Question

Find the area of the of the rectangle whose base is 40cm and diagonal is 41cm. Solve it step by step​

Answer 1

Solution

Given :-

• Base of rectangle = 40 cm
• Diagonal of Rectangle = 41 cm

Find :-

• Area of rectangle

Explanation

Formula

$\dag\boxed{\underline{\tt{\red{\: Diagonal_{rectangle}\:=\:\sqrt{(L²+B²)}}}}}$

where,

• L = Length
• B = Base

So, Now

==> Diagonal of rectangle = √(L² + B²)

==> 41 = √(L² + 40²)

Squaring both side

==> 41² = L² + 1600

==> L² = 1681 - 1600

==> L² = 81

==> L = √81

==> L = 9

Now, Calculate area of rectangle

$\dag\boxed{\underline{\tt{\orange{\:Area_{rectangle}\:=\:(L\times B)}}}}$

keep all above Values

==> Area of rectangle = 40 × 9

==> Area of rectangle = 360 cm²

.Hence

• Area of rectangle will be = 360 cm²

________________

Answer 2

Given:-

• length of rectangle is 40 cm.
• diagonal of rectangle is 41 cm.

To Find:-

• Area of rectangle.

Solution:-

Formula to find diagonal of rectangle:-

$\huge \boxed{\sqrt{ {l}^{2} \times {b}^{2} } }$

Where,

• l = length
• b = breadth

Here we go:-

• Diagonal = 41 cm
• Length = 40 cm

${41}^{2} = {l}^{2} + {b}^{2} \\ \\ {41}^{2} = {40}^{2} + {b}^{2} \\ \\ 1681 = 1600 + {b}^{2} \\ \\ {b}^{2} = 1681 - 1600 \\ \\ {b}^{2} = 81 \\ \\ b = \sqrt{81} = 9$

Area of rectangle:-

$\huge \boxed {\rm{length \times breadth}}$

Keep all above values:-

• Area = 40×9
• Area = 360 cm²

Hence the area of rectangle is 360 cm².