm
(C) 8 cm
(D) 20 cm

Answer:

Answer : [B]

Here ABCD is a rhombus and we know that diagonals of a rhombus are perpendicular bisector of each other
Given: $AC=16 \; cm$ and $BD=12 \; cm$
$\Rightarrow AO=\frac{AC}{2}=\frac{16}{2}=8\; cm$
also $BO=\frac{BD}{2}=\frac{12}{2}=6\; cm$ and $\angle AOB=90^{o}$
Using Pythagoras theorem in $\Delta AOB$ we get
$AB^{2}=AO^{2}+OB^{2}$
$AB^{2}=8^{2}+6^{2}$
$AB^{2}=64+36$
$AB^{2}=100$
$AB=\sqrt{100}$
$AB=10\; cm$
Hence option (B) is correct

Question:3

If $\Delta ABC \sim \Delta EDF$ and $\Delta ABC$ is not similar to $\Delta DEF$ , then which of the following is not true?
(A) $BC . EF = A C. FD$ (B) $AB . EF = AC . DE$
(C) $BC . DE = AB . EF$ (D) $BC . DE = AB . FD$

Answer:

Answer : [C]
Given : $\Delta ABC\sim \Delta EDF$
$\therefore \frac{AB}{ED}=\frac{BC}{DF}=\frac{AC}{EF}$
Taking the first two terms we get
$\therefore \frac{AB}{ED}=\frac{BC}{DF}$
by cross multiply we get
$AB.DF=BC.ED$
Hence option D is correct
Now taking the last two terms we get
$\frac{BC}{DF}=\frac{AC}{EF}\Rightarrow BC.EF=DF.AC$ {By cross multiplication}
Hence option (A) is also correct
Now taking first and last terms, we get
$\frac{AB}{ED}=\frac{AC}{EF}\Rightarrow AB.EF=AC.ED$ {By cross multiplication}
Hence option (B) is also correct
By using congruence properties we conclude that only option C is not true.

Question:4

If in two triangles ABC and PQR, $\frac{AB}{QR}=\frac{BC}{PR}=\frac{CA}{PQ}$ , then
(a) $\Delta PQR\sim \Delta CAB$
(b) $\Delta PQR\sim \Delta ABC$
(c) $\Delta CBA\sim \Delta PQR$
(d) $\Delta BCA\sim \Delta PQR$

Answer:

Answer: [A]
In $\Delta ABC$ and $\Delta PQR$
$\frac{AB}{QR}=\frac{BC}{PR}=\frac{CA}{PQ}\; \; \; \; \; \; \; .......(1)$
(A) If $\Delta PQR \sim \Delta CAB$

Here; $\frac{CA}{PQ}=\frac{CB}{PR}=\frac{AB}{PQ}$
It matches the equation (1) Hence option A is correct.
(B) If $\Delta PQR \sim \Delta ABC$

Here; $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$
It does not match equation (1)
Therefore option (B) is not correct.
(c) $\Delta CBA\sim \Delta PQR$

Here; $\frac{CB}{PQ}=\frac{BA}{QR}=\frac{CA}{PR}$
It does not match equation (1) Hence option (C) is not correct
(D) $\Delta BCA \sim \Delta PQR$

Here; $\frac{BC}{PQ}=\frac{CA}{QR}=\frac{BA}{PR}$
It does not match the equation (1). Hence option (D) is not correct.

Question:5

In Fig., two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, $\angle APB=50^{o}$ and $\angle CDP=30^{o}$ . Then, $\angle PBA$ is equal to

(A) 50° (B) 30° (C) 60° (D) 100°

Answer:

Answer: [D]
In $\Delta APB$ and $\Delta CPD$ .
$\frac{AP}{PD}=\frac{6}{5}$
$\frac{BP}{CP}=\frac{3}{2.5}=\frac{30}{25}=\frac{6}{5}$
$\angle APB=\angle CPD=50^{o}$ {vertically opposite angles}
$\therefore \Delta APB \sim \Delta DPC$ {By SAS similarity criterion}
$\therefore \angle A=\angle D=30^{o}$ {corresponding angles of similar triangles}
In $\Delta APB$
$\angle A+\angle B+\angle APB=180^{o}$ {Sum of interior angles of a triangle is 180°}
$30^{o}+\angle B+50^{o}=180^{o}$
$\angle B=180^{o}-50^{o}-30^{o}$
$\angle B=100^{o}$
$\Rightarrow \angle PBA=100^{o}$
Hence option D is correct.

Question:6

If in two triangles DEF and PQR, $\angle D = \angle Q \; \text {and} \; \angle R = \angle E$ , then which of the following is not true?
(a) $\frac{EF}{PR}=\frac{DF}{PQ}$
(b) $\frac{DE}{PQ}=\frac{EF}{RP}$
(c) $\frac{DE}{QR}=\frac{DF}{PQ}$
(d) $\frac{EF}{RP}=\frac{DE}{QR}$

Answer:

Answer : [B]
In $\Delta DEF$ and $\Delta PQR$
$\angle D=\angle Q$ and $\angle R=\angle E$
Now draw two triangles with the help of given conditions:-

Here $\angle D=\angle Q$ and $\angle E=\angle R$ {Given}
$\therefore \; \; \; \; \Delta DEF\sim \Delta QRP$ {by AA similarity}
$\Rightarrow \angle F=\angle P$ {corresponding angles of similar triangles}
$\therefore \frac{DF}{QP}=\frac{ED}{RQ}=\frac{FE}{PR}\; \; \; \; \; \; \; \; ....(1)$
Here option (A) $\frac{EF}{PR}=\frac{DF}{PQ}$ is true because both the terms are derived from equation (1)
Here option (B) $\frac{DE}{PQ}=\frac{EF}{RP}$ is not true because the first term is not derived from equation (1)
Here option (C) $\frac{DE}{QR}=\frac{DF}{PQ}$ is true because both the terms are derived from equation (1)
Here option D i.e. is also true because both terms are derived from equation (1)
Hence option (B) is not true.

Question:7

In triangles $ABC \; \text {and} \; DEF, \angle B = \angle E, \angle F = \angle C \; \text {and}\; AB = 3 DE$ . Then, the two triangles are
(A) congruent but not similar (B) similar but not congruent
(C) neither congruent nor similar (D) congruent as well as similar

Answer:

Answer : [B]
In $\Delta ABC$ and $\Delta DEF$
$\angle B=\angle E\; \text {and}\; \angle F=\angle C\; \text {also}\; AB=3DE$

In $\Delta ABC$ and $\Delta DEF$
$\angle B=\angle E$ (given)
$\angle F=\angle C$ (given)
$\angle A=\angle D$ (third angle)
Therefore $\Delta ABC\sim \Delta DEF$ (by AAA similarity)
Also, it is given that $AB=3DE$
$\Rightarrow AB\neq DE$
Hence $\Delta ABC$ and $\Delta DEF$ is not congruent because congruent figures have the same shape and same size.
Therefore the two triangles are similar but not congruent.

Question:8

It is given that $\Delta ABC\sim \Delta PQR$ , with $\frac{BC}{QR}=\frac{1}{3}.$ then $\frac{ar\left ( PRQ \right )}{ar\left ( BCA \right )}$ is equal to
(a) $9$
(b) $3$
(c) $\frac{1}{3}$
(d) $\frac{1}{9}$

Answer:

Answer : [A]
Given : $\Delta ABC\sim \Delta PQR$
$\frac{BC}{QR}=\frac{1}{3}$

It is given that $\Delta ABC\sim \Delta PQR$
$\therefore \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}\; \; \; \; \; \; ...(1)$ $\therefore \frac{ar(PRQ)}{ar(BCA)}=\frac{PR^{2}}{AC^{2}}$
[Q If triangles are similar their areas are proportional to the squares of the corresponding sides.]
$\Rightarrow \frac{ar(PRQ)}{ar(BCA)}=\frac{QR^{2}}{BC^{2}}\; \; \; \; \; \; \; ...(2)$ $\text{{using equation (1)}}$
$\frac{QR}{BC}=\frac{3}{1} \; \; \; \; \; \; \; \; \; \left \{ \because \frac{BC}{QR}=\frac{1}{3} \right \}$
$\text{Put}$ $\frac{QR}{BC}=\frac{3}{1}$ $\text{in (2) we get}$
$\therefore \frac{ar(PRQ)}{ar(BCA)}=\frac{3^{2}}{1^{2}}=\frac{9}{1}=9$
Hence option A is correct.

