Multiple choice questions for triangles are given beneath for students to upgrade their problem-solving capacities. Triangles are the most well-known idea whose applications are seen oftentimes in everyday life. Address the MCQ Questions here and become familiar with triangles. Get significant questions for class 10 Maths here too.

Class 10 Maths MCQ Questions of triangles with answers are given here online to students, who are preparing for the board exams. These MCQs of triangles are introduced here with prepared with full clarifications, by which students can without much of a stretch score great marks. Students are encouraged to tackle the Triangles MCQ Questions for Class 10 Maths to know various ideas. Practicing the Objective type Questions will support your confidence in this way assisting you with scoring admirably in the test.

Below are the Multiple Choice Questions of Triangles: –

1. In a $$\Delta ABC$$ , AD is the bisector of ∠BAC . If AB = 8 cm, BD = 6 cm and DC = 3 cm. Find AC

(a) 4 cm
(b) 6 cm
(c) 3 cm
(d) 8 cm

2. A square and a rhombus are always

(a) similar
(b) congruent
(c) similar but not congruent
(d) neither similar nor congruent

3. Which geometric figures are always similar?​

(a) Circles
(b) Circles and all regular polygons
(c) Circles and triangles
(d) Regular

4. D and E are respectively the midpoints on the sides AB and AC of a triangle ABC and BC = 6 cm. If DE || BC, then the length of DE (in cm) is

(a) 2.5
(b) 3
(c) 5
(d) 6

5. The lengths of the diagonals of a rhombus are 16 cm and 12cm. Then, the length of the side of the rhombus is

(a) 9 cm
(b) 10 cm
(c) 8 cm
(d) 20 cm

6. The ratio of the areas of two similar triangles is equal to the:​

(a) square of the ratio of their corresponding sides.
(b) the ratio of their corresponding sides
(c) square of the ratio of their corresponding angles
(d) None of the above

7. Two congruent triangles are actually similar triangles with the ratio of corresponding sides as.​

(a) 1:2
(b) 1:1
(c) 1:3
(d) 2:1

8. Which of the following triangles have the same side lengths?

(a)Scalene
(b)Isosceles
(c)Equilateral
(d)None of these

9. Corresponding sides of two similar triangles are in the ratio of 2:3. If the area of small triangle is 48 sq.cm, then the area of large triangle is:

(a)230 sq.cm.
(b)106 sq.cm
(c)107 sq.cm.
(d)108 sq.cm

10. If perimeter of a triangle is 100cm and the length of two sides are 30cm and 40cm, the length of third side will be:

(a)30cm
(b)40cm
(c)50cm
(d)60cm

11. If triangles ABC and DEF are similar and AB=4cm, DE=6cm, EF=9cm and FD=12cm, the perimeter of triangle is:

(a)22cm
(b)20cm
(c)21cm
(d)18cm

12. The height of an equilateral triangle of side 5cm is:

(a)4.33
(b)3.9
(c)5
(d)4

13. Areas of two similar triangles are 36 $$cm^2$$ and 100 $$cm^2$$. If the length of a side of the larger triangle is 20 cm, then the length of the corresponding side of the smaller triangle is:

(a) 12cm
(b) 13cm
(c) 14cm
(d) 15cm

14. If ABC and DEF are two triangles and AB/DE=BC/FD, then the two triangles are similar if

(a)∠A=∠F
(b)∠B=∠D
(c)∠A=∠D
(d)∠B=∠E

15. Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio

(a)2: 3
(b)4: 9
(c)81: 16
(d)16: 81

16. The length of each side of a rhombus whose diagonals are of lengths 10 cm and 24 cm is

(a) 25 cm
(b) 13 cm
(c) 26 cm
(d) 34 cm

17. ΔABC ~ ΔDEF, ar (ΔABC) = 9 cm$$^2$$, ar (ΔDEF) = 16 cm$$^2$$. If BC = 2.1 cm, then the measure of EF is

(a) 2.8 cm
(b) 4.2 cm
(c) 2.5 cm
(d) 4.1 cm

18. In an isosceles triangle ABC, if AB = AC = 25 cm and BC = 14 cm, then the measure of altitude from A on BC is

(a) 20 cm
(b) 22 cm
(c) 18 cm
(d) 24 cm

19. If in triangles ABC and DEF,$$\frac{AB}{DE}=\frac{BC}{FD}$$ , then they will be similar, if

(a) ∠B = ∠E
(b) ∠A = ∠D
(c) ∠B = ∠D
(d) ∠A = ∠F

20. If ΔABC ~ ΔDEF and ΔABC is not similar to ΔDEF then which of the following is not true?

(a) BC.EF = AC.FD
(b) AB.ED = AC.DE
(c) BC.DE = AB.EE
(d) BC.DE = AB.FD

​​Explanation: Given AD is the bisector of ∠BAC. AB = 8 cm, DC = 3 cm and BD = 6 cm.

