Question

Manisha has a garden in the shape of rhombus the perimeter of the garden is 40 m and its diagonal is 16m she wants to divide it into two equal parts in used parts in rotation find the area of each part of garden

Answer 1

$\huge\bold{Answer}$

Perimeter of a garden = 40 m

We All know, rhombus has all sides equal

=$\frac{Perimeter}{4}$

=$\frac{40}{4}$

=10 m.

First diagonal = 16 m

$\frac{1}{2}$ = height of ∆OB.

Diagonal of rhombus are perpendicular
bisector of each other .

Bo^2 + Ao^2 = AB^2
Bo^2 + 8^2 = 10^2.
Bo ^2 = 10^2-8^2= 6^2
Bo= 6×2 => 12m

Area of ∆ BoA [1st part ]
$\frac{1}{2}$ × base × height .

$\frac{1}{2}$×Ao×Bo.
6×8 => 48m ^2

Area of 2nd part.

$\frac{1}{2}$×base×height

$\frac{1}{2}$12×8

6× 8

48m^2

$\huge\bold{Same\:for\:2nd\:part}$

=>She divided the garden into 2 equal part in 48m^2

Answer 2

$\mathbb{\red{\bold{SOMEONES.JOKER}}}$
Perimeter of a garden = 40 m
We All know, rhombus has all sides equal 

=perimeter/4

=40/4

=10 m.

First diagonal = 16 m

1/2 = height of ∆OB.

Diagonal of rhombus are perpendicular 
bisector of each other .

Bo^2 + Ao^2 = AB^2
Bo^2 + 8^2 = 10^2.
Bo ^2 = 10^2-8^2= 6^2
Bo= 6×2 => 12m

Area of ∆ BoA [1st part ]
1/2 × base × height .

1/2×Ao×Bo.
6×8 => 48m ^2

Area of 2nd part.

1/2 ×base×height

1/2* 12×8 

6× 8

48m^2
for 2 part

=>She divided the garden into 2 equal part in 48m^2
$<marquee>$
$\mathbb{\red{\bold{KUHU}}}$