Download App

Maths M.C.Q (1 Marks)

P-2 Similarity · Maharashtra Board STD 10 (MSBSHSE - English Medium). 10 questions.

Questions

M.C.Q (1 Marks)

🎯

Test yourself on this topic

10 questions · timed · auto-graded

Question 11 Mark
Select the appropriate alternative.
In figure 1.71, seg XY || seg BC, then which of the following statements is true?

$A.\frac{A B}{A C}=\frac{A X}{A Y}$
$B.\frac{A X}{X B}=\frac{A Y}{A C}$
$C.\frac{A X}{Y C}=\frac{A Y}{X B}$
$D.\frac{A B}{Y C}=\frac{A C}{X B}$
Answer
∵ segXY||segBC
⇒∠ AXY≅ ∠ ABC
And, ∠ XAY≅ ∠ BAC (Common)
⇒ Δ AXY~ Δ ABC (By AA Test)
$\Rightarrow \frac{ AX }{ AB }=\frac{ AY }{ AC }=\frac{ XY }{ BC } \text { (corresponding sides are proportional) }$
$\Rightarrow \frac{ AB }{ AC }=\frac{ AX }{ AY }$
(B) doesn’t match the solution.
(C) doesn’t match the solution.
(D) doesn’t match the solution.
View full question & answer→
MCQ 21 Mark
Select the appropriate alternative.
ΔABC and ΔDEF are equilateral triangles, A(ΔABC) : A(ΔDEF) = 1 : 2. If AB = 4 then what is length of DE?

 
  • A
    $2 \sqrt2$
  • B
    4
  • C
    8
  • ✓
    $4 \sqrt2$
Answer
Correct option: D.
$4 \sqrt2$
Solution: We know that, all the angles of an equilateral triangles are equal, i.e., 60°.
⇒ Δ ABC~Δ DEF ……(By AAA Similarity Test)
$\Rightarrow \frac{ A (\triangle ABC )}{ A (\triangle DEF )}=\frac{ AB ^2}{ DE ^2} $
$ \text { And, } \frac{ A (\triangle ABC )}{ A (\Delta DEF )}=\frac{1}{2} \text { (Given) }$
$ \Rightarrow \frac{ AB ^2}{ DE ^2}=\frac{1}{2}$
$\Rightarrow D E^2=2 \times 4^2(\because A B=4) $
$ \Rightarrow D E=\sqrt{32} $
$\Rightarrow D E=4 \sqrt2$
(A) doesn’t match the solution.
(B) doesn’t match the solution.
(C) doesn’t match the solution.
View full question & answer→
Question 31 Mark
Select the appropriate alternative.
In Δ and ΔDEF ∠B = ∠E, ∠F = ∠C and AB = 3DE then which of the statements regarding the two triangles is true?

A. The triangles are not congruent and not similar
B. The triangles are similar but not congruent.
C. The triangles are congruent and similar.
D. None of the statements above is true.
Answer
In Δ ABC & Δ DEF
∠ B≅ ∠ E and ∠ C≅ ∠ F (Given)
⇒ Δ ABC ~ Δ DEF (By AA Test)
⇒ The triangles are similar.
And, Δ ABC ≅ Δ DEF, if AB = DE.
But, given that - AB = 3DE.
⇒ The triangles are not congruent.
(A) doesn’t match the solution.
(C) doesn’t match the solution.
(D) doesn’t match the solution.
View full question & answer→
Question 41 Mark
Select the appropriate alternative.
If in ΔDEF and ΔPQR, ∠D ≅ ∠Q, ∠R ≅ ∠E then which of the following statements is false?

$A.\frac{E F}{P R}=\frac{D F}{P Q}$
$B.\frac{D E}{P Q}=\frac{E F}{R P}$
$C.\frac{D E}{Q R}=\frac{D F}{P Q}$
$D.\frac{E F}{R P}=\frac{D E}{Q R}$
Answer
In Δ DEF & Δ PQR
∠ D≅ ∠ Q and ∠ R≅ ∠ E (Given)
$\Rightarrow$ Δ DEF ~ Δ PQR
$\Rightarrow \frac{D E}{P Q}=\frac{E F}{Q R}=\frac{F D}{R P}$ (corresponding sides are proportional)
(A) is matching the solution, hence can’t be false.
(C) is matching the solution, hence can’t be false.
(D) is matching the solution, hence can’t be false.
View full question & answer→
Question 51 Mark
Select the appropriate alternative.
In Δ ABC and ΔPQR, in a one to one correspondence $\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}$ then