Question:9

It is given theat $\Delta ABC\sim \Delta DFE,\angle A=30^{o},\angle C=50^{o},AB=5\; cm,AC=8\; cm$ and $DF=7.5\; cm.$ then the following is true ?
$(A)$ $DE=12\; cm,\angle E=50^{o}$
$(B)$ $DE=12\; cm,\angle F=100^{o}$
$(C)$ $EF=12\; cm,\angle D=100^{o}$
$(D)$ $EF=12\; cm,\angle D=30^{o}$

Answer:

Answer: [A]
$\Delta ABC \sim \Delta DEF$
$\angle A=30^{o},\angle C=50^{o}$
$AB=5\; cm,AC=8\; cm,DF=7.5\; cm$

$\Delta ABC \sim \Delta DEF$
$\therefore \angle A=\angle D=30^{o}$
$\angle C=\angle E=100^{o}$
$\frac{AB}{DF}=\frac{BC}{FE}=\frac{AC}{DE}\; \; \; \; \; \; \; \; ....(1)$
$In$ $\Delta ABC$
$\angle A+\angle B+\angle C=180^{o}$
$30+\angle B+100^{o}=180^{o}$
$\angle B=180^{o}-100^{o}-30^{o}$
$\angle B=50^{o}$
$\Rightarrow \angle B=\angle E=50^{o}$ $\left \{ \because \Delta ABC\sim \Delta DFE \right \}$
Now equate first and the last term of equation (1) we get
$\frac{AB}{DF}=\frac{AC}{DE}$
$\frac{5}{7.5}=\frac{8}{DE}$
$DE=\frac{8 \times 75}{50}=12\; cm$

Question:10

If in triangle ABC and DEF , $\frac{AB}{DE}=\frac{BC}{FD},$ then they will be similar when
$\text{(A)}$ $\angle B=\angle E$
$\text{(B)}$ $\angle A=\angle D$
$\text{(C)}$ $\angle B=\angle D$
$\text{(D)}$ $\angle A=\angle D$

Answer:

Answer : [C]
Given :
$\frac{AB}{DE}=\frac{BC}{FD}$
Here corresponding sides of given triangles ABC and DEF are equal in ratio, therefore, the triangles are similar
Also, we know that if triangles are similar then their corresponding angles are equal
$\Rightarrow \angle A=\angle E$
$\angle B=\angle D$
$\angle C=\angle F$
Hence option C is correct.

Question:11

$\text{If }\Delta ABC\sim \Delta QRP,\frac{ar(ABC)}{ar(PQR)}=\frac{9}{4},AB=18\; cm$ and BC = 15 cm then PR is equal to ?
$\text{(A)}$ $10\; cm$
$\text{(B)}$ $12\; cm$
$\text{(C)}$ $\frac{20}{3}\; cm$
$\text{(D)}$ $8\; cm$

Answer:

Answer : [A]

$\Delta ABC \sim \Delta QRP \; \text {and}\; AB = 18 \; cm, BC = 15 \; cm$
$\therefore \frac{AC}{PQ}=\frac{AB}{QR}=\frac{BC}{RP}$
$\frac{ar(ABC)}{ar\left ( PQR \right )}=\frac{BC^{2}}{RP^{2}}\; \; \; \; \; \; \; \; ....(1)$
[Q If triangles are similar, their areas are proportional to the squares of the corresponding sides.]
$\frac{ar(ABC)}{ar\left ( PQR \right )}=\frac{9}{4}$
[Using equation (1)]
$\Rightarrow \frac{BC^{2}}{RP^{2}}=\frac{9}{4}$
$\frac{(15)^{2}}{RP^{2}}=\frac{9}{4}$
$RP^{2}=\frac{225\times 4}{9}=100$
$RP=\sqrt{100}$
$RP=10\; cm$

Question:12

If S is a point on side PQ of a triangle PQR such that PS = QS = RS, then
$(A)$ $PR .QR = RS^{2}$ $(B)$ $QS^{2} + RS^{2} = QR^{2}$
$(C)$ $PR^{2} + QR^{2} = PQ^{2}$ $(D)$ $PS^{2} + RS^{2} = PR^{2}$

Answer:

Answer : [C]
Given: In triangle PQR
$PS = QS = RS$
$\Rightarrow PS = RS$
$\therefore \angle 1=\angle 2 ...(1)$
(Q If a right angle triangle have equal length of base and height then their acute angles are also equal)
$\text{Similarly}$ $\angle 3=\angle 4 ...(2)$
( Q corresponding angles of equal sides are equal)
$\text{In }\Delta PQR$
$\angle P+\angle Q+\angle R=180^{o}$
$\because$ sum of angles of a triangle is 180°
$\angle 2+\angle 4+\angle 1+\angle 3=180^{o} ....(3)$
Using equation (1) and (2) in (3)
$\angle 1+\angle 3+\angle 1+\angle 3=180$
$2\left ( \angle 1+\angle 3 \right )=180^{o}$
$\Rightarrow \angle R=90^{o}$
$\text{Here }\angle R=90^{o}\therefore \Delta PQR$ $\text{is right angle triangle }$
$PR^{2} + QR^{2} = PQ^{2}$

NCERT Exemplar Class 10 Maths Solutions Chapter 6: Exercise-6.2

Question:1

Is the triangle with sides 25 cm, 5 cm and 24 cm a right triangle? Give reasons for your answer.

Answer:

Answer : [False]
Sides of the triangle are 25cm, 5 cm and 24 cm
According to Pythagoras theorem, the square of the hypotenuse of a right-angle triangle is equal to the sum of the square of the remaining two sides.
$\Rightarrow \left ( 25 \right )^{2}=\left ( 5 \right )^{2}+\left ( 24 \right )^{2}$
$625=25+576$
$625 \neq 601$
Hence the given statement is false.
The triangle with sides 25 cm, 5 cm and 24 cm is not a right-angle triangle.

Question:2

$\text{It is given that}$ $\Delta DEF\sim \Delta RPQ$ . $\text{is it true to say that}$ $\angle D = \angle R$ $\text{and}$ $\angle F = \angle P$ $? Why?$

Answer:

Answer : [False]
$Given :$ $\Delta DEF \sim \Delta RPQ$
If two triangles are similar, then their corresponding angles are equal and their corresponding sides are in the same ratio

$i.e.$
$\frac{DE}{RP}=\frac{EF}{PQ}=\frac{DF}{RQ}$
and their angles are also equal that is
$\angle D=\angle R$
$\angle E=\angle P$
$\angle F=\angle Q$
$\text{In the given statement}$ $\angle D = \angle R$ $\text{and}$ $\angle F = \angle P$
$\text{But according to properties of a similar triangle}$ $\angle F \neq \angle P$ $\text{that is }$ . $\text{Hence the given statement is false. }$

Question:3

A and B are respectively the points on the sides PQ and PR of a triangle PQR such that PQ = 12.5 cm, PA = 5 cm, BR= 6 cm and PB = 4 cm. Is AB || QR?
Give reasons for your answer.

Answer:

Answer : [True]

Given: PQ = 12.5 cm, PA = 5 cm, BR = 6 cm and PB = 4 cm
QA = QP – PA = 12.5 – 5 = 7.5 cm
$\frac{PA}{AQ}=\frac{5}{7.5}=\frac{50}{75}=\frac{2}{3}$
$\frac{PB}{PR}=\frac{4}{6}=\frac{2}{3}$
Here
$\frac{PA}{AQ}=\frac{PB}{PR}$
According to Converse of basic proportionality theorem if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
$\therefore AB\parallel QR$

Question:4

In Fig 6.4, BD and CE intersect each other at the point P. Is $\Delta$ PBC $\sim$ $\Delta$ PDE? Why?

Answer:

Answer : [True]
Given:- BD and CE intersect each other at point P.
$\text{Here, }$ $\angle BPC=\angle EPD$ $\text{[Vertically opposite angle] }$
$\frac{BP}{PD}=\frac{5}{10}=\frac{1}{2}$
$\frac{PC}{PE}=\frac{6}{12}=\frac{1}{2}$
$\frac{BP}{PD}=\frac{PC}{PE}$
Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio.
$\text{Also in }$ $\Delta PBC$ $\text{ and }$ $\Delta PDE$
$\frac{BP}{PD}=\frac{PC}{PE}\; and\; \angle BPC=\angle EPD$
$\therefore \Delta PBC\sim \Delta PDE$
Hence given statement is true.

Question:5

In triangles PQR and MST, $\angle$ P = 55°, $\angle$ Q = 25°, $\angle$ M = 100° and $\angle$ S = 25°. Is $\Delta$ QPR $\sim$ $\Delta$ TSM? Why?

Answer:

Answer: [False]

$\Delta PQR, \angle P + \angle Q + \angle R = 180^{o}$ $\text{ ({Interior angle of triangle}) }$
$55+25+\angle R=180$
$\angle R=180-80$
$\angle R=100^{o}$
$\text{In }$ $\Delta TSM, \angle T + \angle S + \angle M = 180^{o}$ $\text{[Interior angle of the triangle] }$
$\angle T + 25^{o}+100^{o}=180^{o}$
$\angle T =180^{o}-125$
$\angle T =55^{o}$
$\text{In }$ $\Delta PQR$ $\text{and }$ $\Delta TSM$
$\angle P=\angle T$
$\angle Q=\angle S$
$\angle R=\angle M$
Also are know that if all corresponding angles of two triangles are equal then triangles are similar.
$\therefore \Delta PQR\sim \Delta TSM$
Hence given statement is false because $\angle$ QPR is not similar to $\angle$ TSM.

Question:6

Is the following statement true? Why? “Two quadrilaterals are similar, if their corresponding angles are equal”.

Answer:

Answer: [False]
Quadrilateral:- A quadrilateral can be defined as a closed, two-dimensional shape which has four straight sides. We can find the shape of quadrilaterals in various things around us, like in a chessboard, a kite etc. The given statement "Two quadrilaterals are similar if their corresponding angles are equal" is not true because. Two quadrilaterals are similar if and only if their corresponding angles are equal and the ratio of their corresponding sides are also equal.

Question:7

Two sides and the perimeter of one triangle are respectively three times the corresponding sides and the perimeter of the other triangle. Are the two triangles similar? Why?