We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

$$\Rightarrow \frac{AB}{AC}=\frac{BD}{DC}$$

$$\Rightarrow \frac{8}{AC}=\frac{6}{3}$$

∴ AC = 4 cm

2. Answer: (d) neither similar nor congruent

3. Answer: (b) Circles and all regular polygons

Explanation: By midpoint theorem,

DE=$$\frac{1}{2}$$ BC

DE =$$\frac{1}{2}$$ of 6

DE=3cm

Explanation: The diagonals of rhombus bisect each other at right angle, so side of rhombus is the hypotenuse for the triangles formed.

Therefore,

By Pythagoras theorem

$$(\frac{16}{2})^2 + (\frac{12}{2})^2 = \mathrm{Side}^2$$

⇒ $$8^2 + 6^2 = \mathrm{Side}^2$$

⇒ $$64 + 36 = \mathrm{Side}^2$$

⇒ Side = 10 cm

6. Answer: (a) square of the ratio of their corresponding sides.

Explanation: Equilateral triangles have all its sides and all angles equal.

Explanation: Let $$A_1$$ and $$A_2$$ are areas of the small and large triangle.

Then,

$$\frac{A_2}{A_1}$$=(side of large triangle/side of small triangle)

$$\frac{A_2}{48}=(\frac{3}{2})^2$$

$$A_2$$=108 sq.cm.

Explanation: Perimeter of triangle = sum of all its sides

P = 30+40+x

100=70+x

x=30cm

Explanation: ABC ~ DEF

AB=4cm, DE=6cm, EF=9cm and FD=12cm

AB/DE = BC/EF = AC/DF

4/6 = BC/9 = AC/12

BC = (4.9)/6 = 6cm

AC = (12.4)/6 = 8cm

Perimeter = AB+BC+AC

= 4+6+8

=18cm

Explanation: The height of the equilateral triangle ABC divides the base into two equal parts at point D.

Therefore,

BD=DC= 2.5cm

In triangle ABD, using Pythagoras theorem,

$$AB^2=AD^2+BD^2$$

$$5^2=AD^2+2.5^2$$

$$AD^2 = 25-6.25$$

$$AD^2=18.75$$

Explanation: Let the side of smaller triangle be x cm.

ar(Larger Triangle)/ar(Smaller Triangle) = (side of larger triangle/side of smaller triangle)$$^2$$

$$\frac{100}{36} = (\frac{20}{x})^2$$

x = √144

X = 12 cm

Explanation: Let ABC and DEF are two similar triangles, such that,

ΔABC ~ ΔDEF

And AB/DE = AC/DF = BC/EF = 4/9

As the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,

∴ Area(ΔABC)/Area(ΔDEF) = $$\frac{{AB}^2}{{DE}^2}$$

∴ Area(ΔABC)/Area(ΔDEF) = $$(\frac{4}{9})^2 = \frac{16}{81} = 16: 81$$

Explanation: Given that

Diagonals of the rhombus

Let $$d_1$$ =10cm and $$d_2$$= 24cm

Diagonals meet at the centre and forms right-angled triangles.

So by using pythagoras theorem

Length of the base = 10/2 = 5cm
Length of the height = 24/2 = 12cm

Hypotenuse$$^2$$ = side$$^2$$+ side$$^2$$

Hypotenuse$$^2$$= 5$$^2$$+ 12$$^2$$

Hypotenuse$$^2$$ = 25 + 144

Hypotenuse$$^2$$ = 169

On taking square root we get,

Hypotenuse = 13{ 13 X 13=169 => √169=13}

Hence the side of the rhombus is 13cm.

Explanation: Given Ar (ΔABC) = 9 cm$$^2$$, ar (ΔDEF) = 16 cm$$^2$$ and BC = 2.1 cm
We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

$$\Rightarrow \frac{ar(\Delta ABC)}{ar(\Delta DEF)}=\frac{{BC}^2}{{EF}^2}$$

$$\Rightarrow \frac{9}{16}=\frac{{2.1}^2}{{EF}^2}$$

$$\Rightarrow \frac{3}{4}=\frac{2.1}{EF}$$

$$\therefore EF=2.8$$cm

Explanation: Given in an isosceles ΔABC, AB = AC = 25 cm and BC = 14 cm

Here altitude from A to BC is AD.

We know in isosceles triangle altitude on non-equal sides is also median.

⇒ BD = CD = BC/2 = 7 cm

Applying Pythagoras Theorem,

⇒ AB2 = BD2 + AD2