Chapter 11.2, Problem 26PSC

To determine

To calculate thearea and perimeter of rhombusABCD.

Answer to Problem 26PSC

The area of rhombus $120$ and perimeter is $52$ .

Explanation of Solution

Given:

  $\begin{array}{l}AC=24,\\ BD=10.\end{array}$

Formula Used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ b=Base,\\ h=Height\end{array}$

Calculation:

  

  $\begin{array}{l}OD=OB=\frac{1}{2}(DB)=\frac{1}{2}\ast 10=5\\ OA=OC=\frac{1}{2}(AC)=\frac{1}{2}\ast 24=12\end{array}$

  $\begin{array}{l}A(\Delta AOB)=\frac{1}{2}*OB*AO\\ =\frac{1}{2}*5*12\\ =30\end{array}$

In rhombus ABCD , all four triangles are congruent to each other. All four triangle are of same area.

  $A(\Delta AOB)=A(\Delta AOD)=A(\Delta COD)=A(\Delta COB)$

  $\begin{array}{l}Areaofrhombus=A(\Delta AOB)+A(\Delta AOD)+A(\Delta COD)+A(\Delta COB)\\ =4*A(\Delta AOB)\\ =4*30=120\end{array}$

In right angled $\Delta AOB$ , we get

  $\begin{array}{l}{\left(AB\right)}^2={\left(AO\right)}^2+{\left(OB\right)}^2\\ $ {\left(AB\right)}^2={(5)}^2+{(12)}^2$=25+144\\ {\left(AB\right)}^2=169\\ AB=\sqrt{169}\\ AB=13\end{array}$

In rhombus ABCD , all four sides are congruent to each other. All four sides are of same length.

  $AB=BC=CD=DA=13$

  $\begin{array}{l}Perimeterofrhombus=AB+BC+CD+DA\\ =4\times AB\\ =4\times 13=52\end{array}$