Question

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

Answer 1

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

Explanation:

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

Answer 2

`A=756cm^2`

Explanation:

.

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

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`