MATH SOLVE

7 months ago

Q:
# ABGF is a square with half the perimeter of square ACDE. GD = 4in. Find the area of the shaded region.

Accepted Solution

A:

the complete question in the attached figure

Let

x---------> length side square ACDE

the perimeter of square ACDE=x+x+x+x--------> 4x

we know that

ABGF is a square with half the perimeter of square ACDE

then

perimeter of square ABGF=4x/2-------> 2x

the length side of square ABGF is 2x/4------> 0.50x

the area of the shaded region=[area square ACDE]-[area square ABGF]

the area of the shaded region=[x*x]-[0.50x*0.50x]---> x²-0.25x²---> 0.75x²

the area of the shaded region= 0.75x²

find the value of x

applying the Pythagorean theorem

4²=(0.5x)²+(0.5x)²------> 16=0.25x²+0.25x²------> 16=0.50x²

x²=32-------> x=4√2 in

the area of the shaded region= 0.75x²----> 0.75*(4√2)²---> 0.75*32=24 in²

the answer is

the area of the shaded region is 24 in²

Let

x---------> length side square ACDE

the perimeter of square ACDE=x+x+x+x--------> 4x

we know that

ABGF is a square with half the perimeter of square ACDE

then

perimeter of square ABGF=4x/2-------> 2x

the length side of square ABGF is 2x/4------> 0.50x

the area of the shaded region=[area square ACDE]-[area square ABGF]

the area of the shaded region=[x*x]-[0.50x*0.50x]---> x²-0.25x²---> 0.75x²

the area of the shaded region= 0.75x²

find the value of x

applying the Pythagorean theorem

4²=(0.5x)²+(0.5x)²------> 16=0.25x²+0.25x²------> 16=0.50x²

x²=32-------> x=4√2 in

the area of the shaded region= 0.75x²----> 0.75*(4√2)²---> 0.75*32=24 in²

the answer is

the area of the shaded region is 24 in²