Question

The perimeter of a rhombus ABCD is 40 cm . Find the area of rhombus of its diagonals BD measures 12 cm. By herons formula

Answer 1

Answer:

$\huge{\pink{\boxed{\green{\sf{AREA=96cm^{2}}}}}}$

Step-by-step explanation:

$\huge{\pink{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}$

$\: \: \: \: \: \: \: \: \: \: {\orange{given}} \\ { \pink{ \boxed{ \green{perimeter = 40cm}}}} \\ { \pink{ \boxed{ \green{d1 = 12cm}}}} \\ \\ \: \: \: \: \: \: \: \: { \blue{to \: find}} \\ { \purple{ \boxed{ \red{area \: of \: rhombus=? }}}}$

☆ According to given question:

$\to perimeter \: of \: rhombus = 4 \times side \\ \to 40 = 4 \times side \\ \to side = \frac{40}{4} \\ { \boxed{\to side = 10cm }}\\ \\ \to {h}^{2} = {p}^{2} + {b}^{2} \\ \to {10}^{2} = {p}^{2} + {6}^{2} \: \: \: \: \: \: \: \: \: (according \: to \: rhombus \: property) \\ \to 100 = {p}^{2} + 36 \\ \to 100 - 36 = {p}^{2} \\ \to {p}^{2} = 64 \\ \to {p} = \sqrt{64} \\ { \boxed{\to p = 8cm}} \\ \\ \to d2 = 2 \times p \\ \to d2 = 2 \times 8 \\ { \boxed{\to d2 = 16cm}}$

☆ We find diagonal 1 and diagonal 2 so we find area :

$\to area \: of \: rhombus = \frac{1}{2} \times d1 \times d2 \\ \to area = \frac{1}{2} \times 12 \times 16 \\ { \pink{ \boxed{ \green{\therefore area = 9 6 {cm}^{2} }}}}$

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Answer 2

Answer:

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Step-by-step explanation:

Given,

Diagonal BD of rhombus = 12cm

Perimeter of rhombus = 40cm

4s = 40

s = 40/4

s = 10cm

Now,

Divide the rhombus into four right angled triangles.

Such that, every triangle has a hypotrneuse of 10cm

12/2 = 6cm

Using pythagorus therom :

= √10² - 6²

= 8cm

Now,

8+8= 16cm (sum of the bases of two triangles = measure of d1)

Area of rhombus = 1/2 × d1 × d2

= 1/2 × 16 × 12

= 96cm²

Therefore, diagonal AC = 16cm and the area of rhombus = 96cm²