Answer:

Answer: True

According to question,
$AB = 3PQ \; \; \; \; \; \; ...(1)$
$AC = 3PR \; \; \; \; \;...(2)$
also the perimeter of $\Delta$ ABC is three times the perimeter of $\Delta$ PQR
$AB + BC + CA = 3(PQ + QR + RP)$

$AB + BC + CA = 3PQ + 3QR + 3RP$

$3PQ + BC + 3PR = 3PQ + 3QR + 3PR$ (using eq. (1) and (2))

$BC=3QR\; \; \; \; \; \; .....(3)$
from equation (1), (2) and (3) we conclude that sides of both the triangles are in the same ratio.
As we know that if the corresponding sides of two triangles are in the same ratio, then the triangles are similar by SSS similarity criterion
Hence given statement is true.

Question:8

If in two right triangles, one of the acute angles of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles will be similar? Why?

Answer:

Answer : [True]
Let two right angle triangles are ABC and PQR.

Given:- One of the acute angle of one triangle is equal to an acute angle of the other triangle.
$\text {In}\Delta ABC \; \text {and}\; \Delta PQR$
$\\\angle B = \angle Q = 90^{o}\\\angle C = \angle R = x^{o}$
In $\Delta ABC$
$\angle A + \angle B + \angle C = 180^{o}$
[Sum of interior angles of a triangle is 180°]
$\angle A + 90^{o}+ x^{o} = 180^{o}$
$\angle A + 90^{o}- x^{o} \; \; \; \; \; \; \; \; \; ....(1)$
Also in $\Delta PQR$
$\angle P + \angle Q + \angle R = 180^{o}$
$\angle P + 90^{o}- x^{o} = 180^{o}$
$\angle P = 90^{o}- x^{o} \; \; \; \; \; \; .....(2)$
from equation (1) and (2)
$\angle A=\angle P \; \; \; \; \; \; .....(3)$
from equation (1), (2) and (3) are observed that corresponding angles of the triangles are equal therefore triangle are similar.

Question:9

The ratio of the corresponding altitudes of two similar triangles is 3/5. Is it correct to say that ratio of their areas is 6/5? Why?

Answer:

Answer: [False]
Given:- Ratio of corresponding altitudes of two similar triangles is 3/5
As we know that the ratio of the areas of two similar triangles is equal to the ratio of squares of any two corresponding altitudes.
$\therefore \frac{\text {Area 1}}{\text {Area 2}}=\left ( \frac{\text {Altitude 1}}{\text {Altitude 2}} \right )^{2}$
$\frac{\text {Area 1}}{\text {Area 2}}=\left ( \frac{3}{5} \right )^{2}$ $\left [ Q\frac{\text {Altitude 1}}{\text {Altitude 2}}=\frac{3}{5} \right ]$
$=\frac{9}{25}$
Hence the given statement is false because ratio of areas of two triangles is 9/25 which is not equal to 6/5

Question:10

D is a point on side QR of $\Delta$ PQR such that PD $\perp$ QR. Will it be correct to say that $\Delta$ PQD ~ $\Delta$ RPD? Why?

Answer:

Answer : [False]

In $\Delta$ PQD and $\Delta$ RPD
$PD = PD$ (common side)
$\angle PDQ = \angle PDR$ (each 90^o)
The given statement $\Delta PQD \sim \Delta RPD$ is false.
Because according to the definition of similarity two triangles are similar, if their corresponding angles are equal and their corresponding sides are in the same ratio

Question:11

In Fig., if $\angle$ D = $\angle$ C, then is it true that $\Delta ADE \sim \Delta ACB$ ? Why?

Answer:

Answer : [True]
In $\Delta$ ADE and $\Delta$ ACB
$\angle D=\angle C$ (given)
$\angle A = \angle A$ (common angle)
And we know that if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criteria.
$\therefore$ $\Delta ADE \sim \Delta ACB$ {by AA similarity criterion}
Hence the given statement is true.

Question:12

Is it true to say that if in two triangles, an angle of one triangle is equal to an angle of another triangle and two sides of one triangle are proportional to the two sides of the other triangle, then the triangles are similar? Give reasons for your answer.

Answer:

Answer. [False]
Given:- (1) An angle of one triangle is equal to an angle of another triangle.
(2) Two sides of one triangle are proportional to the two sides of the other triangle.
According to SAS similarity criterion if one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio, then the triangles are similar.
But here, the given statement is false because one angle and two sides of two triangles are equal but these sides not including equal angle.

NCERT Exemplar Class 10 Maths Solutions Chapter 6: Exercise-6.3

Question:1

In a D PQR, PR²–PQ² = QR² and M is a point on side PR such that QM perpendicular to PR.
Prove that : QM2 = PM × MR.
Answer:

Given : $PR^{2} - PQ^{2} = QR^{2} \; and\; QM \perp PR$
To prove : $QM^{2} = PM \times MR$
Proof : $PR^{2} - PQ^{2} = QR^{2}$ (given)
$PR^{2} = PQ^{2} + QR^{2}$
Because $\Delta$ PQR holds Pythagoras theorem, therefore, $\Delta$ PQR is right-angled triangle right angle at Q.
$In\; \Delta QMR \; and \; \Delta PMQ$
$\angle M = \angle M$ {each angle is 90°}
$\angle MQR = \angle QPM$ [each equal to 90° – angle R]
As we know that if the two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion.
$\therefore \Delta QMR \sim \Delta PMQ$
Now using the property of area of similar triangles
$\frac{ar(\Delta QMR)}{ar(\Delta PMQ)}=\frac{(QM)^{2}}{(PM)^{2}}$
$\frac{\frac{1}{2}\times RM \times QM}{\frac{1}{2}\times PM \times QM}=\frac{(QM)^{2}}{(PM)^{2}}$ $\left \{ \text {Q area of triangle}=\frac{1}{2}\times base \times height \right \}$
$\Rightarrow QM^{2} = PM \times RM$

Question:2

Find the value of x for which DE || AB in Fig.

Answer:

Answer : [x=2]
Given: DE || AB
According to basic proportionality theorem. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct point, then the other two sides are divided in the same ratio.
$\\\therefore \frac{CD}{AD}=\frac{CE}{BE}\\\frac{x+3}{3x+19}=\frac{x}{3x+4}\\\left ( x+3 \right )\left ( 3x+4 \right )=x\left ( 3x+19 \right )$
$3x^{2}+ 4x + 9x + 12 = 3x^{2} + 19x$
$19x - 13x = 12$
$6x = 12$
$x=\frac{12}{6}=2$
Hence the required value of x is 2.

Question:3

In Fig., if angle1 = angle 2 and $\Delta$ NSQ $\cong$ $\Delta$ MTR, then prove that $\Delta$ PTS $\sim$ $\Delta$ PRQ.

Answer:

Given :
$\Delta NSQ \cong \Delta MTR \; and\; \angle 1 = \angle 2$
To prove:-
$\Delta PTS \sim \Delta PRQ$
Proof:-
$\text {It is given that }\Delta NSQ \cong \Delta MTR$
$\therefore SQ=TR\; \; \; \; \; ...(1)$
$\angle 1=\angle 2$
We know that sides opposite to equal angles are also equal
$\therefore PT=PS \; \; \; \; \; \; \; .....(1)$
from equation (1) and (2)
$\frac{PS}{SQ}=\frac{PT}{TR}$
According to Converse of basic proportionality theorem if a line divides any two sides of a triangle in the same ratio, then the line is parallel to third side.
$\therefore ST \parallel QR$
$\angle 1=\angle PQR$ $\text{(corresponding angle)}$
$\angle 2=\angle PRQ$ $\text{ (corresponding angle)}$
In $\Delta PTS$ and $\Delta PRQ$
$\angle P=\angle P$ $\text{ (corresponding angle)}$
$\angle 1=\angle PQR$
$\angle 2=\angle PRQ$
We know that if the corresponding angles of two triangles are equal, then the triangles are similar by AAA similarity criterion

$\therefore \Delta PTS \sim \Delta PRQ$ $\text{ (by AAA similarity criterion) }$

Hence proved

Question:4

Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3 RS. Find the ratio of the areas of triangles POQ and ROS.

Answer:

Answer: 9:1
Given:- PQRS is a trapezium in which PQ || RS and PQ = 3 RS.
$\Rightarrow \frac{PQ}{RS}=\frac{3}{1}$

$In\; \Delta POQ \; and\; ROS$
$\angle SOR = \angle QOP$ (vertically opposite angles)
$\angle SRP = \angle RPQ$ (alternate angle)
As we know that if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion
$\therefore \Delta POQ\sim \Delta ROS$
By the property of area of similar triangle
$\frac{ar\left ( \Delta POQ \right )}{ar\left ( \Delta SOR \right )}=\frac{\left ( PQ \right )^{2}}{\left ( RS \right )^{2}}=\left ( \frac{PQ}{RS} \right )^{2}=\left ( \frac{3}{1} \right )^{2}$
$\frac{ar\left ( \Delta POQ \right )}{ar\left ( \Delta SOR \right )}=\frac{9}{1}$
Hence the required ratio is 9: 1.

Question:5

In Fig., if AB || DC and AC and PQ intersect each other at the point O, prove that OA . CQ = OC . AP.

Answer:

Given:- AB || DC and AC and PQ intersect each other at the point O.
To prove:- OA.CQ = OC.AP
Proof:-
$In\; \Delta AOP \; and \; \Delta COQ$
$\angle AOP = \angle COQ$ $\text{(vertically opposite angles)}$
Since AB || OC and PQ is transversal
$\therefore \angle APO = \angle CQO$ $\text{(alternate angles)}$
As we know that if the two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion
$\therefore \Delta AOP\sim \Delta COQ$

Then

$\frac{OA}{OC}=\frac{AP}{CQ}$ $\text{[Corresponding sides are proportional]}$
By cross multiply we get
OA.CQ = AP.OC
Hence proved.

