MCQ 11 Mark
One side of a square is $10 cm$ then its diagonal is _______ .
- A$10$
- B$20$
- ✓$10 \sqrt{2}$
- D$\frac{10}{\sqrt{2}}$
Answer
View full question & answer→Correct option: C.
$10 \sqrt{2}$
$10 \sqrt{2}$
Questions
M.C.Q (1 Marks)
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45 questions · timed · auto-graded
MCQ 21 Mark
Two polygons having same number of sides are similar, if
- Atheir corresponding sides are proportional
- Btheir corresponding angles are equal
- ✓both (a) and (b)
- Dnone of these
Answer
View full question & answer→Correct option: C.
both (a) and (b)
(c) : both (a) and (b)
MCQ 31 Mark
Which of the following pairs shows similar figures?
- A
- B
- ✓
- D
Answer
View full question & answer→Correct option: C.
(c) :
MCQ 41 Mark
If $\triangle A B C \sim \triangle P Q R$, then what are the values of $x$ and $y$ ?
- ✓$\frac{21}{4}, \frac{15}{2}$
- B$\frac{15}{2}, \frac{17}{2}$
- C$\frac{21}{2}, \frac{15}{4}$
- DNone of these
Answer
View full question & answer→Correct option: A.
$\frac{21}{4}, \frac{15}{2}$
(a) : Since, $\triangle A B C \sim \triangle P Q R$
$
\Rightarrow \frac{A B}{P Q}=\frac{B C}{Q R} \Rightarrow \frac{12}{9}=\frac{7}{x} \Rightarrow x=\frac{7 \times 9}{12}=\frac{21}{4}
$
Also, $\frac{A B}{P Q}=\frac{A C}{P R} \Rightarrow \frac{12}{9}=\frac{10}{y} \Rightarrow y=10 \times \frac{9}{12}=\frac{15}{2}$
$
\Rightarrow \frac{A B}{P Q}=\frac{B C}{Q R} \Rightarrow \frac{12}{9}=\frac{7}{x} \Rightarrow x=\frac{7 \times 9}{12}=\frac{21}{4}
$
Also, $\frac{A B}{P Q}=\frac{A C}{P R} \Rightarrow \frac{12}{9}=\frac{10}{y} \Rightarrow y=10 \times \frac{9}{12}=\frac{15}{2}$
MCQ 51 Mark
If $\triangle \text{A B C} \sim \triangle \text{P Q R}$, then $y+z$ equals
- A$2+\sqrt{3}$
- ✓$4+3 \sqrt{3}$
- C$4+\sqrt{3}$
- D$3+4 \sqrt{3}$
Answer
View full question & answer→Correct option: B.
$4+3 \sqrt{3}$
Given, $\triangle \text{A B C} \sim \triangle \text{P Q R}$
$\Rightarrow \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P} $
$\Rightarrow \frac{z}{3}=\frac{8}{6}=\frac{4 \sqrt{3}}{y}$
$\Rightarrow \frac{z}{3}=\frac{8}{6}$ and $\frac{8}{6}=\frac{4 \sqrt{3}}{y}$
$\Rightarrow z=\frac{24}{6}=4$ and $y=\frac{24 \sqrt{3}}{8}=3 \sqrt{3}$
$\Rightarrow y+z=4+3 \sqrt{3} .$
$\Rightarrow \frac{A B}{P Q}=\frac{B C}{Q R}=\frac{C A}{R P} $
$\Rightarrow \frac{z}{3}=\frac{8}{6}=\frac{4 \sqrt{3}}{y}$
$\Rightarrow \frac{z}{3}=\frac{8}{6}$ and $\frac{8}{6}=\frac{4 \sqrt{3}}{y}$
$\Rightarrow z=\frac{24}{6}=4$ and $y=\frac{24 \sqrt{3}}{8}=3 \sqrt{3}$
$\Rightarrow y+z=4+3 \sqrt{3} .$
MCQ 61 Mark
In the given figure, $\triangle \text{A C B} \sim \triangle \text{A P Q}$. If $\text{A B}=6\ cm , \text{B C}=8 \ cm$ and $\text{P Q}=4 \ cm$, then $\text{A Q}$ is equal to
- A$2 \ cm$
- B$2.5 \ cm$
- ✓$3 \ cm$
- D$3.5 \ cm$
Answer
View full question & answer→Correct option: C.
$3 \ cm$
Given, $\triangle \text{A C B} \sim \triangle \text{A P Q}$
$\Rightarrow \frac{B C}{P Q}=\frac{A B}{A Q}$
$ \Rightarrow \frac{8}{4}=\frac{6}{A Q}$
$ \Rightarrow A Q=\frac{6 \times 4}{8}$
$\Rightarrow A Q=3 \ cm$
$\Rightarrow \frac{B C}{P Q}=\frac{A B}{A Q}$
$ \Rightarrow \frac{8}{4}=\frac{6}{A Q}$
$ \Rightarrow A Q=\frac{6 \times 4}{8}$
$\Rightarrow A Q=3 \ cm$
MCQ 71 Mark
If triangle $\text{ABC}$ is similar to triangle $\text{DEF}$ such that $\text{3AB=DE}$ and $\ce{BC=9\ cm}$, then $\text{EF}$ is equal to
- ✓$27 \ cm$
- B$3 \ cm$
- C$6 \ cm$
- D$9 \ cm$
Answer
View full question & answer→Correct option: A.
