Important Questions Class 10 Maths Chapter 6 Triangles

A basic triangle is a two-dimensional geometric shape with three vertices and three edges. Two triangles are called congruent triangles if they have similar shape and size and if only, shape is similar we call them similar triangles. Hence, we can understand two congruent triangles can be similar but similar triangles can’t be congruent. Two triangles can be similar when their corresponding angles are equal and their sides are of similar proportion or ratio. In the world of triangles, Pythagoras’ Theorem plays an important role while performing mathematical calculations.

Important questions for class 10 maths Chapter 6 Triangles can be from topic :

• Congruent and similar triangles or polygons
• Pythagoras’ theorem: Description and calculations
• Area of similar triangles
• Criteria of triangles to be similar
• Similar and non-similar polygons 
• Similar figures

Important Questions Chapter 6 – Triangles

Q1. Sides of triangles are given below. Determine which of them are right triangles.

In case of a right triangle, write the length of its hypotenuse.

(i) 7 cm, 24 cm, 25 cm

(ii) 3 cm, 8 cm, 6 cm

Solution:

(i) Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.

Squaring the lengths of the sides of the, we will get 49, 576, and 625.

49 + 576 = 625

(7)^2 + (24)^2 = (25)^2

Therefore, the above equation satisfies the Pythagoras theorem. Hence, it is a right-angled triangle.

Length of Hypotenuse = 25 cm

(ii) Given, sides of the triangle are 3 cm, 8 cm, and 6 cm.

Squaring the lengths of these sides, we will get 9, 64, and 36.

Clearly, 9 + 36 ≠ 64

Or, 3^2 + 6^2 ≠ 8^2

Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.

Hence, the given triangle does not satisfy the Pythagoras theorem.

Q2. Given ΔABC ~ ΔPQR, if AB/PQ = ⅓, then find (ar ΔABC)/(ar ΔPQR).

Solution:

Given, 

ΔABC ~ ΔPQR

And

AB/PQ = ⅓, 

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

(ar ΔABC)/(ar ΔPQR) = AB^2/PQ^2 = (AB/PQ)^2 = (⅓)^2 = 1/9

Therefore, (ar ΔABC)/(ar ΔPQR) = 1/9

Or

(ar ΔABC) : (ar ΔPQR) = 1 : 9

Q3. The sides of two similar triangles are in the ratio 7 : 10.  Find the ratio of areas of these triangles.

Solution:

Given,

The ratio of sides of two similar triangles = 7 : 10

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

The ratio of areas of these triangles = (Ratio of sides of two similar triangles)^2

= (7)^2 : (10)^2

= 49 : 100

Therefore, the ratio of areas of the given similar triangles is 49 : 100.

Q4. The areas of two similar triangles are respectively  9 cm^2 and 16 cm^2 . Then ratio of the corresponding sides are

• 3:4  

• 4:3  

• 2:3 

• 4:5 

Solution:

(a)  3:4 

Ratio of area of triangle is equal to the square the ratio of the corresponding sides

Q5. The side of square who’s diagonal is  16 cm is

• 16 cm 

• 8√2 cm 

• 5√2 cm 

• None of these

Solution:

8√2 cm

Length of diagonal = sum of both sides of square.

 (16)^2=s^2+s^2 

 2s^2=16∗16 

 s^2=8(16) 

 s=8√2 

Q6. A certain right-angled triangle has its area numerically equal to its perimeter. The length of each side is an even integer, what is the perimeter?

• 24 units

• 36 units

• 32 units

• 30 units

Solution:

24 units

Length of each side is even integer amd also area is numerically equal its perimeter.

 ∴(6,8,10)  make a Pythagorean triplet.

 8^2+6^2=10^2 

Sum of all sides =perimeter

 6+8+10=24

Q7. The hypotenuse of a right triangle is  6 m  more than twice of the shortest side. If the third side is  2 m  less than the hypotenuse. Find the side of the triangle.

Solution:

Let shortest side be  Xm  in length

Then hypotenuse  =(2x+6)m 

And third side  =(2x+4)m 

We have,

 (2x+6)^2=x^2+(2x+4)^2 

 ⇒4x^2+24x+36=x^2+4x^2+16+16x 

 ⇒x^2−8x−20=0 

 ⇒x=10 or x=−2  

 ⇒x=10 

Hence, the sides of triangle are 10m, 26m and 24m.

Q8. In triangles PQR and TSM, ∠P = 55°, ∠Q = 25°, ∠M = 100°, and ∠S = 25°. Is ∆QPR ~ ∆TSM? Why?

Solution:

Since, ∠R = 180° – (∠P + ∠Q)

= 180° – (55° + 25°) = 100° = ∠M

∠Q = ∠S = 25° (Given)

∆QPR ~ ∆STM

i.e., . ∆QPR is not similar to ∆TSM.

Q9. If ABC and DEF are similar triangles such that ∠A = 47° and ∠E = 63°, then the measures of ∠C = 70°. Is it true? Give reason.

Solution:

Since ∆ABC ~ ∆DEF

∴ ∠A = ∠D = 47°

∠B = ∠E = 63°

∴ ∠C = 180° – (∠A + ∠B) = 180° – (47° + 63°) = 70°

∴ Given statement is true.

Q10. ABC is an isosceles triangle right-angled at C. Prove that AB^2 = 2AC^2.

Solution:

∆ABC is right-angled at C.

∴ AB^2 = AC^2 + BC^2 [By Pythagoras theorem]

⇒ AB^2 = AC^2 + AC^2

[∵ AC = BC]

⇒ AB^2 = 2AC^2