Triangles(10th Grade > Mathematics ) Questions and answers for exam Preparation

10th Grade > Mathematics

TRIANGLES MCQs

Total Questions : 58 | Page 3 of 6 pages

Question 21. Refer to the following figure. Three squares are constructed on each side of the triangle as shown, with the length of each square equal to the side on which it is constructed. If the largest side is 13, sum of the areas of the three squares is __. The angle opposite to the blue colour square is the right angle.

Discuss Question

:
Let the other two sides of the triangle be a and b. Since the largest side is the hypotenuse, we have, by Pythagoras theorem,

{H y p o t e n u s e}^{2}

a^{2}

b^{2}

13^{2}

a^{2}

b^{2}

= 169 ------------------- (I)
Now,
Sum of the areas of the three squares =

{(s i d e o f y e l l o w s q u a r e)}^{2}

{(s i d e o f b r o w n i s h s q u a r e)}^{2}

{(s i d e o f b l u e s q u a r e)}^{2}

Sum of the areas of the three squares =

a^{2}

b^{2}

13^{2}

13^{2}

13^{2}

(from (I))
Sum of the areas of the three squares = 169 + 169 = 338.

Question 22. The areas of two similar triangles are

49 {cm}^{2}

and

64 {cm}^{2}

respectively. The ratio of their corresponding sides is ____.

49:64
7:8
16:49
None of the above

Discuss Question

Answer: Option B. -> 7:8
:
B
The ratio of areas of two similar triangles is equal to ratio of squares of corresponding sides.

\frac{49}{64} = \frac{(Sideof first 1st triangle)^{2}}{(Corresponding side of 2nd triangle)^{2}}

\frac{7}{8} = \frac{Sideof first 1st triangle}{Corresponding side of 2nd triangle}

\Rightarrow

Ratio of correspondingsides

= 7 : 8

Question 23. If the sides of two similar triangles are in the ratio of 4 : 9, then the areas of these triangles are in the ratio ____.

2 : 3
4 : 9
81 : 16
16 : 81

Discuss Question

Answer: Option D. -> 16 : 81
:
D
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
Given, the sides of two triangles are in the ratio of 4 : 9.
Therefore, the ratio of areas of these triangles is equal to

(\frac{4}{9})^{2}

, which is,

\frac{16}{81}

Question 24.

In The Given Figure, △ABC∼△PQR. Then, area of △...

In the given figure, △ A B C \sim △ P Q R . Then, \frac{a r e a o f △ A B C}{a r e a o f △ P Q R} equals

AB2PQ2
BC2QR2
AC2PR2
All of the above.

Discuss Question

Answer: Option D. -> All of the above.
:
D
We are given two triangles ABC and PQR such that

△ A B C \sim △ P Q R

.
For finding the areas of the two triangles, we draw altitudes AM and PN of the triangles ABC and PQR respectively, as shown below.

Now,

area of △ A B C = \frac{1}{2} \times B C \times A M and area of △ P Q R = \frac{1}{2} \times Q R \times P N

So,

\frac{a r e a o f △ A B C}{a r e a o f △ P Q R} = \frac{\frac{1}{2} \times B C \times A M}{\frac{1}{2} \times Q R \times P N} = \frac{B C \times A M}{Q R \times P N} \dots (1)

Now, in

△ A B M

and

△ P Q N

∠ B = ∠ Q

(As

△ A B C \sim △

PQR)
and

∠ A M B = ∠ P N Q = 90^{\circ}

.
So,

△ A B M \sim △ P Q N

( By AA similarity criterion)

∴ \frac{A M}{P N} = \frac{A B}{P Q} \dots (2)

Also,

△ A B C \sim △ P Q R

(Given)
So,

\frac{A B}{P Q} = \frac{A C}{P R} = \frac{B C}{Q R} \dots (3)