$27 \ cm$
Given, $\triangle A B C \sim \triangle D E F$
$\Rightarrow \frac{A B}{D E}=\frac{B C}{E F}$
$\Rightarrow \frac{A B}{3 A B}=\frac{9}{E F}[\because 3 A B=D E]$
$\Rightarrow E F=27 \ cm$
$\Rightarrow \frac{A B}{D E}=\frac{B C}{E F}$
$\Rightarrow \frac{A B}{3 A B}=\frac{9}{E F}[\because 3 A B=D E]$
$\Rightarrow E F=27 \ cm$
MCQ 81 Mark
In a $\triangle A B C$, it is given that $A B=6 cm$, $A C=8 cm$ and $A D$ is the bisector of $\angle A$. Then, $B D: D C$ is equal to
- ✓$3: 4$
- B$9: 16$
- C$4: 3$
- DNone of these
Answer
View full question & answer→Correct option: A.
$3: 4$
(a) : We know that the bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle.
$
\Rightarrow B D: D C=A B: A C=6: 8=3: 4
$
$
\Rightarrow B D: D C=A B: A C=6: 8=3: 4
$
MCQ 91 Mark
If $\triangle \text{A B C} \sim \triangle \text{P Q R}$, perimeter of $\triangle \text{A B C} =20 \ cm$, perimeter of $\triangle \text{P Q R} =40 \ cm$ and $\text{P R}=8 \ cm$, then the length of $\text{A C}$ is
- A$8 \ cm$
- B$6 \ cm$
- ✓$4 \ cm$
- D$5 \ cm$
Answer
View full question & answer→Correct option: C.
$4 \ cm$
Since$, \triangle \text{A B C} \sim \triangle \text{P Q R}$
$\Rightarrow \frac{A C}{P R}=\frac{\text { Perimeter of } \triangle \text{A B C}}{\text { Perimeter of } \triangle \text{P Q R}}$
$\Rightarrow \frac{A C}{8}=\frac{20}{40} $
$\Rightarrow A C=\frac{20 \times 8}{40}=4 \ cm $
$\Rightarrow \frac{A C}{P R}=\frac{\text { Perimeter of } \triangle \text{A B C}}{\text { Perimeter of } \triangle \text{P Q R}}$
$\Rightarrow \frac{A C}{8}=\frac{20}{40} $
$\Rightarrow A C=\frac{20 \times 8}{40}=4 \ cm $
MCQ 101 Mark
In $\triangle A B C$, it is given that $A B=9 \ cm$, $B C=6 \ cm$ and $C A=7.5 \ cm$. Also, $\triangle D E F$ is given such that $E F=8 \ cm$ and $\triangle D E F \sim \triangle A B C$. Then, perimeter of $\triangle D E F$ is
- A$22.5 \ cm$
- B$25 \ cm$
- C$27 \ cm$
- ✓$30 \ cm$
Answer
View full question & answer→Correct option: D.
$30 \ cm$
Perimeter of $\triangle A B C =(9+6+7.5) \ cm$
$ =22.5 \ cm$
Let the perimeter of $\triangle D E F$ be $p cm$.
Given $\triangle D E F \sim \triangle A B C$
$\Rightarrow \frac{\text { Perimeter of } \triangle D E F}{\text { Perimeter of }$
$\triangle A B C}=\frac{E F}{B C}$
$\Rightarrow \frac{p}{22.5}=\frac{8}{6}$
$\Rightarrow p=\frac{22.5 \times 8}{6}$
$=\frac{225 \times 8}{60}$
$=30 \ cm$
$ =22.5 \ cm$
Let the perimeter of $\triangle D E F$ be $p cm$.
Given $\triangle D E F \sim \triangle A B C$
$\Rightarrow \frac{\text { Perimeter of } \triangle D E F}{\text { Perimeter of }$
$\triangle A B C}=\frac{E F}{B C}$
$\Rightarrow \frac{p}{22.5}=\frac{8}{6}$
$\Rightarrow p=\frac{22.5 \times 8}{6}$
$=\frac{225 \times 8}{60}$
$=30 \ cm$
MCQ 111 Mark
The perimeters of two similar triangles $\text{ABC}$ and $\text{PQR}$ are $60 \ cm$ and $36 \ cm$ respectively. If $P Q=9 \ cm$, then $A B$ equals
- A$6 \ cm$
- B$10 \ cm$
- ✓$15 \ cm$
- D$24 \ cm$
Answer
View full question & answer→Correct option: C.
$15 \ cm$
Given, $\triangle A B C \sim \triangle P Q R$
$\Rightarrow \frac{A B}{P Q}=\frac{\text { Perimeter of } \triangle A B C}{\text { Perimeter of } \triangle P Q R}$
$\Rightarrow \frac{A B}{9}=\frac{60}{36}$
$\Rightarrow A B=15 \ cm$
$\Rightarrow \frac{A B}{P Q}=\frac{\text { Perimeter of } \triangle A B C}{\text { Perimeter of } \triangle P Q R}$
$\Rightarrow \frac{A B}{9}=\frac{60}{36}$
$\Rightarrow A B=15 \ cm$
MCQ 121 Mark
In the given figure, if $P Q \| B C$. Find $A Q$.
- A$3.5 cm$
- ✓$4.5 cm$
- C$9 cm$
- D$9.5 cm$
Answer
View full question & answer→Correct option: B.
$4.5 cm$
(b): In $\triangle A B C$, we have $P Q \| B C$ $\Rightarrow \quad \frac{A Q}{Q C}=\frac{A P}{P B}$ (By Thales theorem)
$\Rightarrow \quad \frac{A Q}{3}=\frac{3}{2} \Rightarrow A Q=\frac{9}{2}=4.5 cm$
$\Rightarrow \quad \frac{A Q}{3}=\frac{3}{2} \Rightarrow A Q=\frac{9}{2}=4.5 cm$
MCQ 131 Mark
If in two triangles, corresponding sides are in the same ratio, then the two triangles are similar. This criterion is known as
- ✓SSS similarity criterion
- BSAS similarity criterion
- CAA similarity criterion
- DAAA similarity criterion.
Answer
View full question & answer→Correct option: A.
