Question

plzzzzz help me out..... ​

Answer 1

⭐ Diagram:- Refer the attachment ⭐

Given:-

Each side of a rhombus = 5cm

One of the diagonals of the rhombus = 8cm.

To Find:-

The length of the other diagonal.

The area of the rhombus.

Solution:-

(i) As we know that:-

For a rhombus :-

$\sf \implies {Side}^{2} = \bigg(\dfrac{ d_{1}}{2} \bigg)^{2} + \bigg(\dfrac{ d_{2}}{2} \bigg)^{2}$

Where, d₁ and d₂ are the diagonals.

$\sf \implies {(5)}^{2} = \bigg(\dfrac{ 8}{2} \bigg)^{2} + \bigg(\dfrac{ d_{2}}{2} \bigg)^{2}$

$\sf \implies {(5)}^{2} = {(4)}^{2} + \dfrac{ (d_{2} )^{2} }{ {(2)}^{2} }$

$\sf \implies 25 = 16 + \dfrac{ (d_{2} )^{2} }{ {4} }$

$\sf \implies \dfrac{ (d_{2} )^{2} }{ {4} } = 25 - 16$

$\sf \implies (d_{2} )^{2} = 9 \times 4$

$\sf \implies (d_{2} )^{2} =36$

$\sf \implies d_{2} = \sqrt{36} = 6$

$\sf \therefore \underline{\:\: \underline{\: Length \: of \: other \: diagonal \: (d_{2})= 6cm\:}\:\:}$

$\sf (ii) \: Area \: of \: rhombus = \dfrac{1}{2}\times d_{1} \times d_{2}$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{1}{2}\times 8 \times 6$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{48}{2}$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 24cm^2$

$\large \underline{\boxed{\sf \therefore Area \: of \: the \: rhombus = 24cm^2}}$

Answer 2

$\sf \purple{Given:-}$

Each side of a rhombus = 5cm

One of the diagonals of the rhombus = 8cm.

$\sf \purple{To \: Find:-}$

The length of the other diagonal.

The area of the rhombus.

$\sf \purple{Solution:-}$

(i) As we know that:-

For a rhombus :-

$\sf \implies {Side}^{2} = \bigg(\dfrac{ d_{1}}{2} \bigg)^{2} + \bigg(\dfrac{ d_{2}}{2} \bigg)^{2}$

Where, d₁ and d₂ are the diagonals.

$\sf \implies {(5)}^{2} = \bigg(\dfrac{ 8}{2} \bigg)^{2} + \bigg(\dfrac{ d_{2}}{2} \bigg)^{2}$

$\sf \implies {(5)}^{2} = {(4)}^{2} + \dfrac{ (d_{2} )^{2} }{ {(2)}^{2} }$

$\sf \implies 25 = 16 + \dfrac{ (d_{2} )^{2} }{ {4} }$

$\sf \implies \dfrac{ (d_{2} )^{2} }{ {4} } = 25 - 16$

$\sf \implies (d_{2} )^{2} = 9 \times 4$

$\sf \implies (d_{2} )^{2} =36$

$\sf \implies d_{2} = \sqrt{36} = 6$

$\sf \therefore \underline{\:\: \underline{\: Length \: of \: other \: diagonal \: (d_{2})= 6cm\:}\:\:}$

$\sf (ii) \: Area \: of \: rhombus = \dfrac{1}{2}\times d_{1} \times d_{2}$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{1}{2}\times 8 \times 6$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{48}{2}$

$\sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 24cm^2$

$\large \underline{\boxed{\sf \therefore Area \: of \: the \: rhombus = 24cm^2}}$