From (1) and (3), we get,

\frac{a r e a o f (A B C)}{a r e a o f (P Q R)} = \frac{A B \times A M}{P Q \times P N} = \frac{A B \times A B}{P Q \times P Q} = \frac{A B^{2}}{P Q^{2}}

Now using (3), we get,

\frac{a r e a o f △ A B C}{a r e a o f △ P Q R} = \frac{A B^{2}}{P Q^{2}} = \frac{B C^{2}}{Q R^{2}} = \frac{A C^{2}}{P R^{2}}

Question 25. In the figure below, if the line segment ST is parallel to line segment QR such that

\frac{P S}{S Q} = \frac{P T}{T R}

. The provided data is not sufficient to prove that triangles PQR and PST are similar.

In The Figure Below, If The Line Segment ST Is Parallel To L...

True
False
4.2 cm
0.525 cm

Discuss Question

Answer: Option B. -> False
:
B

\frac{P S}{S Q} = \frac{P T}{T R}

...... (given)

S T ∥ Q R

...... (converse of BPT)
Therefore,

∠ P S T = ∠ R Q R a n d ∠ P T Q = ∠ P T R

Therefore

△ P S T \sim △ P Q R

...... (AA similarity)
Therefore, the information provided in the question is sufficient to prove that the triangles are similar.

Question 26. The areas of two similar triangles are 12

c m^{2}

and 48

c m^{2}

. If the height of the smaller triangle is 2.1

c m

, then the corresponding height of the bigger triangle is _____.

4.41 cm
8.4 cm
4.2 cm
0.525 cm

Discuss Question

Answer: Option C. -> 4.2 cm
:
C
The ratio of the area of two similar triangles isequal to the ratio of the squares of their corresponding heights
So,

\frac{(H e i g h t o f s m a l l e r △)^{2}}{(H e i g h t o f b i g g e r △)^{2}} = \frac{12}{48} = \frac{1}{4}

Let the height of bigger triangle be x

∴

\frac{(2.1)^{2}}{x^{2}} = \frac{1}{4} \Leftrightarrow x = \sqrt{4 \times (2.1)^{2}}

= (2 \times 2.1) = 4.2 c m

Question 27. In given figure

△

ABC and

△

DEF are similar, BC=3cm, EF=4cm, and area of triangle ABC=54

c m^{2}

find the area of

△

DEF

In Given Figure △ ABC And △ DEF Are Similar, BC=3cm, E...

c m^{2}

Discuss Question

:
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
So

\frac{9}{16}

\frac{54}{a r e a o f △ D E F}

area of

△

DEF = 96

c m^{2}

Question 28. All congruent figures are similar but the similar figures need not be congruent.

True
False
4.2 cm
0.525 cm

Discuss Question

Answer: Option A. -> True
:
A
All congruent figures are similar but the similar figures need not be congruent as in case of similar figures only shape is considered whereas, in the case of congruent figures, both shape and sizes are considered. Hence, the statement is correct.

Question 29.

$△ A B C$ is such that $A B = 3 c m, B C = 2 c m$ and $C A = 2.5 c m$ . $△ D E F$ is similar to $△ A B C$ . If $E F = 4 c m$ , then the perimeter of $△ D E F$ is ____.

7.5 cm
15 cm
22.5 cm
30 cm

Discuss Question

Answer: Option B. -> 15 cm
:
B

$\frac{A B}{D E}$ = $\frac{A C}{D F}$ = $\frac{B C}{E F}$ = $\frac{2}{4}$ = $\frac{1}{2}$

$D E = 2 \times A B = 6 c m, D F = 2 \times A C = 5 c m$ .

$∴$ $Perimeter of Δ D E F = (D E + E F + D F)$
$= 6 + 4 + 5 = 15 c m$ .

Question 30.

$A B C$ is a triangle and $D E$ is drawn parallel to $B C$ cutting the other sides at $D$ and $E$ . If $A B = 3.6 c m, A C = 2.4 c m$ and $A D = 2.1 c m$ , then $A E$ = ______.