Geometry Problem 1543. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

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240 using coordinate geometry

ReplyDeleteGeometry Solution

ReplyDeleteExtend 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

I would be grateful for more detailed explanations of your demonstration

Deletewhat is the step you cannot understand? do let me know

DeleteFind formula of circle Centre (20,0)

ReplyDeleteFind 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

Trigonometry Solution

ReplyDeleteExtend 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

2nd Geometry Solution

ReplyDeleteExtend 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