Chapter 9, Problem 14CT

To determine

To find the dimensions of a rectangle with area $36$$m^2$ and diagonal is $5\sqrt{3}$$m$ .

Answer to Problem 14CT

The length is $4\sqrt{3}$$m$ and breadth is $3\sqrt{3}$$m$ .

Explanation of Solution

Given:

Area of rectangle $=$$36$$m^2$

Diagonal of rectangle $=$$5\sqrt{3}$$m$

Formula Used:

Area of rectangle $=$$l\times b$

Diagonal of rectangle $=$$\sqrt{l^2+b^2}$

Calculation:

Given:

Area of rectangle $=$$36$$m^2$

Diagonal of rectangle $=$$5\sqrt{3}$$m$

Using the above formula, we have:

Area of rectangle $=$$l\times b$

  $36$$=$$l\times b$

  $b=\frac{36}{l}$

And, Diagonal of rectangle $=$$\sqrt{l^2+b^2}$

  $5\sqrt{3}$ $=$$\sqrt{l^2+{\frac{36}{l^2}}^2}$

  $\begin{array}{l}{\left(5\sqrt{3}\right)}^2=l^2+\frac{36}{l^2}\\ 75=\frac{l^4+36}{l^2}\\ l^4-75l^2+1296=0\\ l^4-48l^2-27l^2+1296=0\\ l^2\left(l^2-48\right)-27\left(l^2-48\right)=0\\ \left(l^2-48\right)\left(l^2-27\right)=0\\ \left(l^2-{\left(4\sqrt{3}\right)}^2\right)\left(l^2-{\left(3\sqrt{3}\right)}^2\right)=0\\ l=4\sqrt{3},-4\sqrt{3},3\sqrt{3},-3\sqrt{3}\end{array}$

Thus, length is $4\sqrt{3}$$m$ .

Breadth is $b=\frac{36}{4\sqrt{3}}$

  $b=3\sqrt{3}$$m$

Thus, the length is $4\sqrt{3}$$m$ and breadth is $3\sqrt{3}$$m$ .

Conclusion:

The length is $4\sqrt{3}$$m$ and breadth is $3\sqrt{3}$$m$ .