Question:6

Find the altitude of an equilateral triangle of side 8 cm.

Answer:

Let ABC be an equilateral triangle of side 8 cm
i.e. AB = BC = AC = 8 cm and AD $\perp$ BC
Then D is the mid-point of BC.
$\therefore BD=DC=\frac{1}{2}BC=\frac{8}{2}=4cm$
Apply Pythagoras theorem in triangle ABD we get
$\left ( AB \right )^{2}=\left ( BD \right )^{2}+\left ( AD \right )^{2}$
$\left ( 8 \right )^{2}=\left ( 4 \right )^{2}+\left ( AD \right )^{2}$
$64=16+\left ( AD \right )^{2}$
$64-16=\left ( AD \right )^{2}$
$48=\left ( AD \right )^{2}$
$\sqrt{48}=AD$
$4\sqrt{3}=AD$
$\text{Hence the altitude of the equilateral triangle is}$ $4\sqrt{3}cm$

Question:7

If $\\ \Delta ABC \sim \Delta DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm \; and \; FD = 12 cm,$ find the perimeter of $\Delta ABC.$

Answer:

Answer: 18 cm
Given : $\Delta ABC \sim \Delta DEF$
and AB = 4cm , DE = 6 cm
EF = 9 cm, FD = 12 cm
Here $\Delta ABC \sim \Delta DEF$ (given)
$\therefore \frac{AB}{ED}=\frac{BC}{EF}=\frac{AC}{DF}$
$\frac{4}{6}=\frac{BC}{9}=\frac{AC}{12}$
Taking first two terms we get
$\frac{4}{6}=\frac{BC}{9}$
$\frac{9 \times 4}{6}=BC$
$BC=6 \; cm$
Taking first and last terms we get
$\frac{4}{6}=\frac{AC}{12}$
$AC=\frac{4 \times 12}{6}=8$
$Perimeter \; of\; \Delta ABC = AB + BC + CA$
$= 4 + 6 + 8 = 18 \; cm.$

Question:8

In Fig., if DE || BC, find the ratio of ar (ADE) and ar (DECB).

Answer:

Answer : 1: 3
Given:- DE || BC and DE = 6 cm, BC = 12 cm
$In$ $\Delta ABC \; and\; \Delta ADE$ $\angle ABC = \angle ADE$ $\text{ (corresponding angle) }$
$\angle A = \angle A$ $\text{ (common angle) }$
As we know that if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion.
$\therefore \; \; \; \; \; \Delta ABC \sim \Delta ADE$
Then
$\frac{ar\left ( \Delta ADE \right )}{ar\left ( \Delta ABC \right )}=\left ( \frac{DE}{BC} \right )^{2}=\left ( \frac{6}{12} \right )^{2}=\left ( \frac{1}{2} \right )^{2}=\frac{1}{4}$
$Let \; ar\left ( \Delta ADE \right )=\text {k then ar}\left ( \Delta ABC \right )=4k$

$Now\; ar(\Delta ECB) = ar(\Delta ABC) = ar(\Delta ADE) = 4k - k = 3k$

$\therefore Required \; ratio = ar(ADE) : ar (DECB)$

$\\k:3k\\1:3$

Question:9

ABCD is a trapezium in which AB ||DC and P and Q are points on AD and BC, respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD.

Answer:

Answer : [AD=60cm]
Given:- ABCD is a trapezium in which AB || DC, P and Q are points on AD and BC. Such that PQ || DC.
PD = 18 cm, BQ = 35, QC = 15 cm
To prove:- Find AD
Proof:-

Construction:- Join BD
In $\Delta$ ABD, PO || AB
And we know that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then according to basic proportionality theorem the other two sides are divided in the same ratio.
$\therefore \frac{DP}{AP}=\frac{DO}{OB}$
$\Rightarrow \frac{DP}{AP}=\frac{OD}{OB}\; \; \; \; \; \; \; \; \; .....(1)$
In $\Delta$ BDC, OQ || DC
Similarly by using basic proportionality theorem.
$\frac{BQ}{QC}=\frac{OB}{OD}$
$\Rightarrow \frac{QC}{BQ}=\frac{OD}{OB}\; \; \; \; \; \; \; \; \; .....(2)$
from equation (1) and (2) we get
$\frac{DP}{AP}=\frac{QC}{BQ}$
$\Rightarrow \frac{18}{AP}=\frac{15}{35}$
$AP=\frac{18 \times 35}{15}$
$AP=42 \; cm$
$\\AD = AP + PD\\ 42 + 18 = 60 cm\\ \therefore AD = 60 cm$

Question:10

Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm², find the area of the larger triangle.

Answer:

Answer : [108 cm²]
Given:- Corresponding sides of two similar triangles are in the ratio of 2 : 3.
Area of smaller triangle = 48 cm²
We know that the ratio of the areas of two similar triangles is equal to the ratio of the square of any two corresponding sides.
i.e $\frac{ar\left ( \text {smaller triangle} \right )}{ar\left ( \text {larger triangle} \right )}=\left ( \frac{2}{3} \right )^{2}$
$\frac{48}{ar\left ( \text {larger triangle} \right )}=\frac{4}{9}$
$ar\left ( \text {larger triangle} \right )=\frac{48 \times 9}{4}=12 \times 9=108\; cm^{2}$

Question:11

In a triangle PQR, N is a point on PR such that QN $\perp$ PR . If PN. NR = QN², prove that $\angle$ PQR = 90° .

Answer:

Given:- In a triangle PQR, N is a point on PR such that QN $\perp$ PR and PN.NR = QN²
To prove:- $\angle$ PQR = 90°
Proof:

We have PN.NR = QN²
$\\\Rightarrow PN.NR=QN.QN\\\frac{PN}{QN}=\frac{QN}{NR}\; \; \; \; \; \; \; ....(1)$
$In \Delta QNP \; and\; \Delta RNQ$
$\frac{PN}{QN}=\frac{QN}{NR}$
and $\angle$ PNQ = $\angle$ RNQ (each equal to 90°)
we know that if one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio, then the triangles are similar by SAS similarity criterion.
$\therefore \Delta QNP \sim \Delta RNQ$
Then $\Delta$ QNP and $\Delta$ RNQ are equiangular.
i.e. $\angle PQN = \angle QRN \; \; \; \; \; ....(2)$
$\angle RQN = \angle QPN \; \; \; \; \; ....(3)$
adding equation (2) and (3) we get
$\angle PQN + \angle RQN = \angle QRN + \angle QPN$
$\Rightarrow \angle PQR = \angle QRN + \angle QPN \; \; \; \; \; ...(4)$
In triangle PQR
$\angle PQR + \angle QPR + \angle QRP = 180^{o}$
$\angle PQR + \angle QPN + \angle QRN = 180^{o}$
$\begin{Bmatrix} \because \angle QPR=\angle QPN \\\angle QRP=\angle QRN \end{Bmatrix}$
$\angle PQR + \angle PQR = 180^{o}$
[Using equation (4)]
$2\angle PQR=180^{o}$
$\angle PQR=\frac{180^{o}}{2}$
$\angle PQR=90^{o}$
Hence proved

Question:12

Answer:

Answer : [12 cm ]
Given:- area of smaller triangle = 36 cm²
area of larger triangle = 100 cm²
length of the side of larger triangle = 20 cm
let the length of the corresponding side of the smaller triangle = x cm
According to the property of area of similar triangles,
$\frac{ar\left ( \text {larger triangle} \right )}{ar\left ( \text {smaller triangle} \right )}=\frac{\left ( \text {side of larger triangle} \right )^{2}}{\left ( \text {Side of smaller triangle} \right )^{2}}$
$\frac{100}{36}=\frac{\left ( 20 \right )^{2}}{x^{2}}$
$x^{2}=\frac{20 \times 20 \times 36}{100}$
$x=\sqrt{144}$
$x=12\; cm$

Question:13

In Fig., if $\angle ACB = \angle CDA$ , AC = 8 cm and AD = 3 cm, find BD.

Answer:

Answer :
$\left [ \frac{55}{3}\; cm\right ]$
$\text{Given : }$ $\angle ACB=\angle CDA$
AC = 8 cm and AD = 3 cm
$\text{In}$ $\Delta ACD\; and\; \Delta ABC$
$\angle A = \angle A$
$\angle ADC = \angle ACB$
We know that if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion.
$\therefore \; \; \; \; \Delta ADC\sim \Delta ACB$
$\Rightarrow \frac{AC}{AD}=\frac{AB}{AC}$ $\text{ [corresponding sides are proportional]}$
$=\frac{8}{3}=\frac{AB}{8}$
$AB=\frac{8 \times 8}{3}=\frac{64}{3}cm$
$BD+AD=\frac{64}{3}\; \; \; \; \; \; \; \; \; \; \; \left ( Q\; AB=BD+AD \right )$
$BD=\frac{64}{3}-AD$
$BD=\frac{64}{3}-3=\frac{64-9}{3}=\frac{55}{3}\; cm$

Question:14

A 15 metres high tower casts a shadow 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.

