Question

find the area of the rombus who's one side measures 5cm and one diagonal as 8cm​

Answer 1

Step-by-step explanation:

Let ABCD be the given rhombus with diagonals AC and BD which intersect each other at O such that AC = 8 cm.

Since diagonals of a rhombus bisect each other, we get

AO=OC=\frac{1}{2} AC=\frac{1}{2} \times 8=4 cmAO=OC=

2

1

AC=

2

1

×8=4cm

Let BO = x and AB = 5 cm (given).

We know that diagonals of a rhombus cut each other at right angles. So ΔAOB is a right angled triangle.

Using Pythagoras theorem in ΔAOB, we get

OA^2+OB^2=AB^2OA

2

+OB

2

=AB

2

4^2+x^2=5^24

2

+x

2

=5

2

16+x^2=2516+x

2

=25

x^2=25-16x

2

=25−16

x^2=9x

2

=9

x=\sqrt{9}x=

9

x = 3 cm

Thus the length of another diagonal BD is,

BD = 2x = 6 cm.

Therefore the area of the rhombus is given by,

Area of rhombus =\frac{1}{2}\times (poduct of diagonals)

2

1

×(poductofdiagonals)

Area(ABCD) = \frac{1}{2}\times AC\times BD

2

1

×AC×BD

= \frac{1}{2}\times 8\times 6

2

1

×8×6

= 24 cm^224cm

2