The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Class 8

The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Class 8

Question 1.

The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Find bounds on the length of the hypotenuse such that they have at least one digit after the decimal point. Class 8

(i) 3

(ii) 4

(iii) 6

(iv) 8

(v) 9

Solution:

(i) When length of each side is 3, then

hypotenuse = [latex]\sqrt{3^2+3^2}[/latex] = [latex]3 \sqrt{2}[/latex]

= 3 (1.414) = 4.242.

So, its bounds can be 4.2 and 4.3.

Thus, 4.2 < [latex]3 \sqrt{2}[/latex] < 4.3.

(ii) When length of each side is 4, then

hypotenuse = [latex]\sqrt{4^2+4^2}[/latex] = [latex]4 \sqrt{2}[/latex]

= 4 (1.414) = 5.656.

So, its bounds can be 5.6 and 5.7.

Thus, 5.6 < [latex]4 \sqrt{2}[/latex] < 5.7.

(iii) If length of each side is 6, then

hypotenuse = [latex]\sqrt{6^2+6^2}[/latex] = [latex]6 \sqrt{2}[/latex]

= 6 (1.414) = 8.484.

So, its bounds can be 8.4 and 8.5.

Thus, 8.4 < [latex]6 \sqrt{2}[/latex] < 8.5.

(iv) For length of side 8,

hypotenuse = [latex]8 \sqrt{2}[/latex]

= 8 (1.414) = 11.312.

So, its bounds can be 11.3 and 11.4.

Thus, 11.3 < [latex]8 \sqrt{2}[/latex] < 11.4.

(v) For length of side 9,

hypotenuse = [latex]9 \sqrt{2}[/latex]

= 9 (1.414) = 12.726.

So, its bounds can be 12.7 and 12.8.

Thus, 12.7 < [latex]9 \sqrt{2}[/latex] < 12.8.