Answer:

Answer : [10 m]

Let BC = 15 m, AB = 24 m in $\Delta$ ABC and $\angle$ A = Q
Again let DE = 16m and $\angle$ EDF = Q in $\Delta$ DEF
In $\Delta$ ABC and $\Delta$ DEF
$\angle A=\angle D=Q$
$\angle B=\angle E=90^{o}$
We know that if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by AA similarity criterion.
$\therefore \; \; \; \; \; \Delta ABC\sim \Delta DEF$
$\text{Then }\frac{AB}{DE}=\frac{BC}{EF}$ $\text{(corresponding sides are proportional)}$
$\frac{24}{16}=\frac{15}{h}$
$h=\frac{15 \times 16}{24}=10$
$h=10$
Hence the height of the point on the wall where the top of the laden reaches is 10 m.

Question:15

Foot of a 10 m long ladder leaning against a vertical wall is 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.

Answer:

Answer : [8 m]

Here AC = 10 m is a ladder
BC = 6 m distance from the base of the wall.
In right-angle triangle ABC use Pythagoras theorem
$\left ( AC \right )^{2}=\left ( AB \right )^{2}+\left ( BC \right )^{2}$
$\left ( 10 \right )^{2}=\left ( AB \right )^{2}+\left ( 6 \right )^{2}$
$100= AB^{2}+36$
$100-36= AB^{2}$
$64= AB^{2}$
$AB=\sqrt{64}$
$AB=8\; m$
Hence, the height of the point on the wall where the top of the ladder reaches is 8 m.

NCERT Exemplar Class 10 Maths Solutions Chapter 6: Exercise-6.4

Question:1

$\\\text{In Fig., if }\angle A = \angle C, AB = 6 \; cm, BP = 15 \; cm, AP = 12 \; cm \; and \; CP = 4cm$
, then find the lengths of PD and CD.

Answer:

Answer: [PD=5cm, CD=2cm]
Given
: $\angle A=\angle C,AB=6\; cm,BP=15\; cm$
AP = 12 cm and CP = 4 cm
$\text{In }$ $\Delta APB\; and\; \Delta CPD$
$\angle A=\angle C$
$\angle APB=\angle DPC$ $\text{(Vertically opposite angle) }$
$\therefore \Delta APB \sim \Delta CPD$ $\text{ [by AA similarity criterion]}$
$\Rightarrow \frac{AP}{CP}=\frac{PB}{PD}=\frac{AB}{CD}$
$\frac{12}{4}=\frac{15}{PD}=\frac{6}{CD}$
on taking first and second term we get
$PD=\frac{15 \times 4}{12}$
$PD=5\; cm$
On taking first and last term we get
$\frac{12}{4}=\frac{6}{CD}$
$CD=\frac{6 \times 4}{12}$
$CD=2\; cm$

Question:2

$\text{It is given that }$ $\Delta ABC \sim \Delta EDF$ such that $AB = 5 \; cm, AC = 7 \; cm, DF= 15 \; cm$ and $DE = 12 \; cm$ .
Find the lengths of the remaining sides of the triangles.

Answer:

Answer:
$[EF=16.8\; cm, BC= 6.25 \; cm]$
Given :
$\Delta ABC \sim \Delta EDF$
$\therefore$ sides of both triangles are in the same ratio
i.e
$\frac{AB}{ED}=\frac{AC}{EF}=\frac{BC}{DF}\; \; \; \; \; \; .....(1)$

$\text{In }$ $\Delta ABC, AB=5\; cm,AC=7\; cm$
$\text{In }$ $\Delta DEF, DF=15\; cm,DE=12\; cm$
Put all these values in equation (1)
$\frac{5}{12}=\frac{7}{EF}=\frac{BC}{15}$
Taking the first and second term
$\frac{5}{12}=\frac{7}{EF}$
$EF=\frac{7 \times 12}{5}=\frac{84}{5}=16.8$
Taking the first and the last term
$\frac{5}{12}=\frac{BC}{15}$
$BC=\frac{15 \times 5}{12}=\frac{75}{12}=6.25$

Question:3

Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then the two sides are divided in the same ratio.

Answer:

Let ABC is a triangle in which line DE parallel to BC which interest line AB and AC at D and E.
To prove:- line DE divides both sides in the same ratio.
$\frac{AD}{DB}=\frac{AE}{EC}$
Construction:-
$\text{Join BE , CD and draw }$ $EF\perp AB$ $and$ $DG\perp AC$
Proof :-
$\frac{ar(\Delta ADE)}{ar(\Delta BDE)}=\frac{\frac{1}{2}\times AD \times EF}{\frac{1}{2}\times DB \times EF}$
$\frac{ar(\Delta ADE)}{ar(\Delta BDE)}=\frac{AD}{DB}\; \; \; \; \; \; \; \; .....(1)$
Similarly
$\frac{ar(\Delta ADE)}{ar(\Delta DEC)}=\frac{\frac{1}{2}\times AE \times GD}{\frac{1}{2}\times EC \times GD}=\frac{AE}{EC}\; \; \; \; \; \; \; \; ...(2)$
Also
$ar\left ( \Delta BDE \right )=ar\left ( \Delta DEC \right )\; \; \; \; \; ....(3)$
$[\Delta BDE\ and \ \Delta DEC\ \text{lie between the same parallel lines DE and BC and on the same base DE ]}$
from equation (1), (2) and (3)
$\frac{AD}{DB}=\frac{AE}{EC}$
Hence proved

Question:4

In Fig, if PQRS is a parallelogram and $AB \parallel PS$ then prove that $OC \parallel SR.$

Answer:

To prove:-

$OC \parallel SR$
$\text{Proof:- In}$ $\Delta OPS \; and \; \Delta OAB$
$\angle POS=\angle AOB$ $\text{(common angle)}$
$\angle OSP=\angle OBA$ $\text{(corresponding angles)}$
$\therefore \Delta OPS \sim \Delta OAB$ $\text{(by AA similarity criterion)}$
$\Rightarrow \frac{OS}{OB}=\frac{PS}{AB}\; \; \; \; \; \; \; \; ....(i)$
$\Rightarrow \frac{QR}{AB}=\frac{CR}{CB}$
$\frac{PS}{AB}=\frac{CR}{CB}\; \; \; \; \; \; \; \; \; \; ....(2)$ $\left [ \because \; \text {PQRS is parallelogram,}\therefore PS=QR \right ]$
from equation (1) and (2)
$\frac{OS}{OB}=\frac{CR}{CB}\; or\; \frac{OB}{OS}=\frac{CB}{CR}$
Subtracting 1 from both sides
$\frac{OB}{OS}-1=\frac{CB}{CR}-1$
$\frac{OB-OS}{OS}=\frac{CB-CR}{CR}$
$\frac{BS}{OS}=\frac{BR}{CR}$
We know that if a line divides any two sides of a triangle in the same ratio then by converse of basic proportionality theorem the line is parallel to the third side.
$\therefore SR\parallel OC$
Hence proved.

Question:5

A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then find the distance by which the top of the ladder would slide upwards on the wall.

Answer:

Answer: [0.8 m]

Here, AC = 5m, BC = 4 m, AD = 1.6 m
Let CE = x m
In $\Delta ABC$ use Pythagoras theorem
$AC^{2}=AB^{2}+BC^{2}$
$\left ( 5 \right )^{2}=\left ( AB \right )^{2}+\left ( 4 \right )^{2}$
$25-16=AB^{2}$
$9=AB^{2}$
$AB=\sqrt{9}$
$AB=3\; m$
$DB=AB-AD$
$=3-1.6=1.4\; m$
Similarly, in triangle EBD use Pythagoras theorem
$ED^{2}=EB^{2}+BD^{2}$
$\left ( 5 \right )^{2}=EB^{2}+\left ( 1.4 \right )^{2}$
$23.04=EB^{2}$
$EB=\sqrt{23.04}$
$EB=4.8$
$EB=EB-BC$
$4.8-4=0.8\; m$
Hence the top of the ladder would slide up words on the wall at distance is 0.8 m.

Question:6

For going to a city B from city A, there is a route via city C such that . $AC \perp CB, AC = 2 \times km \; and \; CB = 2 (x + 7) km$ . It is proposed to construct a 26 km highway which directly connects the two cities A and B. Find how much distance will be saved in reaching city B from city A after the construction of the highway.

Answer:

Answer: [8 km]
Given :-
$AC \perp CB, AC = 2x\; km, CB = 2 (x + 7) km$
$AB=26\; km$

In $\Delta ABC$ use, Pythagoras theorem
$AB^{2}=AC^{2}+BC^{2}$
$\left ( 26 \right )^{2}=\left ( 2x \right )^{2}+\left ( 2\left ( x+7 \right ) \right )^{2}$
$676=4x^{2}+4\left ( x^{2}+49+14x \right )$ $(using \left ( a+b \right )^{2}=a^{2}+b^{2}+2ab)$
$676=4x^{2}+4x^{2}+196+56x$
$676=8x^{2}+56x+196$
$8x^{2}+56x-480=0$
Dividing by 8 we get
$x^{2}+7x-60=0$
$x^{2}+12x-5x-60=0$
$x\left ( x+12 \right )-5\left ( x+12 \right )=0$
$\left ( x+12 \right )\left ( x-5 \right )=0$
$x=-12 \; or\; x=5$
x = – 12 is not possible because distance cannot be negative.
$\therefore x=5$
Now
$AC = 2x = 2 \times 5 = 10 \; km$
$BC = 2(x + 7) = 2(5 + 7) = 2 \times 12 = 24 \; km$
Distance covered to reach city B from A via city C = AC + CB
$=10+24=34\; km$
Distance covered to reach city B from city A after the construction of highway
$BA = 26\; cm$
Saved distance $= 34 - 26 = 8 km.$

Question:7

A flagpole 18 m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.

