Question

find the perimeter of the rectangle whose length is 40 cm and a diagonal is 41 cm

Answer 1

Hey there,

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In the given figure---
Diagonal = 41
Length = 9

So applying Pythagoras theorem, we get-

x²+40²=41²
x²=41²-40²
x²=1681-1600
x²=81
x=√81=9

So the breadth of triangle = 9
 
So the perimeter=(l+b)
=2(9+41)
=2(50)
=100 units

Regards
07161020
Ace

Answer 2

Given:-

• Diagonal of the rectangle is 41 cm
• Length of rectangle is 40 cm.

To Find:-

• Area of rectangle

Solution:-

According to the Pythagoras theorem:-

$\dag{ \boxed{ \underline{ \tt { \pink{hypotenuse² = length² + breadth²}}}}}$

$\Longrightarrow {41}^{2} = {40}^{2} + {b}^{2} \\ \\ \Longrightarrow 1681 = 1600 + {b}^{2} \\ \\ \Longrightarrow1681 - 1600 = {b}^{2} \\ \\ \Longrightarrow81 = {b}^{2} \\ \\ \Longrightarrow \sqrt{81} = b \\ \\ \Longrightarrow \dag{ \boxed{ \underline{ \tt{ \green{b = 9}}}}}$

Area of rectangle:-

$\dag{ \boxed{ \underline{ \tt{ \blue{length \times breadth}}}}}$

$\Longrightarrow40 \times 9 \\ \\ \Longrightarrow \dag{ \boxed { \underline{ \tt{ \red{{360 \: cm}^{2} }}}}}$

Hence, the area of rectangle is 360 cm².