Question

the perimeter of a rhombus is 26 cm and one of its diagonal is 66 cm find the area of the rhombus and its other diagonal. ​

Answer 1

Step-by-step explanation:

Given:-

Perimeter of a Rhombus =26cm

Length of its one diagonal =66cm

To find:-

• Area of Rhombus
• Length of other diagonal

Solution:-

Let the side = a

As we know that in a Rhombus

$\boxed {\sf Perimeter =4a}$

• Substitute the values

$\\\qquad\quad\sf {:}\hookrightarrow 4a=26$

$\\\qquad\quad\sf {:}\hookrightarrow a=\dfrac {26}{4}$

$\\\qquad\quad\sf {:}\hookrightarrow a=6.5cm$

• Side =6.5cm

$\rule {200}{2}$

Let the other diagonal be x

We know that in a Rhombus

$\boxed {\purple {\sf Side=\dfrac {1}{2}\sqrt{(d_1)^2+(d_2)^2}}}$

• Substitute the values

$\\\qquad\quad\sf {:}\hookrightarrow 6.5=\dfrac {1}{2}\sqrt{(66)^2+(x)^2}$

$\\\qquad\quad\sf {:}\hookrightarrow 6.5=\dfrac {1}{2}\sqrt {4356+x^2}$

$\\\qquad\quad\sf {:}\hookrightarrow \sqrt {4356+x^2}=6.5\times 2$

$\\\qquad\quad\sf {:}\hookrightarrow \sqrt {4356+x^2}=13$

$\\\qquad\quad\sf {:}\hookrightarrow x^2+4356=(13)^2$

$\\\qquad\quad\sf {:}\hookrightarrow x^2+4356=169$

$\\\qquad\quad\sf {:}\hookrightarrow x^2=4356-169$

$\\\qquad\quad\sf {:}\hookrightarrow x^2=4187$

$\\\qquad\quad\sf {:}\hookrightarrow x=\sqrt {4187}$

$\\\qquad\quad\sf {:}\hookrightarrow x=64.7cm$

• Length of Other diagonal=64.7cm

$\rule {200}{2}$

As we know that in a Rhombus

$\boxed{\pink {\sf Area=\dfrac {1}{2}\times d_1{}^2\times d_2 {}^2}}$

• Substitute the values

$\\\qquad\quad\sf {:}\hookrightarrow Area=\dfrac {1}{2}\times (66)^2\times (64.7)^2$

$\\\qquad\quad\sf {:}\hookrightarrow Area=\dfrac {1}{2}\times 4356\times 4187$

$\\\qquad\quad\sf {:}\hookrightarrow Area=\dfrac {4356×4187}{2}$

$\\\qquad\quad\sf {:}\hookrightarrow Area=9119cm^2$