Chapter 8, Problem 14CT

To determine

The area of the quadrilateral shown in the figure

  

Answer to Problem 14CT

Solution:

The area of the quadrilateral is $54.153$ square units.

Explanation of Solution

Given information:

A quadrilateral having length of sides are $5,7,8,11$ and one angle is ${72}^0$

  

Draw a line joining the points $BD$ as shown below

  

Now, there are two triangles $△ABD$ and $△BCD$

The area of the quadrilateral is the sum of the area of these two triangles.

In triangle $ABD$, $AB=5,AD=11$ and angle $A={72}^0$

By using area of an SAS triangle formula $K=\frac{1}{2}bd\sin A$

  $\Rightarrow K=\frac{1}{2}\cdot 5\cdot 11\cdot \sin \left(72\right)$

  $\Rightarrow K=26.1541$

Thus, area of triangle $ABD=26.1541$ sq. units

Now, to find the length $BD$ use law of cosines $a^2=b^2+d^2-2bd\cos \left(A\right)$

  $\Rightarrow a^2={11}^2+5^2-2\cdot 11\cdot 5\cdot \cos \left(72\right)$

  $\Rightarrow a^2=121+25-33.9919$

  $\Rightarrow a^2=112.0081$

  $\Rightarrow a=10.5834$

Thus, $BD=10.5834$

Now, in triangle $BCD$, $BC=7,CD=8$ and $BD=10.5834$

By using Heron’s formula $K=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}$, where $s=\frac{1}{2}\left(a+b+c\right)$

  $\Rightarrow s=\frac{1}{2}\left(7+8+10.5834\right)$

  $\Rightarrow s=12.7917$

  $\Rightarrow K=\sqrt{\left(12.7917\right)\left(12.7917-7\right)\left(12.7917-8\right)\left(12.7917-10.5834\right)}$

  $\Rightarrow K=\sqrt{\left(12.7917\right)\left(5.7917\right)\left(4.7917\right)\left(2.2083\right)}$

  $\Rightarrow K=\sqrt{783.9385}$

  $\Rightarrow K=27.9989$

Thus, area of triangle $BCD=27.9989$

Now, adding the area of both triangles

Area of quadrilateral $=26.1541+27.9989=54.153$

Thus, the area of the quadrilateral is $54.153$ square units.