Posted by **marie darling** on Friday, April 26, 2013 at 5:04pm.

The four sequential sides of a quadrilateral have lengths a=3.6, b=5.3, c=8.4, and d=10.2 (all measured in yards). The angle between the two smallest sides is alpha = 117°.

What is the area of this figure?

- precalculus -
**Reiny**, Friday, April 26, 2013 at 9:17pm
join the ends of the two smallest sides

let its length be x

x^2 = 3.6^2 + 5.3^2 - 2(3.6)(5.3)cos117°

= 58.374..x = 7.64 ( I stored the entire number)

area of the triangle formed by x and the the two smaller sides

= (1/2)(5.3)(3.6)sin117 = 8.5

In the larger triangle, let the angle opposite side x be Ø

7.64^ = 8.4^2 + 10.2^ - 2(8.4)(10.2)cosØ

171.36cosØ = 116.2304

cosØ = .67828..

Ø = 47.29°

area of larger triangle = (1/2)(10.2)(8.4)sin47.29°

= 31.4789

total area = 31.4789 + 8.5 = 39.979 units^2

check my arithmetic

- precalculus -
**marie darling**, Saturday, April 27, 2013 at 12:06am
thank you for your help! These problems get really confusing when they don't provide a picture.

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