+0

I keep dividin this wrong

+1
305
2
+394

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 ___

Nov 29, 2017

#1
+556
0

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

Is it $$30+5.5\sqrt{26}$$?

Nov 29, 2017
#2
+95883
+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

Nov 29, 2017
edited by CPhill  Nov 29, 2017