SSS similarity criterion
(a) : SSS similarity criterion
MCQ 141 Mark
If in two similar triangles $A B C$ and $D E F$, $\frac{A B}{D E}=\frac{B C}{E F}$, then
- ✓$\angle B=\angle E$
- B$\angle A=\angle E$
- C$\angle B=\angle D$
- D$\angle A=\angle F$
Answer
View full question & answer→Correct option: A.
$\angle B=\angle E$
(a) : We have, $\triangle A B C \sim \triangle D E F$
Also, $\frac{A B}{D E}=\frac{B C}{E F}$ (Given)
By SAS similarity criterion, $\angle B=\angle E$.
Also, $\frac{A B}{D E}=\frac{B C}{E F}$ (Given)
By SAS similarity criterion, $\angle B=\angle E$.
MCQ 151 Mark
In two triangles $\text{A B C}$ and $\text{D E F}, \angle A=\angle E$ and $\angle B=\angle F$. Then, $\frac{A B}{A C}$ is equal to
- A$\frac{D E}{D F}$
- B$\frac{E D}{E F}$
- ✓$\frac{E F}{E D}$
- D$\frac{E F}{D F}$
Answer
View full question & answer→Correct option: C.
$\frac{E F}{E D}$
In $\triangle \text{A B C}$ and $\triangle \text{E F D}$,
$\angle A=\angle E$ and $\angle B=\angle F\ ($Given$)$
$\Rightarrow \triangle \text{A B C} \sim \triangle \text{E F D}\ ($By $AA$ similarity criterion$)$
$\Rightarrow \frac{A B}{E F}=\frac{A C}{E D} $
$\Rightarrow \frac{A B}{A C}=\frac{E F}{E D}$
$\angle A=\angle E$ and $\angle B=\angle F\ ($Given$)$
$\Rightarrow \triangle \text{A B C} \sim \triangle \text{E F D}\ ($By $AA$ similarity criterion$)$
$\Rightarrow \frac{A B}{E F}=\frac{A C}{E D} $
$\Rightarrow \frac{A B}{A C}=\frac{E F}{E D}$
MCQ 161 Mark
In the figure, find $x$ in terms of $a, b$ and $c$.
- A$\frac{a b}{a+c}$
- ✓$\frac{a c}{b+c}$
- C$\frac{b c}{a+b}$
- D$\frac{a c}{a+b}$
Answer
View full question & answer→Correct option: B.
$\frac{a c}{b+c}$
In $\triangle K N P$ and $\triangle K M L$
$\angle K N P=\angle K M L=35^{\circ} ($ Given $)$
$\angle K=\angle K($ Common $)$
$\Rightarrow \triangle K N P \sim \triangle K M L ($By $AA$ similarity criterion$)$
$\Rightarrow \frac{P N}{L M}=\frac{K N}{K M}$
$\Rightarrow \frac{x}{a}=\frac{c}{K N+N M}=\frac{c}{c+b}$
$\Rightarrow x=\frac{a c}{b+c}$
$\angle K N P=\angle K M L=35^{\circ} ($ Given $)$
$\angle K=\angle K($ Common $)$
$\Rightarrow \triangle K N P \sim \triangle K M L ($By $AA$ similarity criterion$)$
$\Rightarrow \frac{P N}{L M}=\frac{K N}{K M}$
$\Rightarrow \frac{x}{a}=\frac{c}{K N+N M}=\frac{c}{c+b}$
$\Rightarrow x=\frac{a c}{b+c}$
MCQ 171 Mark
Ratio of areas of two similar triangles is $2: 3$. The ratio of their corresponding sides is
- A$\sqrt{3}: \sqrt{2}$
- B$\sqrt{6}: 1$
- C$1: \sqrt{6}$
- ✓$\sqrt{2}: \sqrt{3}$
Answer
View full question & answer→Correct option: D.
$\sqrt{2}: \sqrt{3}$
(d) : Since, ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
$\therefore \quad$ Square of ratio of corresponding sides $=\frac{2}{3}$.
$\Rightarrow \quad$ Ratio of the corresponding sides $=\frac{\sqrt{2}}{\sqrt{3}}$
$\therefore \quad$ Square of ratio of corresponding sides $=\frac{2}{3}$.
$\Rightarrow \quad$ Ratio of the corresponding sides $=\frac{\sqrt{2}}{\sqrt{3}}$
MCQ 181 Mark
The areas of two similar triangles are $25 cm ^2$ and $36 cm ^2$. If the median of the smaller triangle is $10 cm$, then the median of the larger triangle is
- ✓$12 cm$
- B$15 cm$
- C$10 cm$
- D$18 cm$
Answer
View full question & answer→Correct option: A.
$12 cm$
(a) : Let the median of the larger triangle be $x cm$. We know, the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
$
\Rightarrow \quad \frac{25}{36}=\left(\frac{10}{x}\right)^2 \Rightarrow x^2=\frac{100 \times 36}{25}=144 \Rightarrow x=12 cm
$
$
\Rightarrow \quad \frac{25}{36}=\left(\frac{10}{x}\right)^2 \Rightarrow x^2=\frac{100 \times 36}{25}=144 \Rightarrow x=12 cm
$
MCQ 191 Mark
The areas of two similar triangles are $121 \ cm ^2$ and $64 \ cm ^2$ respectively. If the median of the first triangle is $12.1 \ cm$, then the corresponding median of the other triangle is
- A$11 \ cm$
- ✓$8.8 \ cm$
- C$11.1 \ cm$
- D$8.1 \ cm$
Answer
View full question & answer→Correct option: B.
$8.8 \ cm$
Let the corresponding median of the other triangle be $x \ cm$.
As the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
$\therefore \frac{121}{64}=\left(\frac{12.1}{x}\right)^2$
$\Rightarrow x^2=\frac{(12.1)^2 \times 64}{121}=77.44$
$\Rightarrow x=8.8 \ cm$
As the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.
