Question

find the area of a rhombus one side of which measures 20cm and one of whose diagonal is 24 cm

Answer 1

Let ABCD Is a rhombus , in which AD and BC are it's two diagonals.



Let AD = 24 cm .


Side AB = AC = CD = BD = 20 cm.



As we know that, diagonals of a rhombus bisect each other at right Angles.

Therefore,


OA = OD = 1/2 × AD = 1/2 × 24 = 12 cm.



In right angled triangle OAB ,


OA = 12 cm and AB = 20 cm.


By Pythagoras theroem,



AB² = (OA)² + (OB)²



( OB)² = (AB)² - (OA)²


( OB )² = (20)² - (12)²


OB² = 256



OB = √256 = 16 cm.



BC = 2 × OB = 2 × 16 = 32 cm.



Therefore,


Area of rhombus ABCD = 1/2 × AD × BC




=> 1/2 × 24 × 32


=> 384 cm².

Answer 2

$\bold{\Huge{\underline{\boxed{\sf{\green{ANSWER\::}}}}}}$

$\bold{\Large{\underline{\sf{\pink{Given\::}}}}}$

In a rhombus each side of which measures 20cm & one of whose diagonal is 24cm.

$\bold{\Large{\underline{\rm{\red{To\:find\::}}}}}$

The area of rhombus.

$\bold{\Large{\underline{\sf{\orange{Explanation\::}}}}}$

We know that formula of the area of rhombus:

→ $\bold{\frac{1}{2} *d1*d2}$

&

$\bold{We\:have\begin{cases}\sf{Length\:of\:side\:of\:rhombus=20cm}\\ \sf{Length\:of\:one\:diagonal=24cm}\end{cases}}}$

In ΔAOB,

Using Pythagoras Theorem:

→ [Hypotenuse]² = [Base]² + [Perpendicular]²

→ AB² = OA² + OB²

→ 20² = 12² + OB²

→ 400 = 144 + OB²

→ OB² = 400 - 144

→ OB² = 256

→ OB = √256

→ OB = 16cm

∴The length of other diagonal,[d2] = 2OB = 2×16 = 32cm

Now,

⇒ Area = $\bold{\frac{1}{2} *d1*d2}$

⇒ Area = $\bold{(\frac{1}{2} *24*32)cm^{2} }$

⇒ Area = $\bold{(\frac{1}{\cancel{2}} *\cancel{24}*32)cm^{2} }$

⇒ Area = (12×32)cm²

⇒ Area = 384cm²

Thus,

The area of a rhombus is 384cm².