+0

# I keep dividin this wrong

+1
45
2
+350

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 ___

Sort:

#1
+505
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
+79846
+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
edited by CPhill  Nov 29, 2017

### 8 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details