Question

Find the area of a rhombus whose each side is 13m and one of its diagonal is 10m

Answer 1

$\huge\star\underline\mathfrak\pink{Answer}$

Given Data:

Side of rhombus= 13m

Diagonal of rhombus= 10m

AC= 10m

AE= 1/2 * AC = 1/2 * 10 = 5m

Apply Pythagoras theorem :

In ∆ABE,

BE= √AB^2- √AE^2

= √(13)^2-5^2

= √169-25

= √144

= 12m

Now,

BD= 2*BE = 2*12 = 24m

Finding area using standard formula :

Area = 1/2*diagonal 1* diagonal 2

= 1/2 *AC * BD

= 1/2 * 10*24

= ${120m}^{2}$

Answer 2

Answer...

Area of rhombus is 4 times the area of smallest right angled triangle formed by the diagonals.

This is because of the property of the rhombus that the diagonals of the rhombus bisect each other i.e they cut each other at midpoints and at right angles.

Hence, the lengths of both the diagonals are be known.

Now one of diagonal is 10 i.e its half distance is 5.

From this distance the other diagonal will cut.

Therefore the small right angled triangle is formed.

In this triangle the length of the side(hypotenuse)=13 and one perpendicular side is 5.

Now by pythagoras theorem the other perpendicular side is known.

Other perpendicular side= square root(13^2-5^2)

= square root (169-25) = square root (144) =12.

Now area of this small triangle,

Area= 1/2 base height

= 1/2(product of prpendicular sides)

=1/2(5*12)

=30

Now the area of rhombus,

Area= 4 ( area of small triangle)

= 4(30)

=120.

ANS- 120.