A. Δ PQR ~ Δ ABC
B. Δ PQR ~ Δ CAB
C. Δ CBA ~ Δ PQR
D. Δ BCA ~ Δ PQR
Answer
$\therefore\frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q}$
⇒ Δ CAB~Δ PQR
(A) doesn’t match the solution.
(C) doesn’t match the solution.
(D) doesn’t match the solution.
View full question & answer→
MCQ 61 Mark
Answer
Correct option: A.
$\frac{ AB }{ AC }=\frac{ AX }{ AY }$
(A)$\frac{A B}{A C}=\frac{A X}{A Y}$
$\begin{array}{l}\because \operatorname{seg} X Y \| \text { segBC } \\
\Rightarrow \angle A X Y \cong \angle A B C \\
\text { And, } \angle X A Y \cong \angle B A C \text { (Common) } \\
\Rightarrow \triangle A X Y \sim \triangle A B C \text { (By AA Test) } \\
\Rightarrow \frac{A X}{ AB }=\frac{ AY }{ AC }=\frac{ XY }{ BC } \text { (corresponding sides are proportional) } \\
\Rightarrow \frac{ AB }{ AC }=\frac{ AX }{ AY }
\end{array}$
(B) doesn't match the solution.
(C) doesn't match the solution.
(D) doesn't match the solution.
View full question & answer→
MCQ 71 Mark
ΔABC and ΔDEF are equilateral triangles, A(ΔABC) : A(ΔDEF) = 1 : 2. If AB = 4 then what is length of DE?
Answer
Correct option: D.
$4 \sqrt{2}$
(D) $4 \sqrt{ } 2$
We know that, all the angles of an equilateral triangles are equal, i.e., $60^{\circ}$.
$\Rightarrow \triangle ABC \sim \triangle DEF$ $\qquad$ (By AAA Similarity Test)
$\Rightarrow \frac{ A (\triangle ABC )}{ A (\triangle DEF )}=\frac{ AB ^2}{ DE ^2}$
And, $\frac{ A (\triangle ABC )}{ A (\triangle DEF )}=\frac{1}{2}$ (Given)
$\begin{array}{l}
\Rightarrow \frac{A^2}{D^2}=\frac{1}{2} \\
\Rightarrow D E^2=2 \times 4^2(\because A B=4) \\
\Rightarrow D E=\sqrt{32} \\
\Rightarrow D E=4 \sqrt{2}
\end{array}$
(A) doesn't match the solution.
(B) doesn't match the solution.
(C) doesn't match the solution.
View full question & answer→
MCQ 81 Mark
Answer
Correct option: B.
(B) The triangles are similar but not congruent.
In Δ ABC & Δ DEF
∠ B≅ ∠ E and ∠ C≅ ∠ F (Given)
⇒ Δ ABC ~ Δ DEF (By AA Test)
⇒ The triangles are similar.
And, Δ ABC ≅ Δ DEF, if AB = DE.
But, given that - AB = 3DE.
⇒ The triangles are not congruent.
(A) doesn’t match the solution.
(C) doesn’t match the solution.
(D) doesn’t match the solution.
View full question & answer→
MCQ 91 Mark
Answer
Correct option: B.
$\frac{D E}{P Q}=\frac{E F}{R P}$
(B)$\frac{D E}{P Q}=\frac{E F}{R P}$
In $\triangle D E F \& \triangle P Q R$
$\begin{array}{l}
\angle D \cong \angle Q \text { and } \angle R \cong \angle E \text { (Given) } \\
\Rightarrow \triangle DEF \sim \triangle PQR \\
\Rightarrow \frac{ DE }{ PQ }=\frac{ EF }{ QR }=\frac{ FD }{ RP } \text { (corresponding sides are proportional) }
\end{array}$
(A) is matching the solution, hence can't be false.
(C) is matching the solution, hence can't be false.
(D) is matching the solution, hence can't be false.
View full question & answer→
MCQ 101 Mark
In Δ ABC and ΔPQR, in a one to one correspondence $\frac{ AB }{ QR }=\frac{ BC }{ PR }=\frac{ CA }{ PQ }$ then
Image
Answer
Correct option: B.
(B)$\triangle P Q R-\triangle C A B$
$\begin{array}{l}\because \frac{A B}{Q R}=\frac{B C}{P R}=\frac{C A}{P Q} \\
\Rightarrow \triangle C A B \sim \triangle P Q R
\end{array}$
(A) doesn't match the solution.
(C) doesn't match the solution.
(D) doesn't match the solution.
View full question & answer→
Generate Paper FreeAll P-2 Similarity topics