Question

The perimeter of a rhombus is 40 cm and the length of one of its diagonals is 16 cm.Find its area.​

Answer 1

First we have to find length of side of rhombus ----

perimeter of rhombus = 4 x sides

40 = 4 x sides

10 cm = side of rhombus

by using Pythagoras theorem -----

(10)-^2 = (8)^2 + (x)^

6 cm = x

12 cm is length of other diagonal.

Now area of rhombus = 1/2 x product of diagonals

= 1/2 x 16 x 12

= 96 sq.cm

Answer 2

$\fbox{ \fbox{ \large{ \bold{Given :- \: }}}}$

$\implies \: \bold{Perimeter = 40 \: cm \: }$

$\implies \: \bold{ Length \: Of \: One \: Diagonal \: = 16 \: cm \: }$

$\fbox{ \fbox{ \large{ \bold{Solution :- \: }}}}$

$\implies \: \bold{Perimeter = 4 × Side} \\ </p><p> \implies \: \bold{40 = 4 × Side } \\ </p><p> \implies \: \bold{Side = \frac{40}{4} } \\ </p><p> \implies \: \bold{Side = 10 \: cm }$

$\bold{By \: Pythagoras \: Theorem \: , \: } \\ \\ \implies \: \bold{ {h}^{2} = {b}^{2} + \: {p}^{2} \: } \\ \implies \: \bold{ {10}^{2} = {8}^{2} + {x}^{2} \: } \\ \implies \: \bold{ {x}^{2} = 36} \\ \implies \: \bold{x = 6 }$

Hence , The Length Of Other Diagonal Is 12 cm

$\bold{ \underline{We \: Know \: That \: , \: }}$

$\fbox{ \fbox{ \bold{ \huge{Area \: Of \: Rhombus = { \frac{Product \: Of \: Diagonals \: }{2} }}}}}$

$\implies \: \bold{Area \: Of \: Rhombus = \frac{12 \: × \: 16 }{2} \: } \\ \\ \implies \: \bold{Area \: Of \: Rhombus = 96 \: \: {cm}^{2} }$