Question

what will be the d2 of rhombus, whose area is 120cm and d1 is 10.​

Answer 1

Given :

• Area = 120 cm²
• $\sf{ D_1 }$ = 10 cm

To Find :

• $\sf{ D_2 }$ = ?

$\\ \qquad{\rule{200pt}{2pt}}$

SolutioN :

✴ Formula Used :

• Area(Rhombus) = 1/2 × $\sf{ D_1 }$ × $\sf{ D_2 }$

✴ Calculating the Diagonal 2 :

$➺ \qquad \;$ Area = 1/2 × Diagonal 1 × Diagonal 2

$\\ ➺ \qquad \;$ 120 = 1/2 × 10 × Diagonal 2

$\\ ➺ \qquad \;$ 120 \times 2 = 1 × 10 × Diagonal 2

$\\ ➺ \qquad \;$ 240 = 1 × 10 × Diagonal 2

$\\ ➺ \qquad \;$ 240 = 10 × Diagonal 2

$\\ ➺ \qquad \;$ 240/10 = Diagonal 2

• Cancelling 240 by 10 :

$\\ ➺ \qquad \; {\pink{\pmb{\underline{\underline{\frak{ D_2 = 24 \; cm }}}}}}$

✴ Therefore :

2nd Diagonal of the Rhombus is 24 cm .

$\\ \qquad{\rule{200pt}{2pt}}$

Answer 2

Answer:

Length of diagonal 2 is 60cm

Step-by-step explanation:

Question:-

What will be the d2 of rhombus, whose area is 120cm^2 and d1 is 10cm

Given:-

Area of rhombus=

$120 {cm}^{2}$

Length of

$diagonal_{1}$

$= 10cm$

d=diagonal

To find:-

The length of

$diagonal_{2}$

Solution:-

According to the question

d1=19cm

Area of rhombus=120cm^2

d2=?

As we know area of rhombus=

$\frac{1}{2} \times d_{1} \times d_{2}$

On substituting the given values we get,

$\frac{1}{2} \times 10cm \times d_{2} = 300 {cm}^{2}$

By cancelling with 2 we get,

$\mapsto \: \frac{1}{1} \times \times 5cm \times d_{2} = 300 {cm}^{2}$

$\mapsto \: 5cm \times d_{2} = 300 {cm}^{2}$

$\mapsto \: d_{2} = \frac{300 {cm}^{2} }{5cm}$

By cancelling with 5 we get,

$d_{2} = 60cm$

Therefore the length of diagonal 2 is 60cm