MATH SOLVE

2 months ago

Q:
# A square is inscribed in a right isosceles triangle, such that two of its vertices lie on the hypotenuse and two other on the legs. Find the length of the side of the square, if the length of the hypotenuse is 3 in.

Accepted Solution

A:

The explanation of the problem is shown in the attached figure.

Let the side length of the square be x.

The triangular regions outside the square are all isosceles right triangles

The length of the hypotenuse is 3 in

So,

The length of each leg = 3/√2

The area of the given triangle = (1/2) * (3/√2)*(3/√2) = 9/4

With the help of figure:

Area1= (1/2) * (x/√2)*(x/√2) = x²/4

Area2= (1/2) * (x)*(x) = x²/2

Area3= (1/2) * (x)*(x) = x²/2

Area of the square = x²

So, Area1 + Area2 + Area3 + square area = 9/4

∴

x²/4 + x²/2 + x²/2 +x² = 9/4

(9/4) x² = 9/4

x² = 1

∴ x = 1

∴ The side of the square which is inscribed in a right isosceles triangle = 1 in

Let the side length of the square be x.

The triangular regions outside the square are all isosceles right triangles

The length of the hypotenuse is 3 in

So,

The length of each leg = 3/√2

The area of the given triangle = (1/2) * (3/√2)*(3/√2) = 9/4

With the help of figure:

Area1= (1/2) * (x/√2)*(x/√2) = x²/4

Area2= (1/2) * (x)*(x) = x²/2

Area3= (1/2) * (x)*(x) = x²/2

Area of the square = x²

So, Area1 + Area2 + Area3 + square area = 9/4

∴

x²/4 + x²/2 + x²/2 +x² = 9/4

(9/4) x² = 9/4

x² = 1

∴ x = 1

∴ The side of the square which is inscribed in a right isosceles triangle = 1 in