Chapter 11.4, Problem 10PSC

To determine

To find:The height of rhombus whose diagonals are 18 and 24.

Answer to Problem 10PSC

The height of rhombus 14.4 cm.

Explanation of Solution

Given information :

The diagonal one of the rhombus $(d_1)=18$

The diagonal two of the rhombus $(d_2)=18$

Formula used :

Area of rhombus = $\frac{1}{2}\times \left(\text{product of diagonals}\right)$

Area of rhombus = $\text{ Base }\times \text{Height}$

Calculation :

Base of the rhombus will be equal to its side

Therefore,

  $\begin{array}{l}$ {\left(\text{Side}\right)}^2={\left(\frac{d_1}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2$$ {\left(\text{Side}\right)}^2={\left(\frac{18}{2}\right)}^2+{\left(\frac{24}{2}\right)}^2$$ {(\text{Side})}^2=9^2+{12}^2$$ {(\text{Side})}^2=81+144$\text{Side}=\sqrt{225}\\ \text{Side}=15\text{ cm}\end{array}$

Therefore, base = 15 cm

Now,

  $\text{Base }\times \text{Height}$ = $\frac{1}{2}\times \left(\text{product of diagonals}\right)$

  $\begin{array}{l}15\times \text{Height=}\frac{1}{2}\times 18\times 24\\ \Rightarrow \text{Height}=\frac{9\times 24}{15}=14.4\text{ cm}\end{array}$