NCERT Exemplar Class 10 Maths Solutions Chapter 6 Triangles

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

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 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:

Answer : [B]

Here ABCD is a rhombus and we know that diagonals of a rhombus are perpendicular bisector 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 $\mathrm{\Delta }ABC$ is 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:

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 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) $\mathrm{\Delta }PQR\sim \mathrm{\Delta }CAB$
(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:

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 the 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:

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:

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 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:

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 $\mathrm{\Delta }DEF$ 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 $\mathrm{\Delta }ABC\sim \mathrm{\Delta }PQR$, 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:

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:

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 first and the last term of equation (1) we get
$\frac{AB}{DF}=\frac{AC}{DE}$
$\frac{5}{7.5}=\frac{8}{DE}$
$DE=\frac{8\times 75}{50}=12cm$

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:

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:

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 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:

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

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:

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

Answer:

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:

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 are know that if all corresponding angles of two triangles are equal then 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:

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 $\mathrm{\Delta }$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 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.
$\text{In}\mathrm{\Delta }ABC\text{and}\mathrm{\Delta }PQR$
$$\begin{gathered} \mathrm{\angle }B=\mathrm{\angle }Q={90}^o \\ \mathrm{\angle }C=\mathrm{\angle }R=x^o \end{gathered}$$
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 equation (1) and (2)
$\mathrm{\angle }A=\mathrm{\angle }P.....(3)$
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.
$\therefore \frac{\text{Area 1}}{\text{Area 2}}={\left(\frac{\text{Altitude 1}}{\text{Altitude 2}}\right)}^2$
$\frac{\text{Area 1}}{\text{Area 2}}={\left(\frac{3}{5}\right)}^2$ $[Q\frac{\text{Altitude 1}}{\text{Altitude 2}}=\frac{3}{5}]$
$=\frac{9}{25}$
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 $\mathrm{\Delta }$PQR such that PD $\perp$ QR. Will it be correct to say that $\mathrm{\Delta }$PQD ~ $\mathrm{\Delta }$RPD? Why?

Answer:

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.
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 $\mathrm{\angle }$D = $\mathrm{\angle }$C, then is it true that $\mathrm{\Delta }ADE\sim \mathrm{\Delta }ACB$ ? Why?

Answer:

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:

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 : $P{R}^2-P{Q}^2=Q{R}^{2}andQM\perp PR$
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 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 AA similarity criterion.
$\therefore \mathrm{\Delta }QMR\sim \mathrm{\Delta }PMQ$
Now using the property of 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:

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.
$$\begin{gathered} \therefore \frac{CD}{AD}=\frac{CE}{BE} \\ \frac{x+3}{3x+19}=\frac{x}{3x+4} \\ (x+3)(3x+4)=x(3x+19) \end{gathered}$$
$3x^2+4x+9x+12=3x^2+19x$
$19x-13x=12$
$6x=12$
$x=\frac{12}{6}=2$
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 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.
$\therefore ST\parallel QR$
$\mathrm{\angle }1=\mathrm{\angle }PQR$ $\text{(corresponding angle)}$
$\mathrm{\angle }2=\mathrm{\angle }PRQ$ $\text{ (corresponding angle)}$
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 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:

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 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
$\therefore \mathrm{\Delta }POQ\sim \mathrm{\Delta }ROS$
By the property of area of 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 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
$\therefore \mathrm{\Delta }AOP\sim \mathrm{\Delta }COQ$

Then

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