Chapter 15, Problem 28CR

To determine

To Find: The altitude of the rhombus.

Answer to Problem 28CR

  $\frac{48}{\sqrt{13}}$

Explanation of Solution

Given:

  $\text{The diagonals of a rhombus are }8\&12.$

Formula Used:

  $Area=\frac{1}{2}bh$

  $\text{Pythagoras Theorem}:h^2=b^2+p^2$

Calculation:

  

  $\begin{array}{l}Let,\\ AD=12\\ BC=8\\ OC=BO=\frac{BC}{2}=\frac{8}{2}=4\\ AO=OD=\frac{AD}{2}=\frac{12}{2}=6\\ A{C}^2=A{O}^2+O{C}^2\\ \text{⇒}A{C}^2=6^2+4^2\\ \text{⇒}A{C}^2=36+16\\ \text{⇒}AC=\sqrt{52}=2\sqrt{13}\\ \text{⇒}\frac{8\times 12}{2}=\frac{2\sqrt{13}\times h}{2}\\ \text{⇒}h=\frac{96}{2\sqrt{13}}=\frac{48}{\sqrt{13}}\end{array}$

Conclusion:

Thus, the altitude is $\frac{48}{\sqrt{13}}$ .