## Friday, June 9, 2023

### Geometry Problem 1543: Calculating the Area of Quadrilateral ABED in a Square with a Side Length of 20 and an Intersecting Arc

Geometry Problem 1543. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

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#### 7 comments:

1. 240 using coordinate geometry

2. Geometry Solution

Extend DM to meet AB extended at N. Draw a perpendicular EF from E to meet AB at F. Let DE = p and AF = t.

Triangles MCD & MBN are congruent and BN = 20. AN is a diameter of the semicircle AECN and p = 4V5
t^2 + (2t)^2 = (2p)^2 = 320 => t = 8

Therefore S(ABED) = S(AED) + S(AEB) = 1/2 X p X 2p+ 1/2 X 2t X 20
= p^2 + 20t = 80 + 160 = 240

Sumith Peiris
Moratuwa
Sri Lanka

1. I would be grateful for more detailed explanations of your demonstration

2. what is the step you cannot understand? do let me know

3. Find formula of circle Centre (20,0)
Find formula of line DM
Use to find coordinates of crossing point with the arc
(8,16)
Divide the yellow area into 2 triangles and 1 rectangle
6x16=96 8x16=128 4x4=16
Add together to make 240

4. Trigonometry Solution

Extend BEM to meet AD extended at N
< AEN = 90 as I showed before
If < MBC = @ then we can show that < EDM = 90 - @

So S(DEM) = 1/2 X 20 X 10 X sin (90-2@) = 100.cos 2@
= (1-tan^2 @)/(1 + tan^2 @) = (1 - (1/2)^2) / (1 + (1/2)^2) = 60

So S(ABED) = S(ABCD) - S(BCM) - S(DEM)
= 400 - 100 - 60 = 240

Sumith Peiris
Moratuwa
Sri Lanka

5. 2nd Geometry Solution
Extend DM to meet AB extended at N. Complete Rectangle EXBY, X on AB & Y on CM

As before < AEM = 90

So considering Triangle ADN,
AE = 20 X 40 / 20V5 = 8V5 and so
DE = 4V5 by using Pythagoras on Triangle ADE and EM = 10V5 - 4V5 = 6V5
Hence MY = 6 and EX = 16
So S(AEB) = 20 X 16 /2 = 160
S(ADE) = 8V5 X 4V5 /2 = 80
Adding S(ABED) = 160+80 = 240

Sumith Peiris
Moratuwa
Sri Lanka