Answer:

Answer: 20.4 m

height of flag pole BC = 18 m
shadow AB = 9.6 m
In triangle BC using Pythagoras theorem
$\\AC^{2}=AB^{2}+BC^{2}\\AC^{2}=\left ( 9.6 \right )^{2}+\left ( 18 \right )^{2}$
$\\AC^{2}=92.16+324\\AC^{2}=416.16\\AC=\sqrt{416.16}$
$AC=\sqrt{416.16}=20.4\; m$
Hence distance of the top of the pole from the far end of the shadow is 20.4 m

Question:8

A street light bulb is fixed on a pole 6 m above the level of the street. If a woman of height 1.5 m casts a shadow of 3m, find how far she is away from the base of the pole.

Answer:

Answer : [9 m]

height of street light bulb = 6 m
woman's height = 1.5 m
woman's shadow = 3m
Let the distance between pole and women = x m
$Here\ CD\parallel AB$
$In\ \Delta CDE$ $and$ $\Delta ABE$
$\angle E=\angle E$ $\text{(common angle)}$
$\angle ABE=\angle CDE$ $(\text{ each angle 90}^0)$
$\therefore \Delta CDE\sim \Delta ABE\text{ (by AA similarity criterion)}$
Then
$\frac{ED}{EB}=\frac{CD}{AB}$
$\\\frac{3}{3+x}=\frac{1.5}{6}\\3 \times 6=1.5\left ( 3+x \right )\\18=4.5+1.5x\\18-4.5=1.5x$
$\\\frac{13.5}{1.5}=x\\9m=x$

Question:9

In Fig., ABC is a triangle right angled at B and $BD \perp AC$ . If AD = 4 cm, and CD = 5 cm, find BD and AB.

Answer:

Answer: $\left [ 2\sqrt{5}\; cm\; and\; 6\; cm \right ]$
Given :- $\angle B=90^{o}$ and $BD\perp AC$
AD = 4 cm and CD = 5 cm
In $\Delta ABD$ and $\Delta BDC$
$\angle ADB = \angle BDC$ (each equal to 90^o)
$\angle BAD = \angle DBC$ (each equal to 90^o-C)
$\therefore \Delta ABD\sim \Delta BDC$ (by AA similarity criterion)
$\Rightarrow \frac{DB}{DA}=\frac{DC}{DB}$
By cross multiply we get
$DB^{2}=DA.DC$
$DB^{2}=4 \times 5$
$DB=\sqrt{20}=2\sqrt{5}cm$
In $\Delta BDC$ use Pythagoras theorem
$BC^{2}=BD^{2}+CD^{2}$
$BC^{2}=\left ( 2\sqrt{5} \right )^{2}+\left ( 5 \right )^{2}$
$BC^{2}=20+25=45$
$BC=\sqrt{45}=3\sqrt{5}$
We know that $\Delta DBA\sim \Delta DBC$
$\therefore \frac{DB}{DC}=\frac{BA}{BC}\Rightarrow \frac{2\sqrt{5}}{5}=\frac{BA}{3\sqrt{5}}$
$BA=\frac{2\sqrt{5}\times 3\sqrt{5}}{5}=\frac{6 \times 5}{5}=6\; cm$

Question:10

In Fig., PQR is a right triangle right angled at Q and $QS \perp PR$ . If PQ = 6 cm and PS = 4 cm, find QS, RS and QR.

Answer:

Given: PQR is a triangle

$\angle Q = 90^{o} \; and\; QS \perp PR$

PQ = 6 cm, PS = 4 cm

In $\Delta SQP \; and \; \Delta SRQ$

$\angle S=\angle S$ (common angles and each angle is 90°)

$\angle SPQ=\angle SQR$ (each equal to $90^{o}-\angle R$ )

$\therefore \Delta SQP \sim \Delta SRQ$ ( by AA similarity criterion)

$\Rightarrow \frac{SQ}{PS}=\frac{SR}{SQ}$
By cross multiply we get
$SQ^{2}=PS.SR \; \; \; \; \; \; \; ....(1)$
In $\Delta PSQ$ use Pythagoras theorem.
$PQ^{2}=PS^{2}+QS^{2}$
$6^{2}=4^{2}+QS^{2}$
$36-16=QS^{2}$
$20=QS^{2}$
$QS=\sqrt{20}=2\sqrt{5}cm$
Put $QS=2\sqrt{5}$ in equation (1)
$\left ( 2\sqrt{5} \right )^{2}=4 \times SR$
$\frac{20}{4}=SR$
$5cm=SR$
In $\Delta QSR$ use Pythagoras theorem
$QR^{2}=QS^{2}+SR^{2}$
$QR^{2}=\left ( 2\sqrt{5} \right )^{2}+\left ( 5 \right )^{2}$
$QR^{2}=20+25$
$QR=\sqrt{45}=3\sqrt{5}cm$

Question:11

$\text{In} \Delta PQR,$ $PD \perp QR$ $\text{such that D lies on QR}$ . $\text{If}$ $PQ = a, PR = b, QD = c$ $\text{and}$ $DR=d$ $\text{, prove that }$ $(a + b) (a - b) = (c + d) (c - d).$

Answer:

$\text{Given:- PQR is a triangle and}$ $PD\perp QR$

$PQ = a, PR = b, QD = C, DR = d$

To prove:-

$(a + b) (a - b) = (c + d) (c - d)$
$\text{Proof :- In}$ $\Delta PQD$ $\text{use Pythagoras theorem}$
$\\PQ^{2}=QD^{2}+PD^{2}$
$\\a^{2}=c^{2}+PD^{2}\\a^{2}-c^{2}=PD^{2} \; \; \; \; \; \; \; \; ....(1)$
$\text{In }$ $\Delta PRD$ $\text{use Pythagoras theorem }$
$\\PR^{2}=PD^{2}+DR^{2}\\b^{2}=PD^{2}+d^{2}\\b^{2}-d^{2}=PD^{2}\; \; \; \; \; \; \; ....(2)$
equate equation (1) and (2) we get
$\\a^{2} - c^{2} = b^{2} - d^{2}\\a^{2}-b^{2}=c^{2}-d^{2}$
$(a - b) (a + b) = (c - d) (c + d)$ $\left [ \because a^{2} - b^{2} = (a - b) (a + b) \right ]$
Hence proved.

Question:12

In a quadrilateral ABCD, $\angle A + \angle D = 90^{o}$ . Prove that $AC^{2} + BD^{2} = AD^{2} + BC^{2}$

Answer:

Given : ABCD is a quadrilateral, $\angle A+\angle D=90^{o}$
To prove :- $AC^{2} + BD^{2} = AD^{2} + BC^{2}$
Proof : In $\Delta ADE$
$\angle A+\angle D=90^{o} \; \; \; \; \; ....(1)$ (given)
for find $\angle E$ use sum of angle of a triangle is equal to 180°
$\angle A+\angle D+\angle E=180^{o}$
$90+\angle E=180^{o}$
$\angle E=180^{o} -90^{o}$
$\angle E=90^{o}$
In $\Delta ADE$ use Pythagoras theorem we get
$AD^{2}=AE^{2}+DE^{2}\; \; \; \; \; \; \; \; ....(2)$
In $\Delta BEC$ use Pythagoras theorem we get
$BC^{2}=BE^{2}+EC^{2}\; \; \; \; \; \; \; \; ....(3)$
Add equation (2) and (3) we get
$AD^{2}+BC^{2}=AE^{2}+DE^{2}+BE^{2}+CE^{2}\; \; \; \; \; \; \; .....(4)$
In $\Delta ACE$ use Pythagoras theorem we get
$AC^{2}=AE^{2}+CE^{2}\; \; \; \; \; \; \; \; ....(5)$
In $\Delta EBD$ use Pythagoras theorem we get
$BD^{2}=BE^{2}+DE^{2}\; \; \; \; \; \; \; \; \; \; \; \; ....(6)$
Now add equation (5) and (6) we get
$AC^{2} + BD^{2} = AE^{2} + CE^{2} + BE^{2} + DE^{2}\; \; \; \; \; \; ....(7)$
From equation (4) and (7) we get
$AC^{2} + BD^{2} = AD^{2} + BC^{2}$
Hence proved.

Question:13

In fig. , l || m and line segments AB, CD and EF are concurrent at point P. Prove tha
t $\frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD}$ .

Answer:

Given: l || m and line segments AB, CD and EF are concurrent at point P
To prove:-
$\frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD}$
Proof :In $\Delta APC \; and\; \Delta DPB$
$\angle APC=\angle DPB$ (vertically opposite angles)
$\angle PAC=\angle PBD$ (alternate angles)
and we know that if two angles of one triangle are equal to the two angles of another triangle then the two triangles are similar by AA similarity criterion
$\therefore \; \; \; \; \; \Delta APC\sim \Delta DPB$
Then, $\frac{AP}{BP}=\frac{AC}{BD}=\frac{PC}{PD}\; \; \; \; \; \; \; \; \; ....(1)$
In $\Delta APE\; and\; \Delta FPB$
$\angle APE=\angle BPF$ (vertically opposite angle)
$\angle PAE=\angle PBF$ (alternate angle)
$\therefore \; \; \; \; \Delta APE\sim \Delta FPB$ (by AA similarity criterion)
Then $\frac{AP}{PB}=\frac{AE}{BF}=\frac{PE}{PF}\; \; \; \; \; \; \; ...(2)$
In $\Delta PEC\; and\; \Delta PFD$
$\angle EPC=\angle FPD$ (vertically opposite angle)
$\angle PCE=\angle PDF$ (alternate angle)
$\therefore \; \; \; \; \Delta PEC\sim \Delta PFD$ (by AA similarity criterion)
Then $\frac{PE}{PF}=\frac{PC}{PD}=\frac{EC}{FD}\; \; \; \; \; \; \; ...(3)$
From equation (1), (2) and (3) we get
$\frac{AP}{BP}=\frac{AC}{BD}=\frac{AE}{BF}=\frac{PE}{PF}=\frac{EC}{FD}$
$\Rightarrow \frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD}$
Hence proved.

