Question 1190321
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In parallelogram ABCD, AB = 13, AD = 14, and the length of diagonal AC is 15. 
What is the area of the parallelogram in square units?
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<pre>
Apply the cosine law and find cosine of the angle B between the sides AB and BC of the parallelogram

(the diagonal AC is the opposite side of the triangle ABC to angle B)


    AC^2 = AB^2 + BC^2 - 2*AB*BC*cos(B)

    15^2 = 13^2 + 14^2 - 2*13*14*cos(B)

    cos(B) = {{{(13^2 + 14^2 - 15^2)/(2*13*14)}}} = {{{140/(2*13*14)}}} = {{{10/26}}}.


Next, find sin(B) = {{{sqrt(1-cos^2(B))}}}= {{{sqrt(1 - (10/26)^2)}}} = {{{24/26}}} = {{{12/13}}}.


Finally, find the area of the parallelogram ABCD


    {{{area[ABCD]}}} = AB*BC*sin(B) = {{{13*14*(12/13)}}} = 14*12 = 168 square units.    <U>ANSWER</U>
</pre>

Solved. &nbsp;&nbsp;&nbsp;&nbsp;// &nbsp;&nbsp;&nbsp;&nbsp;An amazing phenomenon is that this answer is a precise integer number &nbsp;(&nbsp;!&nbsp;)


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Another way to solve the problem is to find the area of the triangle &nbsp;ABC &nbsp;using

the &nbsp;Heron's formula &nbsp;(the lengths of the sides of the triangle are given &nbsp;(&nbsp;!&nbsp;) ).


After finding the area of the triangle &nbsp;ABC &nbsp;simply double it, 
and you will get the area of the parallelogram &nbsp;ABCD.