The Baudhayana-Pythagoras Theorem Class 8 Short Question Answer

The Baudhayana-Pythagoras Theorem Class 8 Short Question Answer

The Baudhayana-Pythagoras Theorem Class 8 Short Question Answer

Question 1.

Find the length of the hypotenuse of an isosceles right triangle of equal sides 3 cm each.

Solution:

Hypotenuse² = 3² + 3² = 9 + 9 = 18.

So, length of hypotenuse = [latex]\sqrt{18}[/latex] cm

=[latex]3 \sqrt{2}[/latex] cm.

Question 2.

The hypotenuse of an isosceles right triangle is 20 cm, find the length of its other two sides.

Solution:

As it is an isosceles right triangle, the other two sides are equal.

Let their length be x cm each.

So, x² + x² = 20² = 400

or 2x² = 400

or x² = 200

So, x = [latex]\sqrt{200}[/latex] = [latex]10 \sqrt{2}[/latex] cm.

Question 3.

If side lengths a and c of right triangle are 1.5 cm and [latex]5\sqrt{0.58}[/latex] cm (where c is the length of the hypotenuse, then find length b of the remaining side.

Solution:

a² + b² = c²

So, 1.5² + 6² = [latex](5 \sqrt{0.58})^2[/latex]

⇒ 2.25 × b² = 25 × 0.58

⇒ 2.25 + b² = 14.50

⇒ b² = 14.50 - 2.25 = 12.25

So, b = [latex]\sqrt{12.25}[/latex] cm = 3.5 cm

Question 4.

25 is an odd square number. Using this fact, find a Baudhayana triple.

Solution:

25 is [latex]\frac{25+1}{2}[/latex] = 13th odd number,

So, we have

1 + 3 + 5 + ... + 25 = 13²

We have

13² - 25 = 169 - 25 = 144 = 12².

(5, 12, 13) is a Baudhayana triple.

Question 5.

Find the area of an equilateral triangle of side lengths 8 cm each.

Solution:

From the figure, A

AD² = AB² - BD²

= 64 - 16

= 48

AD = [latex]\sqrt{48}[/latex]

= [latex]4 \sqrt{3}[/latex] cm

So, area of the triangle ABC

= [latex]\frac{1}{2}[/latex] BC × AD

= [latex]\frac{1}{2}[/latex] × 8 × [latex]4 \sqrt{3}[/latex] cm²

= [latex]16 \sqrt{3}[/latex] cm².