$\therefore \frac{121}{64}=\left(\frac{12.1}{x}\right)^2$
$\Rightarrow x^2=\frac{(12.1)^2 \times 64}{121}=77.44$
$\Rightarrow x=8.8 \ cm$
MCQ 201 Mark
$\triangle ABC \sim \triangle PQR$ such that $\operatorname{ar}(\triangle ABC)=4 \operatorname{ar}(\triangle P Q R)$. If $B C=12 \ cm$, then $Q R=$
- A$9 \ cm$
- B$10 \ cm$
- ✓$6 \ cm$
- D$8 \ cm$
Answer
View full question & answer→Correct option: C.
$6 \ cm$
Since $\triangle A B C \sim \triangle P Q R$
$\therefore \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle P Q R)}=\frac{B C^2}{Q R^2}$
$\Rightarrow \frac{4 \operatorname{ar}(\triangle P Q R)}{\operatorname{ar}(\triangle P Q R)}=\frac{12^2}{Q R^2}$
$[\because$ Given, $\operatorname{ar}(\triangle A B C)=4 \operatorname{ar}(\triangle P Q R)]$
$\Rightarrow Q R^2=\frac{144}{4}=36$
$\Rightarrow Q R=6 \ cm $
$\therefore \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle P Q R)}=\frac{B C^2}{Q R^2}$
$\Rightarrow \frac{4 \operatorname{ar}(\triangle P Q R)}{\operatorname{ar}(\triangle P Q R)}=\frac{12^2}{Q R^2}$
$[\because$ Given, $\operatorname{ar}(\triangle A B C)=4 \operatorname{ar}(\triangle P Q R)]$
$\Rightarrow Q R^2=\frac{144}{4}=36$
$\Rightarrow Q R=6 \ cm $
MCQ 211 Mark
If in two triangles $A B C$ and $P Q R$, $\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}$, then
- ✓$\triangle P Q R \sim \triangle C A B$
- B$\triangle P Q R \sim \triangle A B C$
- C$\triangle C B A \sim \triangle P Q R$
- D$\triangle B C A \sim \triangle P Q R$
Answer
View full question & answer→Correct option: A.
$\triangle P Q R \sim \triangle C A B$
(a): Given, in triangles $A B C$ and $P Q R$, $\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}$, which shows that sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are also equal. So, by SSS similarity, triangles are similar i.e., $\triangle C A B \sim \triangle P Q R$.
MCQ 221 Mark
If in two triangles $\text{DEF}$ and $\text{PQR}, \angle D=\angle Q$ and $\angle R=\angle E$, then which of the following is not true?
- A$\frac{E F}{P R}=\frac{D F}{P Q}$
- ✓$\frac{D E}{P Q}=\frac{E F}{R P}$
- C$\frac{D E}{Q R}=\frac{D F}{P Q}$
- D$\frac{E F}{R P}=\frac{D E}{Q R}$
Answer
View full question & answer→Correct option: B.
$\frac{D E}{P Q}=\frac{E F}{R P}$
Given, in $\triangle D E F$ and $\triangle P Q R$
$\angle D=\angle Q, \angle R=\angle E$
$\therefore \triangle D E F \sim \triangle Q R P$
$($ By $AA$ similarity criterion$)$
$\therefore \frac{D F}{Q P}=\frac{E D}{R Q}=\frac{F E}{P R}$
$\angle D=\angle Q, \angle R=\angle E$
$\therefore \triangle D E F \sim \triangle Q R P$
$($ By $AA$ similarity criterion$)$
$\therefore \frac{D F}{Q P}=\frac{E D}{R Q}=\frac{F E}{P R}$
MCQ 231 Mark
In triangles $A B C$ and $D E F, \angle B=\angle E, \angle F=\angle C$ and $A B=3 D E$. Then, the two triangles are
- Acongruent but not similar
- ✓similar but not congruent
- Cneither congruent nor similar
- Dcongruent as well as similar
Answer
View full question & answer→Correct option: B.
similar but not congruent
(b): In $\triangle A B C$ and $\triangle D E F, \angle B=\angle E, \angle F=\angle C$ and $A B$ $=3 DE$
We know that, if in two triangles corresponding two angles are same, then they are similar by AA similarity criterion.
Since, $A B \neq D E$
Therefore, $\triangle A B C$ and $\triangle D E F$ are not congruent.
We know that, if in two triangles corresponding two angles are same, then they are similar by AA similarity criterion.
Since, $A B \neq D E$
Therefore, $\triangle A B C$ and $\triangle D E F$ are not congruent.
MCQ 241 Mark
If in triangles $A B C$ and $D E F, \frac{A B}{D E}=\frac{B C}{F D}$, then they will be similar, when
- A$\angle B=\angle E$
- B$\angle A=\angle D$
- ✓$\angle B=\angle D$
- D$\angle A=\angle F$
Answer
View full question & answer→Correct option: C.
$\angle B=\angle D$
(c) : Given, in $\triangle A B C$ and $\triangle E D F, \frac{A B}{D E}=\frac{B C}{F D}$
So, $\triangle A B C \sim \triangle E D F$
if $\angle B=\angle D$
(By SAS similarity criterion)
So, $\triangle A B C \sim \triangle E D F$
if $\angle B=\angle D$
(By SAS similarity criterion)
MCQ 251 Mark
$D$ is a point on side $\text{BC}$ of a $\triangle ABC$ such that $\angle ADC=\angle BAC$, then $CA^2=$
- A$B C \times A D$
- B$B C^2$
- C$A B^2$
- ✓$B C \times C D$
Answer
View full question & answer→Correct option: D.
