Question

the area of a rhombus is 24 sq cm and one of its diagonal is 8 cm . Find its perimeter

Answer 1

area of rhombus = 1/2* diagonal 1 * diagonal 2
24 = 1/2*8 * diagonal 2.
24 *2/8 = diagonal 2
diagonal = 6 cm
according to Pythagoras theorem ..
( diagonal /2)^2 + ( diagonal/2 ) ^2 = ( side )^2
= (8/2 )^2 + (6/2)^2= side ^2
16 + 9 = side ^2
side = √25
= 5cm
perimeter = 4* side
= 4*5= 20 sq. cm.

Answer 2

Heya User ✌

Here's your answer friend,

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✔ GIVEN : Area of a rhombus = 24 cm²

==> and, One of the diagonal is 8 cm

Let other diagonal be x cm.

$since$

Area of Rhombus =

1/2 ( diagonal 1 X diagonal 2)

==> 1/2 (8 X diagonal 2)

==> 1/2 (8 X x)

==> 1/2(8 X x) = 24.........{ from given }

==> 1/2(8x) = 24

==> 4x = 24

==> x = 6 cm

Is the required value of other diagonal.

NOW,

✔PERIMETER OF RHOMBUS

==> 4a

where a = side of the rhombus

$therefore$

PERIMETER OF THE GIVEN RHOMBUS

By using the formula

==> Area(A) = (d1 X d2)/2

==> and perimeter = 4(side of the rhombus)

Let d1(diagonal 1) be p and d2(diagonal 2) be q

and side of the rhombus(a) = (√p² + √q²)/2

NOW,

PERIMETER = ²√p² + 4(A/p)²

==> ²√8² + 4(24/8)²

==> 20 cm.

HOPE IT HELPS YOU :)

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