Question

the length of the diagonals of a rhombus was 24m and 10m.then find (1) perimeter and (2) area .​

Answer 1

$\tt \blue{Solution:- }$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{Given:- }$

$\footnotesize\tt{d1 = 10cm }$

$\footnotesize\tt{d2 = 24cm }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Diagonals meet at the center and forms right angle triangles.

Now, By Pythagoras theorem

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{Length \: of \: the \: base = \: \frac{10}{2}cm }$

$\footnotesize\tt{Length \: of \: the \: base = \: \cancel{ \frac{10}{2}}cm }$

$\footnotesize\tt{Length \: of \: the \: base = \: 5cm }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{Length \: of \: the \: height = \: \frac{24}{2}cm }$

$\footnotesize\tt{Length \: of \: the \: height = \: \cancel\frac{24}{2}cm }$

$\footnotesize\tt{Length \: of \: the \: height = \: 12cm }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{{Hypotenuse}^{2} = \: {side}^{2} + {side}^{2} }$

$\footnotesize\tt{{Hypotenuse}^{2} = \: {5}^{2} + {12}^{2} }$

$\footnotesize\tt{{Hypotenuse}^{2} = \: 25 + 144 }$

$\footnotesize\tt{{Hypotenuse}^{2} = \: 169}$

$\footnotesize\tt{Hypotenuse = \: \sqrt{169} }$

$\footnotesize\tt{Hypotenuse = \: 13}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{Perimeter \: of \: square = 4 \times side}$

$\footnotesize\tt{Perimeter \: of \: square = 4 \times 13cm}$

$\footnotesize\tt{Perimeter \: of \: square = 52cm}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{Area \: of \: square = \: \frac{1}{2} \times d1 \times d2}$

$\footnotesize\tt{Area \: of \: square = \: \frac{1}{2} \times10 \times 24}$

$\footnotesize\tt{Area \: of \: square = \: \frac{1}{ \cancel2} \times \cancel{10} \times 24}$

$\footnotesize\tt{Area \: of \: square = \: 5 \times 24}$

$\footnotesize\tt{Area \: of \: square = \: {120cm}^{2} }$