Question:14

In Fig., PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. Find PQ, QR and RS.

Answer:

Answer: $\left [ PQ=8_{cm},QR=12\; cm,RS=16\; cm\right ]$
Given :- PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm
We know that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
$\therefore PA \parallel QB \parallel RC \parallel SD$ (parallel lines)
Then according to basic proportionality theorem
$PQ : QR : RS = AB : BC : CD = 6 : 9 : 12$
$\text {Let}\\PQ = 6x\\ QR = 9x\\ RS = 12x$
length of PS = 36 (given)
$\therefore \; \; \; PQ + QR + RS = 36\; \; \; \Rightarrow \; \; \; 6x + 9x + 12x = 36 \; \; \; \; \; \Rightarrow \; \; \; \; \; 27x = 36$
$x=\frac{36}{27}=\frac{4}{3}$
$PQ=6x=6 \times\frac{4}{3}=8\; cm$
$QR=9x=\frac{9\times 4}{3}=12\; cm$
$RS=12x=\frac{12\times 4}{3}=16\; cm$

Question:15

O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO.

Answer:

ABCD is a trapezium and O is the point of intersection of the diagonals AC and BD.
$AB\parallel DC$

Proof :- $\text {In} \Delta ABD \; \text {and }\; \Delta POD$
$\angle D=\angle D$ (Common angle)
$\angle ABD=\angle POD$ (corresponding angles)
$\therefore \Delta ABD\sim \Delta POD$ (by AA similarity criterion)
Then $\frac{OP}{AB}=\frac{PD}{AD}\; \; \; \; \; \; \; \; \; \; ...(1)$
$\text {In} \Delta ABC \; \text {and }\; \Delta OQC$
$\angle C=\angle C$ (Common angle)
$\angle BAC=\angle QOC$ (corresponding angles)
$\therefore \Delta ABC\sim \Delta OQC$ (by AA similarity criterion)
Then $\frac{OQ}{AB}=\frac{QC}{BC}\; \; \; \; \; \; \; \; \; \; ...(2)$
$\text {In}\Delta ADC$
$OP\parallel DC$
We know that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
$\therefore \frac{AP}{PD}=\frac{OA}{OC}\; \; \; \; \; \; \; \; \; ...(3)$
$\text {Also In}\; \Delta ABC$
$OQ\parallel AB$
$\therefore \frac{BQ}{QC}=\frac{OA}{OC}\; \; \; \; \; \; \; \; \; ...(4)$ (by basic proportionality theorem)
from equation (3) and (4)
$\frac{AP}{PD}=\frac{BQ}{QC}$
Add 1 on both sides we get
$\frac{AP}{PD}+1=\frac{BQ+QC}{QC}$
$\frac{AP}{PD}=\frac{BC}{QC}$
$\Rightarrow \frac{PD}{AD}=\frac{QC}{BC}\; \; \; \; \; \; \; ...(5)$
$\frac{OP}{AB}=\frac{QC}{BC}$ (use equation (1))
$\Rightarrow \frac{OP}{AB}=\frac{OQ}{AB}$ (use equation (2))
$\Rightarrow OP=OQ$
Hence proved

Question:16

In Fig., line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and $\angle AEF=\angle AFE$ .Prove that $\frac{BD}{CD}=\frac{BF}{CE}$ .

Answer:

Given: Line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and $\angle AEF=\angle AFE$
To prove :-
$\frac{BD}{CD}=\frac{BF}{CE}$
Construction:- Take point G on AB such that $CG\parallel DF$

Proof:- E is mid-point of CA (given)
$\therefore \; \; \; \; \; \; CE = AE$
In $\Delta ACG, CG\parallel EF$ and E is mid-point of CA
Then according to mid-point theorem
$CE = GF \; \; \; \; \; \; \; ....(1)$
In $\Delta BCG \; and\; \Delta BDF \; \; CG || DF$
According to basic proportionality theorem.
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.
$\therefore \frac{BC}{CD}=\frac{BG}{GF}$
$\frac{BC}{CD}=\frac{BF-GF}{GF}$
$\frac{BC}{CD}=\frac{BF}{GF}=-1$
$\frac{BC}{CD}+1=\frac{BF}{GF}$
$\frac{BC}{CD}+1=\frac{BF}{CE}$ (use equation (1))
$\frac{BC+CD}{CD}=\frac{BF}{CE}$
$\frac{BD}{CD}=\frac{BF}{CE}$
Hence proved.

Question:17

Prove that the area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.

Answer:

Let PQR is a right angle triangle which is right angle at point Q.
$PQ = b, QR = a$
Three semicircles are drawn on the sides of $\Delta PQR$ having diameters PQ, QR and PR respectively.
Let $x_{1}, x_{2}$ and $x_{3}$ are the areas of semicircles respectively.
To prove:- $x_{3}=x_{1}+x_{2}$
Proof: In $\Delta PQR$ use Pythagoras theorem we get
$PR^{2}=PQ^{2}+QR^{2}$
$PR^{2}=a^{2}+b^{2}$
$PR^{2}=\sqrt{a^{2}+b^{2}}$
Now area of semi-circle drawn on side PR is
$x_{3}=\frac{\pi }{2}\left ( \frac{PR}{2} \right )^{2}$ $\left [ \therefore \text {area of semicircle=}\frac{\pi r^{2}}{2} \right ]$
$=\frac{\pi}{2}\left ( \frac{\sqrt{a^{2}+b^{2}}}{2} \right )^{2}$ $\left ( \therefore PR=\sqrt{a^{2}+b^{2}} \right )$
$=\frac{\pi}{2}\times\frac{\left ( a^{2}+b^{2} \right )}{4}$
$x_{3}=\frac{\pi}{8}\left ( a^{2}+b^{2} \right )$
area of semi-circle drawn a side QR is
$x_{2}=\frac{\pi }{2}\left ( \frac{QR}{2} \right )^{2}$
$=\frac{\pi }{2}\left ( \frac{a}{2} \right )^{2}$
$=\frac{\pi}{2}\times \frac{a^{2}}{4}$
$x_{2}=\frac{\pi}{8}a^{2}\; \; \; \; \; \; \; ....(2)$
area of semicircle drawn on side PQ is
$x_{1}=\frac{\pi }{2}\left ( \frac{PQ}{2} \right )^{2}$
$x_{1}=\frac{\pi }{2}\left ( \frac{b^{2}}{4} \right )\Rightarrow \frac{\pi}{8}b^{2}\; \; \; \; \; \; \; \; ....(3)$
add equation (2) and (3) we get
$x_{2}+x_{1}=\frac{\pi }{8}a^{2}+\frac{\pi}{8}b^{2}$
$x_{2}+x_{1}=\frac{\pi }{8}\left ( a^{2}+b^{2} \right )=x_{3}$
Hence $x_{2}+x_{1}=x_{3}$
Hence proved

Question:18

Prove that the area of the equilateral triangle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle.

Answer:

Let PQR is a right angle triangle which is a right angle at point R.
$PR = b, RQ = a$
Three equilateral triangles are drawn on the sides of triangle PQR that is PRS, RTQ and PUQ
Let $x_{1},x_{2}$ and $x_{3}$ are the areas of equilateral triangles respectively
To prove:- $x_{1}+x_{2}=x_{3}$
Using Pythagoras theorem in $\Delta PQR$ we get
$PQ^{2}+RQ^{2}+RP^{2}$
$PQ^{2}=a^{2}+b^{2}$
$PQ=\sqrt{a^{2}+b^{2}}$
The formula of area of an equilateral triangle is
$=\frac{\sqrt{3}}{4}\left ( side \right )^{2}$
$\therefore$ Area of equilateral $\Delta RTQ$
$x_{1}=\frac{\sqrt{3}}{4}\left ( a^{2} \right )\; \; \; \; \; \; \; \; ....(1)$
Area of equilateral $\Delta RSP$
$x_{2}=\frac{\sqrt{3}}{4} b^{2} \; \; \; \; \; \; \; \; ....(2)$
Area of equilateral $\Delta PQU$
$x_{3}=\frac{\sqrt{3}}{4}\left ( \sqrt{a^{2}+b^{2}} \right )^{2}$ $\left ( Q\; PQ=\sqrt{a^{2}+b^{2}} \right )$
$x_{3}=\frac{\sqrt{3}}{4} \left ( a^{2}+b^{2} \right )$
add equation (1) and (2) we get
$x_{1}+x_{2}=\frac{\sqrt{3}}{4}a^{2}+\frac{\sqrt{3}}{4}b^{2}$
$x_{1}+x_{2}=\frac{\sqrt{3}}{4}\left ( a^{2}+b^{2} \right )=x_{3}$
Hence $x_{1}+x_{2}=x_{3}$
Hence proved

NCERT Exemplar Class 10 Maths Chapter 6 Solutions Important Topics:

In this chapter, all the conditions required to prove “any two triangles are similar” is discussed.
In this chapter, it is also discussed that if we know the ratio of sides of similar triangles, what relation in their area will be followed.
Class 10 Maths NCERT exemplar chapter 6 solutions discusses some triangular theorems such as Pythagoras Theorem.