$B C \times C D$
In $\triangle A B C$ and $\triangle D A C$,
$\angle B A C=\angle A D C ($ Given$)$
$\angle B C A=\angle A C D($Common$)$
$\therefore \triangle A B C \sim \triangle D A C$
$($By $AA$ similarity criterion$)$
$\Rightarrow \frac{B C}{C A}=\frac{C A}{C D}$
$\Rightarrow C A^2=B C \times C D$
$\angle B A C=\angle A D C ($ Given$)$
$\angle B C A=\angle A C D($Common$)$
$\therefore \triangle A B C \sim \triangle D A C$
$($By $AA$ similarity criterion$)$
$\Rightarrow \frac{B C}{C A}=\frac{C A}{C D}$
$\Rightarrow C A^2=B C \times C D$
MCQ 261 Mark
If $\triangle A B C \sim \triangle P Q R, A B=6.5 \ cm , P Q=10.4 \ cm$ and perimeter of $\triangle A B C=60 \ cm$, then find the perimeter of $\triangle P Q R$.
- ✓$96 \ cm$
- B$100 \ cm$
- C$12 \ cm$
- D$60 \ cm$
Answer
View full question & answer→Correct option: A.
$96 \ cm$
$\triangle ABC \sim \triangle PQR$
$\therefore \frac{A B}{P Q}=\frac{\text { Perimeter of } \triangle A B C}{\text { Perimeter of } \triangle P Q R}$
$\Rightarrow \frac{6.5}{10.4}=\frac{60}{\text { Perimeter of } \triangle P Q R}$
$\Rightarrow$ Perimeter of $\triangle P Q R=\frac{60 \times 10.4}{6.5}$
$=\frac{60 \times 104}{65}$
$=\frac{60 \times 8}{5}$
$=96 \ cm $
$\therefore \frac{A B}{P Q}=\frac{\text { Perimeter of } \triangle A B C}{\text { Perimeter of } \triangle P Q R}$
$\Rightarrow \frac{6.5}{10.4}=\frac{60}{\text { Perimeter of } \triangle P Q R}$
$\Rightarrow$ Perimeter of $\triangle P Q R=\frac{60 \times 10.4}{6.5}$
$=\frac{60 \times 104}{65}$
$=\frac{60 \times 8}{5}$
$=96 \ cm $
MCQ 271 Mark
In the given figure, $D E \| A B, A D=2 x$, $D C=x+3, B E=2 x-1$ and $C E=x$. Then, find the value of $x$.
- A$5 / 3$
- ✓$3 / 5$
- C$3 / 2$
- D$2 / 3$
Answer
View full question & answer→Correct option: B.
$3 / 5$
$\operatorname{In} \triangle A B C, D E \| A B$
$\therefore \frac{C D}{D A}=\frac{C E}{E B}$
$\Rightarrow \frac{x+3}{2 x}=\frac{x}{2 x-1}$
$\Rightarrow(x+3)(2 x-1)=(2 x)(x)$
$\Rightarrow 2 x^2-x+6 x-3=2 x^2$
$\Rightarrow 5 x-3=0 \Rightarrow x=3 / 5$
$\therefore \frac{C D}{D A}=\frac{C E}{E B}$
$\Rightarrow \frac{x+3}{2 x}=\frac{x}{2 x-1}$
$\Rightarrow(x+3)(2 x-1)=(2 x)(x)$
$\Rightarrow 2 x^2-x+6 x-3=2 x^2$
$\Rightarrow 5 x-3=0 \Rightarrow x=3 / 5$
MCQ 281 Mark
In the given figure, $B A \| Q R$, and $C A \| S R$.
Then, $\frac{Q B}{B P}=$
Then, $\frac{Q B}{B P}=$
- A$\frac{A B}{A C}$
- B$\frac{A P}{R A}$
- ✓$\frac{S C}{C P}$
- D$\frac{P C}{S C}$
Answer
View full question & answer→Correct option: C.
$\frac{S C}{C P}$
(c) : In $\triangle P R Q, A B|| R Q$
$\therefore \quad \frac{Q B}{B P}=\frac{R A}{A P} \quad$ (By Thales theorem)$\ldots(i)$
In $\triangle P R S, C A|| S R$
$\therefore \quad \frac{R A}{A P}=\frac{S C}{C P} \quad$ (By Thales theorem)$\ldots(ii)$
From (i) and (ii), we have $\frac{Q B}{B P}=\frac{S C}{C P}$
$\therefore \quad \frac{Q B}{B P}=\frac{R A}{A P} \quad$ (By Thales theorem)$\ldots(i)$
In $\triangle P R S, C A|| S R$
$\therefore \quad \frac{R A}{A P}=\frac{S C}{C P} \quad$ (By Thales theorem)$\ldots(ii)$
From (i) and (ii), we have $\frac{Q B}{B P}=\frac{S C}{C P}$
MCQ 291 Mark
$D$ and $E$ are points on the sides $A B$ and $A C$ of a $\triangle A B C . A B=12 cm , A D=8 cm , A E=12 cm$ and $A C=18 cm$. Then
- A$B D \| C E$
- B$A D \perp A E$
- C$D E \perp BC$
- D$D E \| B C$
Answer
(d) : $B D=A B-A D=12-8=4 cm$
$
C E=A C-A E=18-12=6 cm
$
Now, $\frac{A D}{B D}=\frac{8}{4}=2$
Again, $\frac{A E}{C E}=\frac{12}{6}=2$
From (i) and (ii), we get $\frac{A D}{B D}=\frac{A E}{C E}$
$\Rightarrow D E \| B C$ (By Converse of Thales theorem)
View full question & answer→(d) : $B D=A B-A D=12-8=4 cm$
$
C E=A C-A E=18-12=6 cm
$
Now, $\frac{A D}{B D}=\frac{8}{4}=2$
Again, $\frac{A E}{C E}=\frac{12}{6}=2$
From (i) and (ii), we get $\frac{A D}{B D}=\frac{A E}{C E}$
$\Rightarrow D E \| B C$ (By Converse of Thales theorem)
MCQ 301 Mark
In the given figure, find $\angle F$.
- ✓$60^{\circ}$
- B$80^{\circ}$
- C$40^{\circ}$
- D$100^{\circ}$
Answer
View full question & answer→Correct option: A.
$60^{\circ}$
In triangles $\text{ABC}$ and $\text{DFE}$, we have
$\frac{A B}{D F}=\frac{3.8}{7.6}=\frac{1}{2}, \frac{B C}{F E}=\frac{6}{12}=\frac{1}{2}, \frac{C A}{E D}=\frac{3 \sqrt{3}}{6 \sqrt{3}}=\frac{1}{2}$
$\therefore \triangle A B C \sim \triangle D F E ($ By $\text{SSS}$ similarity criterion$)$
$\Rightarrow \angle A=\angle D, \angle B=\angle F \text { and } \angle C=\angle E$
$($Corresponding angles of similar triangles$)$
Hence, $\angle F=60^{\circ}$
$\frac{A B}{D F}=\frac{3.8}{7.6}=\frac{1}{2}, \frac{B C}{F E}=\frac{6}{12}=\frac{1}{2}, \frac{C A}{E D}=\frac{3 \sqrt{3}}{6 \sqrt{3}}=\frac{1}{2}$
$\therefore \triangle A B C \sim \triangle D F E ($ By $\text{SSS}$ similarity criterion$)$
$\Rightarrow \angle A=\angle D, \angle B=\angle F \text { and } \angle C=\angle E$
$($Corresponding angles of similar triangles$)$
Hence, $\angle F=60^{\circ}$
MCQ 311 Mark
If $\triangle O A Q \sim \triangle O B P, P B$ and $Q A$ are perpendiculars to segment $A B$ and $P O=5 \ cm$, $Q O=6 \ cm$ and $\operatorname{ar}(\triangle P O B)=125 \ cm ^2$, then the area of $\triangle Q O A$ is
- A$100 \ cm ^2$
- B$150 \ cm ^2$
- C$200 \ cm ^2$
- ✓$180 \ cm ^2$
Answer
View full question & answer→Correct option: D.
$180 \ cm ^2$
$\frac{\operatorname{ar}(\triangle O A Q)}{\operatorname{ar}(\triangle O B P)}=\frac{O Q^2}{O P^2} \quad(\because \triangle O A Q \sim \triangle O B P)$
$\Rightarrow \frac{\operatorname{ar}(\triangle O A Q)}{125}=\frac{6^2}{5^2} \quad\left(\because \operatorname{ar}(\triangle O B P)=125 \ cm ^2\right)$
$\Rightarrow \operatorname{ar}(\triangle O A Q)=\frac{36}{25} \times 125=180 \ cm ^2$
$\Rightarrow \frac{\operatorname{ar}(\triangle O A Q)}{125}=\frac{6^2}{5^2} \quad\left(\because \operatorname{ar}(\triangle O B P)=125 \ cm ^2\right)$
$\Rightarrow \operatorname{ar}(\triangle O A Q)=\frac{36}{25} \times 125=180 \ cm ^2$
MCQ 321 Mark
In the given figure, $D$ and $E$ are two points lying on side $A B$, such that $A D=B E$. If $D P \| B C$ and $E Q \| A C$, then
- ✓$P Q \| A B$
- B$P Q=A B$
- C$P Q \| C D$
- D$P Q=A C$
Answer
View full question & answer→Correct option: A.
$P Q \| A B$
In $\triangle A B C$, we have $D P \| B C$ and $E Q \| A C$
$\therefore$ By Thales theorem,
$\frac{A D}{D B}=\frac{A P}{P C}$ and $\frac{B E}{E A}=\frac{B Q}{Q C}$
$\Rightarrow \frac{A D}{D B}=\frac{B Q}{Q C}$
$(\because \text{EA=ED+DA=ED+BE=BD})$
$\Rightarrow \frac{A P}{P C}=\frac{B Q}{Q C}$
$\therefore P Q \| A B ($ By converse of Thales theorem$)$
$\therefore$ By Thales theorem,
$\frac{A D}{D B}=\frac{A P}{P C}$ and $\frac{B E}{E A}=\frac{B Q}{Q C}$
$\Rightarrow \frac{A D}{D B}=\frac{B Q}{Q C}$
$(\because \text{EA=ED+DA=ED+BE=BD})$
$\Rightarrow \frac{A P}{P C}=\frac{B Q}{Q C}$
$\therefore P Q \| A B ($ By converse of Thales theorem$)$
MCQ 331 Mark
In the given figure, $D E \| B C$. If $D E=4 \ cm$, $B C=8 \ cm$ and area of $\triangle A D E=25 sq . \ cm$. The area of $\triangle A B C$ is
- A$150 \ cm ^2$
- ✓$100 \ cm ^2$
- C$200 \ cm ^2$
- D$80 \ cm ^2$
Answer
View full question & answer→Correct option: B.
$100 \ cm ^2$
$\therefore \angle A D E=\angle A B C$ and $\angle A E D=\angle A C B$
$($Corresponding angles$)$
$\therefore \triangle A D E \sim \triangle A B C$
$($By $AA$ similarity criterion$)$
$\Rightarrow \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle A D E)}=\frac{B C^2}{D E^2} $
$\Rightarrow \frac{\operatorname{ar}(\triangle A B C)}{25}=\frac{64}{16} $
$ \Rightarrow \operatorname{ar}(\triangle A B C)=\frac{64 \times 25}{16}$
$=100 \ cm ^2$
$($Corresponding angles$)$
$\therefore \triangle A D E \sim \triangle A B C$
$($By $AA$ similarity criterion$)$
$\Rightarrow \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle A D E)}=\frac{B C^2}{D E^2} $
$\Rightarrow \frac{\operatorname{ar}(\triangle A B C)}{25}=\frac{64}{16} $
$ \Rightarrow \operatorname{ar}(\triangle A B C)=\frac{64 \times 25}{16}$
$=100 \ cm ^2$
MCQ 341 Mark
If $\triangle A B C \sim \triangle P Q R$ such that $\angle B=\angle Q$ and $\angle C=\angle R$, then find $\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle P Q R)}$ if $B C=3.5 m$ and $Q R=7 \ cm$.
- A$5$
- B$25$
- ✓$2500$
- D$50$
Answer
View full question & answer→Correct option: C.
$2500$
Given, $\triangle A B C \sim \triangle P Q R$
We know, areas of similar triangles are proportional to the squares of corresponding sides.
$\therefore \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle P Q R)}=\left(\frac{B C}{Q R}\right)^2$
$=\left(\frac{3.5 m }{7 \ cm }\right)^2$
$=\left(\frac{350 \ cm }{7 \ cm }\right)^2$
$=(50)^2$
$=2500$
We know, areas of similar triangles are proportional to the squares of corresponding sides.
$\therefore \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle P Q R)}=\left(\frac{B C}{Q R}\right)^2$
$=\left(\frac{3.5 m }{7 \ cm }\right)^2$
$=\left(\frac{350 \ cm }{7 \ cm }\right)^2$
$=(50)^2$
$=2500$
MCQ 351 Mark
In given figure, $P Q \| B C$. Find $A B$ $($ in $cm).$
- A$2$
- B$5$
- C$4$
- ✓$6$
Answer
View full question & answer→Correct option: D.
$6$
Given, $P Q|| A C$
$\therefore \frac{A P}{B P}=\frac{A Q}{C Q} ($ By Thales theorem $)$
$\Rightarrow \frac{2.4}{B P}=\frac{2}{3}$
$\Rightarrow B P=\frac{2.4 \times 3}{2}=3.6 \ cm$
$\text{AB=AP+BP}$
$=2.4+3.6$
$=6 \ cm$
$\therefore \frac{A P}{B P}=\frac{A Q}{C Q} ($ By Thales theorem $)$
$\Rightarrow \frac{2.4}{B P}=\frac{2}{3}$
$\Rightarrow B P=\frac{2.4 \times 3}{2}=3.6 \ cm$
$\text{AB=AP+BP}$
$=2.4+3.6$
$=6 \ cm$
MCQ 361 Mark
In the adjoining figure, $D E \| B C$. If $A D= 3.4 \ cm , A B=8.5 \ cm$ and $A C=13.5 \ cm$, find $A E$.
- ✓$5.4 \ cm$
- B$4.8 \ cm$
- C$5.8 \ cm$
- D$3.5 \ cm$
Answer
View full question & answer→Correct option: A.
$5.4 \ cm$
Since $D E \| B C$,
so we have $\frac{A D}{A B}=\frac{A E}{A C}$
$\therefore \frac{3.4}{8.5}=\frac{A E}{13.5}$
$\Rightarrow A E=\frac{3.4 \times 13.5}{8.5}$
$\Rightarrow A E=5.4 \ cm$
so we have $\frac{A D}{A B}=\frac{A E}{A C}$
$\therefore \frac{3.4}{8.5}=\frac{A E}{13.5}$
$\Rightarrow A E=\frac{3.4 \times 13.5}{8.5}$
$\Rightarrow A E=5.4 \ cm$
MCQ 371 Mark
What values of $x$ will make $D E \| A B$ in the figure?
- A$1$
- ✓$2$
- C$3$
- D$4$
Answer
View full question & answer→Correct option: B.
$2$
$D E \| A B$ if $\frac{C D}{D A}=\frac{C E}{E B}$
$($By converse of Thales theorem$)$
$\therefore \frac{x+3}{3 x+19}=\frac{x}{3 x+4}$
$\Rightarrow (x+3)(3 x+4)=x(3 x+19)$
$\Rightarrow 3 x^2+4 x+9 x+12=3 x^2+19 x$
$\Rightarrow 12=6 x$
$\Rightarrow x=2$
Hence, $x=2$ will make $D E \| A B$.
$($By converse of Thales theorem$)$
$\therefore \frac{x+3}{3 x+19}=\frac{x}{3 x+4}$
$\Rightarrow (x+3)(3 x+4)=x(3 x+19)$
$\Rightarrow 3 x^2+4 x+9 x+12=3 x^2+19 x$
$\Rightarrow 12=6 x$
$\Rightarrow x=2$
Hence, $x=2$ will make $D E \| A B$.
MCQ 381 Mark
In the figure, $A D$ is the bisector of $\angle A$. If $B D=4 \ cm , D C=2 \ cm$ and $A B=6 \ cm$, determine $A C$.
- A$2 \ cm$
- B$5 \ cm$
- ✓$3 \ cm$
- D$8 \ cm$
Answer
View full question & answer→Correct option: C.
$3 \ cm$
In $\triangle A B C, A D$ is the bisector of $\angle A$.
$\therefore \frac{B D}{D C}=\frac{A B}{A C} $
$\Rightarrow \frac{4}{2}=\frac{6}{A C}$
$\Rightarrow 4 A C=12 $
$\Rightarrow A C=3 \ cm$
$\therefore \frac{B D}{D C}=\frac{A B}{A C} $
$\Rightarrow \frac{4}{2}=\frac{6}{A C}$
$\Rightarrow 4 A C=12 $
$\Rightarrow A C=3 \ cm$
MCQ 391 Mark
In $\triangle L M N, P Q \| M N$ such that $L P=2 \ cm$ and $P M=6 \ cm$. Also $M N=20 \ cm$. Find $P Q ($ in $ cm).$
- A$2$
- B$3$
- C$4$
- ✓$5$
Answer
View full question & answer→Correct option: D.
