Answer:
Step-by-step explanation:
From the information given,
ABCD and EFGH are squares
JH = 4
JC = 9
The diagonals of a square are perpendicular and bisect each other.
Considering triangle JEH,
angle J = 90 degrees
EJ = JH = 4
Triangle JEH is a right triangle. We would find EH by applying the Pythagorean theorem which is expressed as
hypotenuse^2 = one leg^2 + other leg^2
hypotenuse = EH
one leg = EJ = 4
other leg = JH = 4
Thus,
EH^2 = 4^2 + 4^2 = 16 + 16 = 32
EH = √32
Recall,
Area of a square = length^2
Length of square EFGH = √32
Area of square EFGH = (√32)^2 = 32
Considering triangle JBC,
angle J = 90 degrees
JC = JB = 9
Triangle JBC is a right triangle. We would find BC by applying the Pythagorean theorem. From triangle JBC,
hypotenuse = BC
one leg = JC = 9
other leg = JB = 9
Thus,
BC^2 = 9^2 + 9^2 = 81 + 81 = 162
BC = √162
Length of square ABCD = √162
Area of square ABCD = (√162)^2 = 162
Area of shaded region = area of square ABCD - area of square EFGH = 162 - 32
Area of shaded region = 130 cm^2