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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.  smiley

                                   

 

 Feb 8, 2020
edited by Dragan  Feb 8, 2020
edited by Dragan  Feb 8, 2020
edited by Dragan  Feb 9, 2020
edited by Dragan  Feb 9, 2020
edited by Dragan  Feb 9, 2020
edited by Dragan  Mar 9, 2020
 #1
avatar+110100 
0

We need a better description or a picture Dragan.  

 Feb 9, 2020
 #2
avatar+1055 
0

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

Dragan  Feb 9, 2020
 #3
avatar+2856 
+1

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

CalculatorUser  Feb 9, 2020
 #4
avatar+110100 
+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. 

 Feb 9, 2020
 #5
avatar+1055 
+5

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      indecision

 

 Feb 13, 2020
edited by Dragan  Feb 14, 2020
 #6
avatar+110100 
+4

I did not do it algebraically I just drew a pic with GeoGebra and got GeoGebra to estimate the area.

 

Your answer looks spot on to me  cool

 

 Feb 17, 2020

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