Find the area of the quadrilateral ABCD, if angles A and C are right angles.

Quadrilateral A B C D has side A B that falls slowly from left to right, side B C that falls quickly from left to right, side C D that falls slowly from right to left, and side A D that rises quickly from left to right. Angles A and C are right angles. The sides are labeled as follows: A B, 5; B C, 11; C D, unlabeled; A D, 12.ABCD 5 12 11

The area of the quadrilateral is ___

ladiikeiii
Nov 29, 2017

#1**0 **

I know CPhil got to this before me, but I want to gve this a try before I see the answer.

Is it \(30+5.5\sqrt{26}\)?

helperid1839321
Nov 29, 2017

#2**+2 **

Draw BD.... and this forms a hypotenuse of triangles ABD and triangles BDC

In triangle ABD , we can find BD using the Pythagorean Theorem

BD = √[ AD^2 + AB^2 ] = √ [ 12^2 + 5^2 ] = √ [ 144 + 25] = √169 = 13

And using the same theorem, in triangle BDC, leg DC can be found as

DC = √ [ BD^2 - BC^2] = √ [ 13^2 - 11^2] = √ [169 - 121] = √48 = √[16 * 3] =

√16 * √3 = 4√3

And the area of each right triangle = (1/2)(product of the legs)

So....the total area = (1/2)(12 * 5) + (1/2) (11 * 4√3) =

(1/2) [ (12*5) + (11 * 4√3) ] =

(1/2) [ 60 + 44√3] =

[30 + 22√3] sq units

CPhill
Nov 29, 2017