The area of the square ABCD is 16 cm². Its area is divided into 4 equal squares. Line EH passes through the center of the large square.

Angle AHG is 90°. Find the area of the yelow-bordered region.

Dragan Feb 8, 2020

#1

#2**0 **

On my PC the picture is clear enough. What part do you not understand? Let me know.

Dragan
Feb 9, 2020

#3**+1 **

um the picture is blocked on my browser. You might have to screenshot it.

CalculatorUser
Feb 9, 2020

#4**+1 **

Yes thanks, we can see the pic now.

I would tackle this question by using coordinate geometry.

A((0,0)

B(0,4)

C((4,4)

D(4,0)

centre(2,2)

work out H (intersection of 2 lines)

then work out E

then look at areas.

Melody Feb 9, 2020

#5**+3 **

AD = 4 cm

AG = 2 cm

**Area of square ABCD = AD² = 16 cm²**

AH = 1.789

Angle ABG = 26.565°

GH = sqrt(AG² - AH²) = 0.894

**Area ABHFD = (AB * AG) - [(AH * GH) /2] = 7.2 cm² **

BG = sqrt( 4² + 2² ) = 4.472

BH = BG - GH = 3.578

HL = cos(26.565°) * 3.578 = 3.2

BL = sqrt[(BH)² - (HL)²] = 1.6

**Area of Δ BHL = (HL * BL) /2 = 2.56 cm²**

Angle ABG = BHE = 45°

Angle EHL = ∠BHE - ∠BHL = 18.435°

EL = tan( 18.435°) * 3.2 = 1.067

**Area of ** **Δ EHL = ( HL * EL) /2 = 1.707 cm²**

**Area CFHE = 16 - ( 7.2 + 2.56 + 1.707 ) = 4.533 cm² ** _{}

Dragan Feb 13, 2020