It's a geometry question stating to find the area of quadrilateral ABCD?

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2 Answers
Feb 2, 2018

Area of Quadrilateral ABCD color(green)(A_q = 756) sq. units

Explanation:

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ABED is a cyclic quafrilateral as /_(BAD) = /_(BFD) = 90^0

Therefore, /_(ADE) = 180 - /_(ABE) = 180 - 135 = 45^0

Construction :

Complete rectangle ABGH with sides AB = GH = 18.

/_(BEG) = 135 - 90 = 45^0

BG = BE * sqrt2

Consider isosceles Triangle BEG.

Consider isosceles triangle DGH

DH = GH = 18 Hence AD = AH + HD = 18#

DG = (GH) / sin 45 = 18 / sin 45 = 18 sqrt2 = 25.4558

:. EG = BE = (36 - 18sqrt2) = 18(2-sqrt2) = 10.5442

Area of Delta BEG = (1/2) * (BE) * (EG)

Delta BEG = (1/2) (18(2-sqrt2))^2 = 55.5896 color(blue)((1))

Similarly, Area of Delta DGH = (1/2) * (GH) * (DH) = (1/2) * 18 * 18 = 162 color(blue)((2))

Area of Delta CDF = (1/2) * CE * DE = (1/2) * 15 * 36 = 270 color(blue)((34))

Consider rectangle ABGH

BG = AH = BE * sqrt 2 = 18 * (2-sqrt2) * sqrt 2 = 14.9117

Area of rectangle ABGH = 18 * 14.9117) = 268.4104 color(blue)((4))

Area of quadrilateral ABCD is

A_q = ABGH + Delta BEG + Delta DGH + Delta CDE

A_q = 268.4104 + 55.5896 + 162 + 270 = color(green)(756) sq. units

Feb 2, 2018

A=756cm^2

Explanation:

.

We can add line BF parallel to AD and line FG parallel to AB.

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Since the sum of the angles in any quadrilateral is 360^@, in quadrilateral ABED we have two 90^@ angles and one 135^@ angle which indicates the measure of angle /_ADE as follows:

/_ADE=360^@-(90^@+90^@+135^@)=360^@-315^@=45^@

Quadrilateral ABFG is a rectangle which means FG=AB=18 cm.

Triangle DeltaFGD is an isosceles right triangle because it has a 45^@ angle. As such, FG=GD=18 cm.

We can use Pythagoras' formula in this triangle to find the measure of DF:

(DF)^2=(FG)^2+(GD)^2=18^2+18^2=2(18)^2

DF=18sqrt2 cm

FE=DE-DF=36-18sqrt2=18(2-sqrt2) cm

Triangle DeltaBEF is also an isosceles right triangle because /_EBF=45^@. Therefore, BE=FE=18(2-sqrt2) cm.

(BF)^2=(BE)^2+(FE)^2=2[18(2-sqrt2)]^2

BF=sqrt2(18(2-sqrt2))=sqrt2(36-18sqrt2)=36sqrt2-36=36(sqrt2-1) cm.

Area of rectangle ABFG=18(36sqrt2-36)cm^2

Area of triangle DeltaBEF=1/2[18(2-sqrt2)*18(2-sqrt2)]=324(3-2sqrt2)cm^2.

Area of triangle DeltaFGD=1/2(18)(18)=162cm^2.

Area of triangle Delta CDE=1/2(36)(15)=270cm^2

Area of quadrilateral ABCD is equal to the sum of these four areas:

A=18(36sqrt2-36)+324(3-2sqrt2)+162+270=

A=648sqrt2-648+972-648sqrt2+162+270

A=cancelcolor(red)(648sqrt2)-648+972cancelcolor(red)(-648sqrt2)+162+270

A=1404-648=756cm^2