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NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

Chapter 1 Real Numbers

Chapter 2 Polynomials

Chapter 3 Pair of Linear Equations in Two Variables

Chapter 4 Quadratic Equations

Chapter 5 Arithmetic Progressions

Chapter 7 Coordinate Geometry

Chapter 8 Introduction to Trigonometry & Its Equations

Chapter 9 Circles

Chapter 10 Constructions

Chapter 11 Areas related to Circles

Chapter 12 Surface Areas and Volumes

Chapter 13 Statistics and Probability

Features of NCERT Exemplar Class 10 Maths Solutions Chapter 6:

These Class 10 Maths NCERT exemplar chapter 6 solutions provide a detailed knowledge of triangles and the similarity of triangles along with the theorems such as Pythagoras theorem. Most of the question types related to Triangles for Class 10 can be practiced by the students through the exemplar.

The NCERT exemplar Class 10 Maths solutions chapter 6 Triangles, due to its plentiful question bank makes it easier for the student to attempt other books such as NCERT Class 10 Maths, A textbook of Mathematics by Monica Kapoor, RD Sharma Class 10 Maths, RS Aggarwal Class 10 Maths et cetera.

Students get a very resourceful feature of NCERT exemplar Class 10 Maths solutions chapter 6 pdf download through which one can view/download the pdf version of these solutions and use them in case of low/no internet connectivity while practicing NCERT exemplar Class 10 Maths chapter 6 solutions.

Check Chapter-Wise Solutions of Book Questions

Chapter No.	Chapter Name
Chapter 1	Real Numbers
Chapter 2	Polynomials
Chapter 3	Pair of Linear Equations in Two Variables
Chapter 4	Quadratic Equations
Chapter 5	Arithmetic Progressions
Chapter 6	Triangles
Chapter 7	Coordinate Geometry
Chapter 8	Introduction to Trigonometry
Chapter 9	Some Applications of Trigonometry
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Areas Related to Circles
Chapter 13	Surface Areas and Volumes
Chapter 14	Statistics
Chapter 15	Probability

Must, Read NCERT Solution Subject Wise

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Frequently Asked Question (FAQs)

1. If two similar triangles have lengths ratio 1:3. What is the ratio of their areas?

We know that if two similar triangles have a sides ratio equal to R then the ratio of their areas will be R2. Therefore, in the given problem the ratio of area of two triangles will be 1:9.

2. What is the condition for two triangles to be similar?

Two triangles are said to be similar if their corresponding angles are equal and the ratio of the corresponding sides is consistent. The basic difference between congruent triangles and similar triangles is that congruent triangles have the same shape and size however similar triangles have different sizes and the same shapes.

3. Is the chapter Triangles important for Board examinations?

The chapter Triangles is extremely important for Board examinations as it holds around 10-13% weightage of the whole paper.

4. What type of questions from the chapter Triangles are expected in the board examination?

Generally, the paper consists of 3-4 questions from the chapter of Triangles, and those are mainly distributed between Very Short, Short, and Long Answer questions. NCERT exemplar Class 10 Maths solutions chapter 6 explores all of the above-mentioned type-based questions in detail.

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Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

hindi 10th class mp board sample paper

Dear aspirant !

Hope you are doing well ! The class 10 Hindi mp board sample paper can be found on the given link below . Set your time and give your best in the sample paper it will lead to great results ,many students are already doing so .

https://www.google.com/url?sa=t&source=web&rct=j&opi=89978449&url=https://school.careers360.com/download/sample-papers/mp-board-10th-hindi-model-paper&ved=2ahUKEwjO3YvJu5KEAxWAR2wGHSLpAiQQFnoECBMQAQ&usg=AOvVaw2qFFjVeuiZZJsx0b35oL1x .

Hope you get it !

Thanking you

Read Complete Answer

Shivanshu 04 February,2024

CBSE class 10th date sheet 11/1/2024 2024

Hello aspirant,

The dates of the CBSE Class 10th and Class 12 exams are February 15–March 13, 2024 and February 15–April 2, 2024, respectively. You may obtain the CBSE exam date sheet 2024 PDF from the official CBSE website, cbse.gov.in.

To get the complete datesheet, you can visit our website by clicking on the link given below.

https://school.careers360.com/boards/cbse/cbse-date-sheet

Thank you

Hope this information helps you.

Read Complete Answer

Sajal Trivedi 12 January,2024

iam in 7 th plz send 7th previous paper

Hello aspirant,

The paper of class 7th is not issued by respective boards so I can not find it on the board's website. You should definitely try to search for it from the website of your school and can also take advise of your seniors for the same.

You don't need to worry. The class 7th paper will be simple and made by your own school teachers.

Thank you

Hope it helps you.

Read Complete Answer

Sajal Trivedi 25 October,2023

my son's dob is 13/02/2009 is hi eligible for 10th board exam in 2024,,from cbse. olz reply me

The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.

Read Complete Answer

Shubhangi 29 July,2022

can Hindi marks be replace with IT 402(additional subject) in CBSE class 10th board

That totally depends on what you are aiming for. The replacement of marks of additional subjects and the main subject is not like you will get the marks of IT on your Hindi section. It runs like when you calculate your total percentage you have got, you can replace your lowest marks of the main subjects from the marks of the additional subject since CBSE schools goes for the best five marks for the calculation of final percentage of the students.

However, for the admission procedures in different schools after 10th, it depends on the schools to consider the percentage of main five subjects or the best five subjects to admit the student in their schools.

Read Complete Answer

Aditi Singh 22 July,2022

View All

Popular CBSE Class 10th Questions

A block of mass 0.50 kg is moving with a speed of 2.00 ms^-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×10⁷J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms⁻² :

Option 1)

2.45×10⁻³ kg

Option 2)

6.45×10⁻³ kg

Option 3)

9.89×10⁻³ kg

Option 4)

12.89×10⁻³ kg

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

A particle is projected at 60⁰ to the horizontal with a kinetic energy $K$ . The kinetic energy at the highest point

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

In the reaction,

$2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, $Mg_{3}(PO_{4})_{2}$ will contain 0.25 mole of oxygen atoms?

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol^-1) is

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t²) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m² , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

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Bio Medical Engineer

The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary.

4 Jobs Available

Ethical Hacker

A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

3 Jobs Available

GIS Expert

GIS officer work on various GIS software to conduct a study and gather spatial and non-spatial information. GIS experts update the GIS data and maintain it. The databases include aerial or satellite imagery, latitudinal and longitudinal coordinates, and manually digitized images of maps. In a career as GIS expert, one is responsible for creating online and mobile maps.

3 Jobs Available

Data Analyst

The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.

Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.

3 Jobs Available

Geothermal Engineer

Individuals who opt for a career as geothermal engineers are the professionals involved in the processing of geothermal energy. The responsibilities of geothermal engineers may vary depending on the workplace location. Those who work in fields design facilities to process and distribute geothermal energy. They oversee the functioning of machinery used in the field.

3 Jobs Available

Database Architect

If you are intrigued by the programming world and are interested in developing communications networks then a career as database architect may be a good option for you. Data architect roles and responsibilities include building design models for data communication networks. Wide Area Networks (WANs), local area networks (LANs), and intranets are included in the database networks. It is expected that database architects will have in-depth knowledge of a company's business to develop a network to fulfil the requirements of the organisation. Stay tuned as we look at the larger picture and give you more information on what is db architecture, why you should pursue database architecture, what to expect from such a degree and what your job opportunities will be after graduation. Here, we will be discussing how to become a data architect. Students can visit NIT Trichy, IIT Kharagpur, JMI New Delhi.

3 Jobs Available

Remote Sensing Technician

Individuals who opt for a career as a remote sensing technician possess unique personalities. Remote sensing analysts seem to be rational human beings, they are strong, independent, persistent, sincere, realistic and resourceful. Some of them are analytical as well, which means they are intelligent, introspective and inquisitive.

Remote sensing scientists use remote sensing technology to support scientists in fields such as community planning, flight planning or the management of natural resources. Analysing data collected from aircraft, satellites or ground-based platforms using statistical analysis software, image analysis software or Geographic Information Systems (GIS) is a significant part of their work. Do you want to learn how to become remote sensing technician? There's no need to be concerned; we've devised a simple remote sensing technician career path for you. Scroll through the pages and read.

3 Jobs Available

Budget Analyst

Budget analysis, in a nutshell, entails thoroughly analyzing the details of a financial budget. The budget analysis aims to better understand and manage revenue. Budget analysts assist in the achievement of financial targets, the preservation of profitability, and the pursuit of long-term growth for a business. Budget analysts generally have a bachelor's degree in accounting, finance, economics, or a closely related field. Knowledge of Financial Management is of prime importance in this career.

4 Jobs Available

Data Analyst

3 Jobs Available

Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

3 Jobs Available

5 Jobs Available

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications.

2 Jobs Available

Welding Engineer

5 Jobs Available

QA Manager

4 Jobs Available

Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party.

4 Jobs Available

Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems.