NCERT Exemplar Class 10 Maths Solutions Chapter 6 - Triangles

Triangles are everywhere in our daily lives—whether it's in hangers, sandwiches, or traffic signs. But what exactly is a triangle? A triangle is a two-dimensional, closed shape with three sides, three angles, and three vertices (corners) and is a type of polygon. The sum of the interior angles of a triangle is equal to ${180}^{\circ }$. A triangle is a polygon that has the least number of sides.

NCERT Class 10 Maths solutions were developed by the experts at Careers360 to learn the proper approach for the questions. These Class 10 Maths NCERT exemplar Chapter 6 solutions provide step-by-step detailed solutions to enhance the learning of concepts based on Triangles.

NCERT Exemplar Class 10 Maths Solutions Chapter 6

Question:1

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

(a) $BD.CD=B{C}^2$
(b) $AB.AC=B{C}^2$
(c) $BD.CD=A{D}^2$
(d) $AB.AC=A{D}^2$

Answer:

Given :- $\mathrm{\angle }BAC={90}^o$
First of all we have find $\mathrm{\angle }DBA$ and $\mathrm{\angle }DAC$ in triangles $ABD$ and $ADC$ respectively
In $\mathrm{\Delta }ABD$ and $\mathrm{\angle }ADC$
$\mathrm{\angle }ADB={90}^o$
$\mathrm{\angle }ADC={90}^o$
In $\mathrm{\Delta }ABC$
$\mathrm{\angle }BAC={90}^o$ (given)
Now, let us find the angle CAD from triangle ADC.
Here $\mathrm{\angle }ADC+\mathrm{\angle }DCA+\mathrm{\angle }CAD={180}^o$
(sum of interior angles of triangle = 180^o)
${90}^o+\mathrm{\angle }C+\mathrm{\angle }CAD={180}^o$
$\mathrm{\angle }CAD={180}^o-{90}^o-\mathrm{\angle }C$
$\mathrm{\angle }CAD={90}^o-\mathrm{\angle }C....(1)$
Now, let us find angle DBA using triangle ABC.
Here $\mathrm{\angle }ABC+\mathrm{\angle }BCA+\mathrm{\angle }CAB={180}^o$ {sum of interior angles of triangle = 180^o}
$\mathrm{\angle }ABC+\mathrm{\angle }C+{90}^o={180}^o$
$\mathrm{\angle }ABC={90}^o-\mathrm{\angle }C$
$\mathrm{\angle }ABC=\mathrm{\angle }DBA={90}^o-\mathrm{\angle }C...(2)$
In $\mathrm{\Delta }ABD$ and $\mathrm{\Delta }ADC$ we get
$\mathrm{\angle }ADB=\mathrm{\angle }ADC$ {90^o each}
$AD=AD$ {common side}
$\mathrm{\angle }CAD=\mathrm{\angle }DBA$ {from equation (1) and (2) it is clear that both is ${90}^o-\mathrm{\angle }C$}
$\therefore \mathrm{\Delta }ABD\sim \mathrm{\Delta }ADC$ (by ASA similarity)
$\therefore \frac{BD}{AD}=\frac{AD}{CD}$
$BD.CD=AD.AD$ {by cross multiplication}
$BD.CD=A{D}^2$
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: $AC=16cm$ and $BD=12cm$
$\Rightarrow AO=\frac{AC}{2}=\frac{16}{2}=8cm$
also $BO=\frac{BD}{2}=\frac{12}{2}=6cm$ and $\mathrm{\angle }AOB={90}^o$
Using Pythagoras' theorem in $\mathrm{\Delta }AOB$ we get
$A{B}^2=A{O}^2+O{B}^2$
$A{B}^2=8^2+6^2$
$A{B}^2=64+36$
$A{B}^2=100$
$AB=\sqrt{100}$
$AB=10cm$
Hence, option (B) is correct

Question:3

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

Answer : [C]
Given : $\mathrm{\Delta }ABC\sim \mathrm{\Delta }EDF$
$\therefore \frac{AB}{ED}=\frac{BC}{DF}=\frac{AC}{EF}$
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) $\mathrm{\Delta }PQR\sim \mathrm{\Delta }CAB$
(b) $\mathrm{\Delta }PQR\sim \mathrm{\Delta }ABC$
(c) $\mathrm{\Delta }CBA\sim \mathrm{\Delta }PQR$
(d) $\mathrm{\Delta }BCA\sim \mathrm{\Delta }PQR$

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

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

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

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

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, $\mathrm{\angle }APB={50}^o$ and $\mathrm{\angle }CDP={30}^o$. Then, $\mathrm{\angle }PBA$ is equal to

(A) 50° (B) 30° (C) 60° (D) 100°

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

Question:6

If in two triangles DEF and PQR, $\mathrm{\angle }D=\mathrm{\angle }Q\text{and}\mathrm{\angle }R=\mathrm{\angle }E$, 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 $\mathrm{\Delta }DEF$ and $\mathrm{\Delta }PQR$
$\mathrm{\angle }D=\mathrm{\angle }Q$ and $\mathrm{\angle }R=\mathrm{\angle }E$
Now draw two triangles with the help of the given conditions:-

Here $\mathrm{\angle }D=\mathrm{\angle }Q$ and $\mathrm{\angle }E=\mathrm{\angle }R$ {Given}
$\therefore \mathrm{\Delta }DEF\sim \mathrm{\Delta }QRP$ {by AA similarity}
$\Rightarrow \mathrm{\angle }F=\mathrm{\angle }P$ {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 $ABC\text{and}DEF,\mathrm{\angle }B=\mathrm{\angle }E,\mathrm{\angle }F=\mathrm{\angle }C\text{and}AB=3DE$. 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 $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DEF$
$\mathrm{\angle }B=\mathrm{\angle }E\text{and}\mathrm{\angle }F=\mathrm{\angle }C\text{also}AB=3DE$

In $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DEF$
$\mathrm{\angle }B=\mathrm{\angle }E$ (given)
$\mathrm{\angle }F=\mathrm{\angle }C$ (given)
$\mathrm{\angle }A=\mathrm{\angle }D$ (third angle)
Therefore $\mathrm{\Delta }ABC\sim \mathrm{\Delta }DEF$ (by AAA similarity)
Also, it is given that $AB=3DE$
$\Rightarrow AB\ne DE$
Hence, $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }DEF$ 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 $\mathrm{\Delta }ABC\sim \mathrm{\Delta }PQR$, 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 : $\mathrm{\Delta }ABC\sim \mathrm{\Delta }PQR$
$\frac{BC}{QR}=\frac{1}{3}$

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

Question:9

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

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

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

Question:10

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

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.
$\Rightarrow \mathrm{\angle }A=\mathrm{\angle }E$
$\mathrm{\angle }B=\mathrm{\angle }D$
$\mathrm{\angle }C=\mathrm{\angle }F$
Hence, option C is correct.

Question:11

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

Answer : [A]

$\mathrm{\Delta }ABC\sim \mathrm{\Delta }QRP\text{and}AB=18cm,BC=15cm$
$\therefore \frac{AC}{PQ}=\frac{AB}{QR}=\frac{BC}{RP}$
$\frac{ar(ABC)}{ar\left(PQR\right)}=\frac{B{C}^2}{R{P}^2}....(1)$
[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)]
$\Rightarrow \frac{B{C}^2}{R{P}^2}=\frac{9}{4}$
$\frac{(15{)}^2}{R{P}^2}=\frac{9}{4}$
$R{P}^2=\frac{225\times 4}{9}=100$
$RP=\sqrt{100}$
$RP=10cm$

Question:12

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

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

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.
$\Rightarrow {\left(25\right)}^2={\left(5\right)}^2+{\left(24\right)}^2$
$625=25+576$
$625\ne 601$
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

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

Answer : [False]
$Given:$ $\mathrm{\Delta }DEF\sim \mathrm{\Delta }RPQ$
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
$\mathrm{\angle }D=\mathrm{\angle }R$
$\mathrm{\angle }E=\mathrm{\angle }P$
$\mathrm{\angle }F=\mathrm{\angle }Q$
$\text{In the given statement}$ $\mathrm{\angle }D=\mathrm{\angle }R$ $\text{and}$ $\mathrm{\angle }F=\mathrm{\angle }P$
$\text{But according to properties of a similar triangle}$ $\mathrm{\angle }F\ne \mathrm{\angle }P$ $\text{that is }$. $\text{Hence the given statement is false. }$

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 $\mathrm{\Delta }$PBC $\sim$ $\mathrm{\Delta }$PDE? Why?

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

Question:5

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

Answer: [False]

$\mathrm{\Delta }PQR,\mathrm{\angle }P+\mathrm{\angle }Q+\mathrm{\angle }R={180}^o$$\text{ ({Interior angle of triangle}) }$
$55+25+\mathrm{\angle }R=180$
$\mathrm{\angle }R=180-80$
$\mathrm{\angle }R={100}^o$
$\text{In }$ $\mathrm{\Delta }TSM,\mathrm{\angle }T+\mathrm{\angle }S+\mathrm{\angle }M={180}^o$$\text{[Interior angle of the triangle] }$
$\mathrm{\angle }T+{25}^o+{100}^o={180}^o$
$\mathrm{\angle }T={180}^o-125$
$\mathrm{\angle }T={55}^o$
$\text{In }$ $\mathrm{\Delta }PQR$ $\text{and }$ $\mathrm{\Delta }TSM$
$\mathrm{\angle }P=\mathrm{\angle }T$
$\mathrm{\angle }Q=\mathrm{\angle }S$
$\mathrm{\angle }R=\mathrm{\angle }M$
Also, we know that if all corresponding angles of two triangles are equal, then the triangles are similar.
$\therefore \mathrm{\Delta }PQR\sim \mathrm{\Delta }TSM$
Hence given statement is false because $\mathrm{\angle }$QPR is not similar to $\mathrm{\angle }$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 $\mathrm{\Delta }$ABC is three times the perimeter of $\mathrm{\Delta }$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.
$\text{In}\mathrm{\Delta }ABC\text{and}\mathrm{\Delta }PQR$
$$\begin{gathered} \mathrm{\angle }B=\mathrm{\angle }Q={90}^o \\ \mathrm{\angle }C=\mathrm{\angle }R=x^o \end{gathered}$$
In $\mathrm{\Delta }ABC$
$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^o$
[Sum of interior angles of a triangle is 180°]
$\mathrm{\angle }A+{90}^o+x^o={180}^o$
$\mathrm{\angle }A+{90}^o-x^o....(1)$
Also in $\mathrm{\Delta }PQR$
$\mathrm{\angle }P+\mathrm{\angle }Q+\mathrm{\angle }R={180}^o$
$\mathrm{\angle }P+{90}^o-x^o={180}^o$
$\mathrm{\angle }P={90}^o-x^o.....(2)$
from equations (1) and (2)
$\mathrm{\angle }A=\mathrm{\angle }P.....(3)$
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.
$\therefore \frac{\text{Area 1}}{\text{Area 2}}={\left(\frac{\text{Altitude 1}}{\text{Altitude 2}}\right)}^2$
$\frac{\text{Area 1}}{\text{Area 2}}={\left(\frac{3}{5}\right)}^2$ $[Q\frac{\text{Altitude 1}}{\text{Altitude 2}}=\frac{3}{5}]$
$=\frac{9}{25}$
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 $\mathrm{\Delta }$PQR such that PD $\perp$ QR. Will it be correct to say that $\mathrm{\Delta }$PQD ~ $\mathrm{\Delta }$RPD? Why?

Answer: [False]

In $\mathrm{\Delta }$PQD and $\mathrm{\Delta }$RPD
$PD=PD$ (common side)
$\mathrm{\angle }PDQ=\mathrm{\angle }PDR$ (each 90^o)
The given statement $\mathrm{\Delta }PQD\sim \mathrm{\Delta }RPD$ 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 $\mathrm{\angle }$D = $\mathrm{\angle }$C, then is it true that $\mathrm{\Delta }ADE\sim \mathrm{\Delta }ACB$ ? Why?

Answer : [True]
In $\mathrm{\Delta }$ ADE and $\mathrm{\Delta }$ ACB
$\mathrm{\angle }D=\mathrm{\angle }C$ (given)
$\mathrm{\angle }A=\mathrm{\angle }A$ (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.
$\therefore$ $\mathrm{\Delta }ADE\sim \mathrm{\Delta }ACB$ {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 : $P{R}^2-P{Q}^2=Q{R}^{2}andQM\perp PR$
To prove : $Q{M}^2=PM\times MR$
Proof : $P{R}^2-P{Q}^2=Q{R}^2$ (given)
$P{R}^2=P{Q}^2+Q{R}^2$
Because $\mathrm{\Delta }$PQR holds Pythagoras theorem, therefore, $\mathrm{\Delta }$PQR is a right-angled triangle right angle at Q.
$In\mathrm{\Delta }QMRand\mathrm{\Delta }PMQ$
$\mathrm{\angle }M=\mathrm{\angle }M$ {each angle is 90°}
$\mathrm{\angle }MQR=\mathrm{\angle }QPM$ [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.
$\therefore \mathrm{\Delta }QMR\sim \mathrm{\Delta }PMQ$
Now, using the property of the area of similar triangles.
$\frac{ar(\mathrm{\Delta }QMR)}{ar(\mathrm{\Delta }PMQ)}=\frac{(QM{)}^2}{(PM{)}^2}$
$\frac{\frac{1}{2}\times RM\times QM}{\frac{1}{2}\times PM\times QM}=\frac{(QM{)}^2}{(PM{)}^2}$ $\{\text{Q area of triangle}=\frac{1}{2}\times base\times height\}$
$\Rightarrow Q{M}^2=PM\times RM$

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.
$$\begin{gathered} \therefore \frac{CD}{AD}=\frac{CE}{BE} \\ \frac{x+3}{3x+19}=\frac{x}{3x+4} \\ (x+3)(3x+4)=x(3x+19) \end{gathered}$$
$3x^2+4x+9x+12=3x^2+19x$
$19x-13x=12$
$6x=12$
$x=\frac{12}{6}=2$
Hence, the required value of x is 2.

Question:3

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

Answer:

Given :
$\mathrm{\Delta }NSQ\cong \mathrm{\Delta }MTRand\mathrm{\angle }1=\mathrm{\angle }2$
To prove:-
$\mathrm{\Delta }PTS\sim \mathrm{\Delta }PRQ$
Proof:-
$\text{It is given that }\mathrm{\Delta }NSQ\cong \mathrm{\Delta }MTR$
$\therefore SQ=TR...(1)$
$\mathrm{\angle }1=\mathrm{\angle }2$
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.
$\therefore ST\parallel QR$
$\mathrm{\angle }1=\mathrm{\angle }PQR$ $\text{(corresponding angle)}$
$\mathrm{\angle }2=\mathrm{\angle }PRQ$ $\text{ (corresponding angle)}$
In $\mathrm{\Delta }PTS$ and $\mathrm{\Delta }PRQ$
$\mathrm{\angle }P=\mathrm{\angle }P$ $\text{ (corresponding angle)}$
$\mathrm{\angle }1=\mathrm{\angle }PQR$
$\mathrm{\angle }2=\mathrm{\angle }PRQ$
We know that if the corresponding angles of two triangles are equal, then the triangles are similar by the AAA similarity criterion

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

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.
$\Rightarrow \frac{PQ}{RS}=\frac{3}{1}$

$In\mathrm{\Delta }POQandROS$
$\mathrm{\angle }SOR=\mathrm{\angle }QOP$ (vertically opposite angles)
$\mathrm{\angle }SRP=\mathrm{\angle }RPQ$ (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.
$\therefore \mathrm{\Delta }POQ\sim \mathrm{\Delta }ROS$
By the property of an area of a similar triangle
$\frac{ar\left(\mathrm{\Delta }POQ\right)}{ar\left(\mathrm{\Delta }SOR\right)}=\frac{{\left(PQ\right)}^2}{{\left(RS\right)}^2}={\left(\frac{PQ}{RS}\right)}^2={\left(\frac{3}{1}\right)}^2$
$\frac{ar\left(\mathrm{\Delta }POQ\right)}{ar\left(\mathrm{\Delta }SOR\right)}=\frac{9}{1}$
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:-
$In\mathrm{\Delta }AOPand\mathrm{\Delta }COQ$
$\mathrm{\angle }AOP=\mathrm{\angle }COQ$ $\text{(vertically opposite angles)}$
Since AB || OC and PQ is transversal
$\therefore \mathrm{\angle }APO=\mathrm{\angle }CQO$ $\text{(alternate angles)}$
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.
$\therefore \mathrm{\Delta }AOP\sim \mathrm{\Delta }COQ$

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 $\mathrm{\Delta }ABC\sim \mathrm{\Delta }DEF,AB=4cm,DE=6cm,EF=9cmandFD=12cm,$ find the perimeter of $\mathrm{\Delta }ABC.$