Chapter 7.4, Problem 9GP

To determine

To find: The area of the given rhombus.

Answer to Problem 9GP

  $52.87$

Explanation of Solution

Given:

A figure is given as shown below.

In the given figure length of the two diagonals is given.

Concept Used:

Area of rhombus $=$

• $\frac{1}{2}{d}_1{d}_2$ , where $d_1$ and $d_2$are the diagonals of the rhombus. (given in the question)
• Distributive law: It states that the product of sum of two number and a third number is equal to the sum of the products of third number and the two numbers separately. For example $a\left(b+c\right)=ab+ac$ and vice versa.

Calculation:

Area of the given rhombus (A) $\begin{array}{l}=\frac{1}{2}\left(8\sqrt{5}-3\sqrt{6}\right)\left(6\sqrt{5}-2\sqrt{3}\right)\\ =\frac{1}{2}\left(8\sqrt{5}·6\sqrt{5}-8\sqrt{5}·2\sqrt{3}-3\sqrt{6}·6\sqrt{5}+3\sqrt{6}·2\sqrt{3}\right)\\ =\frac{1}{2}\left(8.6\left(\sqrt{5}\times \sqrt{5}\right)\right)-8·2·\sqrt{5}·\sqrt{3}-3·6·\sqrt{6}·\sqrt{5}+3·2·\sqrt{6}·\sqrt{3}\\ =\frac{1}{2}\left(48\times 5-16·\sqrt{5}·\sqrt{3}-18·\sqrt{3}·\sqrt{2}·\sqrt{5}+6·\sqrt{3}·\sqrt{2}·\sqrt{3}\right)\end{array}$

Put $\sqrt{2}=1.41,\sqrt{3}=1.73,\sqrt{5}=2.23$

  $\begin{array}{l}=\frac{1}{2}\left(240-16\times 2.23\times 1.73-18\times 1.73\times 2.23\times 1.41+6\times 3\times 1.41\right)\\ =\frac{1}{2}\left(240-61.73-97.91+25.38\right)\\ =\frac{1}{2}\times 105.74\\ =52.87\end{array}$

Thus, Area of the given rhombus $=52.87$ square units.