NCERT Exemplar Class 10 Maths Solutions Chapter 6 - Triangles

NCERT Exemplar Class 10 Maths Solutions Chapter 6

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 the 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 : [B]

Here, ABCD is a rhombus, and we know that the diagonals of a rhombus are perpendicular bisectors 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 $\Delta ABC$ are 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 : [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 multiplying, 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 the 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: [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 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: [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 : [B]
In $\Delta DEF$ and $\Delta PQR$
$\angle D=\angle Q$ and $\angle R=\angle E$
Now draw two triangles with the help of the 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 : [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 $\Delta DEF$ are not congruent because congruent figures have the same shape and 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 : [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: [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 the first and the last terms 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 : [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 : [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 the 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 : [C]
Given: In triangle PQR
$PS = QS = RS$
$\Rightarrow PS = RS$
$\therefore \angle 1=\angle 2 ...(1)$
(Q If a right triangle has an 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)
$\text{In }\Delta PQR$
$\angle P+\angle Q+\angle R=180^{o}$
$\because$ the sum of angles of a triangle is 180°
$\angle 2+\angle 4+\angle 1+\angle 3=180^{o} ....(3)$
Using equations (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 (Page no. 63, Total questions - 12)

Question:1

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

Answer : [False]
The 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 squares 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 : [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 : [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 the equations, the Converse of the basic proportionality theorem is that 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 point P. Is $\Delta$PBC $\sim$ $\Delta$PDE? Why?

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: [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, we know that if all corresponding angles of two triangles are equal, then the 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: [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 is 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: True

According to the 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 equations (1), (2) and (3), we conclude that the sides of both the triangles are in the same ratio.
As we know, if the corresponding sides of two triangles are in the same ratio, then the triangles are similar by the SSS similarity criterion.
Hence, the 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 : [True]
Let two right-angle triangles be ABC and PQR.

Given: One of the acute angles 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 equations (1) and (2)
$\angle A=\angle P \; \; \; \; \; \; .....(3)$
from equations,,ideasinclude (1), (2) and (3) are observed that corresponding angles of the triangles are equal; therefore triangles are similar.

Question:9

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

Answer: [False]
Given: The ratio of corresponding altitudes of two similar triangles is 3/5
As we know, 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 the ratio of the 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: [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.
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 : [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. [False]
Given- (1) An angle of one triangle is equal to the angle of another triangle.
(2) Two sides of one triangle are proportional to the two sides of the other triangle.
According to the 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 ideasladder, that do not include the equal angle.

NCERT Exemplar Class 10 Maths Solutions Chapter 6: Exercise-6.3 (Page no. 66, Total questions - 15)

Question:1

In a D PQR, PR2–PQ2 = QR2 and M is a point on side PR such that QM is 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 a 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 the, AA similarity criterion.
$\therefore \Delta QMR \sim \Delta PMQ$
Now, using the property of the 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: [x=2]
Given: DE || AB
According to the 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.
$\\\therefore \frac{CD}{AD}=\frac{CE}{BE}\\\frac{x+3}{3x+19}=\frac{x}{3x+4}\\\left ( x+3 \right )\left ( 3x+4 \right )=x\left ( 3x+19 \right )$
$3x^{2}+ 4x + 9x + 12 = 3x^{2} + 19x$
$19x - 13x = 12$
$6x = 12$
$x=\frac{12}{6}=2$
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 equations (1) and (2)
$\frac{PS}{SQ}=\frac{PT}{TR}$
According to the Converse of the 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.
$\therefore ST \parallel QR$
$\angle 1=\angle PQR$ $\text{(corresponding angle)}$
$\angle 2=\angle PRQ$ $\text{ (corresponding angle)}$
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 the 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: 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, if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by the AA similarity criterion.
$\therefore \Delta POQ\sim \Delta ROS$
By the property of an area of a 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, if the two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by the AA similarity criterion.
$\therefore \Delta AOP\sim \Delta COQ$

Then

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

Question:6

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

Answer:

Let ABC be an equilateral triangle of side 8 cm.
i.e. AB = BC = AC = 8 cm and AD $\perp$ BC
D is the midpoint of BC.
$\therefore BD=DC=\frac{1}{2}BC=\frac{8}{2}=4cm$
Applying 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 the first two terms, we get
$\frac{4}{6}=\frac{BC}{9}$
$\frac{9 \times 4}{6}=BC$
$BC=6 \; cm$
Taking the 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 : 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, if two angles of one triangle are equal to the two angles of another triangle, then the two triangles are similar by the 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: [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, 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 the basic proportionality theorem.
$\frac{BQ}{QC}=\frac{OB}{OD}$
$\Rightarrow \frac{QC}{BQ}=\frac{OD}{OB}\; \; \; \; \; \; \; \; \; .....(2)$
From equations (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

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

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 squares 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 the 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 equations (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

The areas of two similar triangles are 36 cm2 and 100 cm2. If the length of a side of the larger triangle is 20 cm, find the length of the corresponding side of the smaller triangle.

Answer : [12 cm ]
Given:- area of smaller triangle = 36 cm²
area of larger triangle = 100 cm²
Length of the side of a larger triangle = 20 cm
Let the length of the corresponding side of the smaller triangle = x cm.
According to the property of the 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:

$\left [ \frac{55}{3}\; cm\right ]$
$\text{Given : }$ $\angle ACB=\angle CDA$
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 the 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-metre-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 : [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 the 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

The 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 : [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 that the top of the ladder reaches is 8 m.

NCERT Exemplar Class 10 Maths Solutions Chapter 6: Exercise-6.4 (Page no. 73, Total questions - 18)

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: [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 terms, we get
$PD=\frac{15 \times 4}{12}$
$PD=5\; cm$
On taking the 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:
$[EF=16.8\; cm, BC= 6.25 \; cm]$
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 be a triangle in which line DE is parallel to BC, which interests lines 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 equations (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 equations (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 the 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: [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 a distance is 0.8 m.

Question:6

For going to 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: [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 the 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: 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 between 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 : [9 m]

Height of street light bulb = 6 m
woman's height = 1.5 m
woman's shadow = 3m
Let the distance between the 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: $\left [ 2\sqrt{5}\; cm\; and\; 6\; cm \right ]$
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 multiplying, 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 multiplying, 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:-

$(a + b) (a - b) = (c + d) (c - d)$
$\text{Proof :- In}$ $\Delta PQD$ $\text{use Pythagoras theorem}$
$\\PQ^{2}=QD^{2}+PD^{2}$
$\\a^{2}=c^{2}+PD^{2}\\a^{2}-c^{2}=PD^{2} \; \; \; \; \; \; \; \; ....(1)$
$\text{In }$ $\Delta PRD$ $\text{use Pythagoras theorem }$
$\\PR^{2}=PD^{2}+DR^{2}\\b^{2}=PD^{2}+d^{2}\\b^{2}-d^{2}=PD^{2}\; \; \; \; \; \; \; ....(2)$
Equating equations (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)
To find $\angle E$, use the sum of the 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
$BC^{2}=BE^{2}+EC^{2}\; \; \; \; \; \; \; \; ....(3)$
Adding equations (2) and (3), we get
$AD^{2}+BC^{2}=AE^{2}+DE^{2}+BE^{2}+CE^{2}\; \; \; \; \; \; \; .....(4)$
In $\Delta ACE$, using 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 equations (5) and (6), and we get
$AC^{2} + BD^{2} = AE^{2} + CE^{2} + BE^{2} + DE^{2}\; \; \; \; \; \; ....(7)$
From equations (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 that
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)
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 the 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 equations (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: $\left [ PQ=8_{cm},QR=12\; cm,RS=16\; cm\right ]$
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 the 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 equations (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 the mid-point of CA (given)
$\therefore \; \; \; \; \; \; CE = AE$
In $\Delta ACG, CG\parallel EF$ and E is mid-point of CA
Then, according to the mid-point theorem
$CE = GF \; \; \; \; \; \; \; ....(1)$
In $\Delta BCG \; and\; \Delta BDF \; \; CG || DF$
According to the 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 be a right triangle with 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, the area of the semi-circle drawn on the side PR is
$x_{3}=\frac{\pi }{2}\left ( \frac{PR}{2} \right )^{2}$ $\left [ \therefore \text {area of semicircle=}\frac{\pi r^{2}}{2} \right ]$
$=\frac{\pi}{2}\left ( \frac{\sqrt{a^{2}+b^{2}}}{2} \right )^{2}$ $\left ( \therefore PR=\sqrt{a^{2}+b^{2}} \right )$
$=\frac{\pi}{2}\times\frac{\left ( a^{2}+b^{2} \right )}{4}$
$x_{3}=\frac{\pi}{8}\left ( a^{2}+b^{2} \right )$
The area of the semi-circle drawn as a side QR is
$x_{2}=\frac{\pi }{2}\left ( \frac{QR}{2} \right )^{2}$
$=\frac{\pi }{2}\left ( \frac{a}{2} \right )^{2}$
$=\frac{\pi}{2}\times \frac{a^{2}}{4}$
$x_{2}=\frac{\pi}{8}a^{2}\; \; \; \; \; \; \; ....(2)$
The area of the semicircle drawn on side PQ is
$x_{1}=\frac{\pi }{2}\left ( \frac{PQ}{2} \right )^{2}$
$x_{1}=\frac{\pi }{2}\left ( \frac{b^{2}}{4} \right )\Rightarrow \frac{\pi}{8}b^{2}\; \; \; \; \; \; \; \; ....(3)$
Adding equations (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 be a right triangle, which is a right angle at point R.
$PR = b, RQ = a$
Three equilateral triangles are drawn on the sides of triangle PQR: 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 for the 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 $\Delta PQU$
$x_{3}=\frac{\sqrt{3}}{4}\left ( \sqrt{a^{2}+b^{2}} \right )^{2}$ $\left ( Q\; PQ=\sqrt{a^{2}+b^{2}} \right )$
$x_{3}=\frac{\sqrt{3}}{4} \left ( a^{2}+b^{2} \right )$
Adding equations (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

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