A square is inscribed in a right triangle with leg lengths 6 and 8, so that they have a common right angle. Find the square side length.

Question

A square is inscribed in a right triangle with leg lengths 6 and 8, so that they have a common right angle. Find the square side length.

Answer 1

Draw the diagram. You have a square and two smaller triangles within the larger triangle.

The two triangles are similar because the three sets of corresponding angles are equal.

Therefore the corresponding sides form equal ratios.

Call each side of the square "x".

Look at the top triangle:

The base is x and the height is 6-x.

Look at the bottom triangle:

The base is 8-x and the height is x.

Form equal ratios.

x/(8-x)=(6-x)/x

Cross multiply.

x^2=(8-x)(6-x)

x^2=48-14x+x^2

Subtract x^2 from both sides.

0=48-14x

14x=48

x=48/14

x=3 6/14 or 3 3/7

 

Another approach is to find the areas of the two triangles as well as the area of the square.

Area of the top triangle is (6-x)(x)/2

Area of the bottom triangle is (8-x)(x)/2

Area of the square is x^2

Add these areas together to get the area of the large triangle which is (1/2)(6)(8)=24

(6-x)(x)/2+(8-x)(x)/2+x^2=24

Multiply both sides by 2.

(6-x)(x)+(8-x)(x)+2x^2=48

6x-x^2+8x-x^2+2x^2=48

Simplify

14x-2x^2+2x^2=48

14x=48

x=48/14

x=3 6/14 or 3 3/7

 

Answer 2

Place the right angle at the origin. Short leg on x-axis.

The forth corner of the square is the point on the hypotenuse where the line y = x intersects. (Draw a diagram!)

In this case the hypotenuse is y = -4x/3 + 8

x = -4x/3 + 8

7x/3 = 8

7x = 24

x = 3.428571429