Important Questions Class 10 Maths Chapter 6 Triangles

# 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 canâ€™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, Pythagorasâ€™ 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 polygonsÂ
• 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 â‰  64

Or, 3^2 + 6^2 â‰  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 Î”ABC ~ Î”PQR, if AB/PQ = â…“, then find (ar Î”ABC)/(ar Î”PQR).

Solution:

Given,Â

Î”ABC ~ Î”PQR

And

AB/PQ = â…“,Â

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 Î”ABC)/(ar Î”PQR) = AB^2/PQ^2 = (AB/PQ)^2 = (â…“)^2 = 1/9

Therefore, (ar Î”ABC)/(ar Î”PQR) = 1/9

Or

(ar Î”ABC) : (ar Î”PQR) = 1 : 9

Q3. The sides of two similar triangles are in the ratio 7 : 10.Â  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 respectivelyÂ  9 cm^2 and 16 cm^2 . Then ratio of the corresponding sides are

• 3:4Â Â
• 4:3Â Â
• 2:3Â
• 4:5Â

Solution:

(a)Â  3:4Â

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

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

• 16 cmÂ
• 8âˆš2 cmÂ
• 5âˆš2 cmÂ
• None of these

Solution:

8âˆš2 cm

Length of diagonal = sum of both sides of square.

Â (16)^2=s^2+s^2Â

Â 2s^2=16âˆ—16Â

Â s^2=8(16)Â

Â s=8âˆš2Â

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.

Â âˆ´(6,8,10)Â  make a Pythagorean triplet.

Â 8^2+6^2=10^2Â

Sum of all sides =perimeter

Â 6+8+10=24

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

Solution:

Let shortest side beÂ  XmÂ  in length

Then hypotenuseÂ  =(2x+6)mÂ

And third sideÂ  =(2x+4)mÂ

We have,

Â (2x+6)^2=x^2+(2x+4)^2Â

Â â‡’4x^2+24x+36=x^2+4x^2+16+16xÂ

Â â‡’x^2âˆ’8xâˆ’20=0Â

Â â‡’x=10 or x=âˆ’2Â Â

Â â‡’x=10Â

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

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

Solution:

Since, âˆ R = 180Â° â€“ (âˆ P + âˆ Q)

= 180Â° â€“ (55Â° + 25Â°) = 100Â° = âˆ M

âˆ Q = âˆ S = 25Â° (Given)

âˆ†QPR ~ âˆ†STM

i.e., . âˆ†QPR is not similar to âˆ†TSM.

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

Solution:

Since âˆ†ABC ~ âˆ†DEF

âˆ´ âˆ A = âˆ D = 47Â°

âˆ B = âˆ E = 63Â°

âˆ´ âˆ C = 180Â° â€“ (âˆ A + âˆ B) = 180Â° â€“ (47Â° + 63Â°) = 70Â°

âˆ´ Given statement is true.

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

Solution:

âˆ†ABC is right-angled at C.

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

â‡’ AB^2 = AC^2 + AC^2

[âˆµ AC = BC]

â‡’ AB^2 = 2AC^2