$5$
In $\triangle L P Q$ and $\triangle L M N, \angle L$ is common and
$\angle L P Q=\angle L M N$ (Corresponding angles $).$
$\therefore \triangle L P Q \sim \triangle L M N ($By $AA$ similarity criterion $)$
$\therefore \frac{L P}{L M}=\frac{P Q}{M N}$
$\Rightarrow \frac{2}{(2+6)}=\frac{P Q}{20}$
$\Rightarrow P Q=\frac{2}{8} \times 20$
$\Rightarrow P Q=5 \ cm$
$\angle L P Q=\angle L M N$ (Corresponding angles $).$
$\therefore \triangle L P Q \sim \triangle L M N ($By $AA$ similarity criterion $)$
$\therefore \frac{L P}{L M}=\frac{P Q}{M N}$
$\Rightarrow \frac{2}{(2+6)}=\frac{P Q}{20}$
$\Rightarrow P Q=\frac{2}{8} \times 20$
$\Rightarrow P Q=5 \ cm$
MCQ 401 Mark
If $\triangle A B C \sim \triangle D E F, B C=4 \ cm , E F=5 \ cm$ and area of $\triangle A B C=64 \ cm ^2$, find the area of $\triangle D E F$.
- ✓$100 \ cm ^2$
- B$75 \ cm ^2$
- C$50 \ cm ^2$
- D$125 \ cm ^2$
Answer
View full question & answer→Correct option: A.
$100 \ cm ^2$
In $\triangle A B C, B C=4 \ cm$ and in $\triangle D E F, E F=5 \ cm$
Now, $\triangle A B C \sim \triangle D E F $
$\Rightarrow \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D E F)}=\frac{B C^2}{E F^2}=\frac{16}{25}$
$\Rightarrow \frac{64}{\operatorname{ar}(\triangle D E F)}=\frac{16}{25}$
$\Rightarrow \operatorname{ar}(\triangle D E F) =64 \times \frac{25}{16} $
$=100 \ cm ^2$
Now, $\triangle A B C \sim \triangle D E F $
$\Rightarrow \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D E F)}=\frac{B C^2}{E F^2}=\frac{16}{25}$
$\Rightarrow \frac{64}{\operatorname{ar}(\triangle D E F)}=\frac{16}{25}$
$\Rightarrow \operatorname{ar}(\triangle D E F) =64 \times \frac{25}{16} $
$=100 \ cm ^2$
MCQ 411 Mark
In $\triangle A B C$ and $\triangle P Q R, \angle B=\angle Q$ and $\frac{A B}{P Q}=\frac{B C}{Q R}$. If $\frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle P Q R)}=\frac{16}{9}$ and $B C=8 cm$, then find $Q R$.
- A$9 \ cm$
- B$8 \ cm$
- C$5 \ cm$
- ✓$6 \ cm$
Answer
View full question & answer→Correct option: D.
$6 \ cm$
In $\triangle A B C$ and $\triangle P Q R$
$\angle B=\angle Q ($Given$)$
$\frac{A B}{P Q}=\frac{B C}{Q R} ($Given$)$
$\therefore \triangle A B C \sim \triangle P Q R ($By $\text{SAS}$ similarity criterion$)$
$\therefore \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle P Q R)}=\frac{B C^2}{Q R^2}$
$\Rightarrow \frac{16}{9}=\frac{8^2}{Q R^2} $
$\Rightarrow Q R^2=\frac{64 \times 9}{16}$
$\Rightarrow Q R^2=36 $
$\Rightarrow Q R=6 \ cm$
$\angle B=\angle Q ($Given$)$
$\frac{A B}{P Q}=\frac{B C}{Q R} ($Given$)$
$\therefore \triangle A B C \sim \triangle P Q R ($By $\text{SAS}$ similarity criterion$)$
$\therefore \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle P Q R)}=\frac{B C^2}{Q R^2}$
$\Rightarrow \frac{16}{9}=\frac{8^2}{Q R^2} $
$\Rightarrow Q R^2=\frac{64 \times 9}{16}$
$\Rightarrow Q R^2=36 $
$\Rightarrow Q R=6 \ cm$
MCQ 421 Mark
In $\triangle ABC , AB =6 \sqrt{3} cm , AC =12 cm$ and $BC =6 cm$ and $BC =6 cm$ then $\angle B =\ldots \ldots \ldots$. :
- A$120^{\circ}$
- B$60^{\circ}$
- ✓$90^{\circ}$
- D$45^{\circ}$
Answer
View full question & answer→Correct option: C.
$90^{\circ}$
$90^{\circ}$
MCQ 431 Mark
Sides of two similar triangles are in the ration $4: 9$ Areas of these triangles are in the ratio....
- A$2: 3$
- B$4: 9$
- C$81: 16$
- ✓$16: 81$
Answer
View full question & answer→Correct option: D.
$16: 81$
$16: 81$
MCQ 441 Mark
In $\triangle ABC , E$ is a midpoint of $AB$ and $D$ is a midpoint of $BC \triangle ABC$ and $\triangle BDE$ are equilateral triangles. The ratio of their areas is .......... .
- A$2: 1$
- B$1: 2$
- ✓$4: 1$
- D$1: 4$
Answer
View full question & answer→Correct option: C.
$4: 1$
$4: 1$
MCQ 451 Mark
Which measure of side given below determine right angle triangle.
- A$6,8,12$
- B$3,7,9$
- C$7,15,17$
- ✓7, 24, 25
Answer
View full question & answer→Correct option: D.
7, 24, 25
7, 24, 25