Chapter 11, Problem 5RP

To determine

To calculate:Thearea of parallelogram ABCD .

Answer to Problem 5RP

The area of parallelogram ABCDis $48\sqrt{3}$ .

Explanation of Solution

Given information:

In a parallelogram ABCD ,

Side AB = 8,

Side BC = 12.

Angle B = ${60}^{\circ }$

Formula used:

Side opposite to ${60}^{\circ }$ angle = $\frac{\sqrt{3}}{2}\times hypotenuse$

Area of parallelogram: $A=b\times h$

b =base of parallelogram

h= height of parallelogram

Calculation:

  

Draw the altitude AE .

ΔABE is ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta$

Side opposite to ${60}^{\circ }$ angle= $\frac{\sqrt{3}}{2}\times hypotenuse=\frac{\sqrt{3}}{2}\times 8=4\sqrt{3}$

  $\begin{array}{l}A(▱ABCD)=BC\times AE\\ A(▱ABCD)=12\times 4\sqrt{3}\\ A(▱ABCD)=48\sqrt{3}\end{array}$