Question

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

Answer 1

Given : -

Length of the Rectangle is 40cm and diagonal is 41cm.

Diagram : -

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\tt\large{A}}\put(7.7,1){ \tt\large{D}}\put(9.5,0.7){\sf{\large{40 cm}}}\put(11.5,1){ \tt\large{C}}\put(8,1){\line(1,0){3.5}}\put(8,1){\line(0,2){2}}\put(11.5,1){\line(0,3){2}}\put(8,3){\line(3,0){3.5}}\put(11.6,2){\sf{\large{ ?}}}\put(9.7,1.7){\sf{\large{ 41 cm}}}\qbezier(8,1)(8,1)(11.5,3)\put(11.5,3){ \tt\large{B}}\put(11.3,1){\line(0,2){0.2}}\put(11.3,1.2){\line(2,0){0.2}}\end{picture}$

To Find : -

Perimeter of the rectangle = ?

Step-by-step Explanation : -

Length of rectangle = 40cm

Diagonal of Rectangle = 41cm

By Using Pythagoras Property : -

In ∆BCD ,

(BD)² = (BC)² + (DC)²

(41)² = (BC)² + (40)²

(41)² - (40)² = (BC)²

1681 - 1600 = (BC)²

81 = (BC)²

BC² = 9²

BC = 9

So , the breadth of rectangle is 9cm..

Now : -

Perimeter of Rectangle = 2(l + b)

2(40 + 9)

2(49)

98

Therefore , the perimeter of rectangle is 98cm..

Answer 2

⋆ DIAGRAM :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\large{A}}\put(9.6,1.7){\sf{\large{(41cm)}}}\put(7.7,1){\large{B}}\put(9.1,0.7){\sf{\large{(40cm)}}}\put(11.1,1){\large{C}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\put(11.1,2){\sf{\large{n}}}\put(8,1){\line(3,2){3}}\put(11.1,3){\large{D}}\put(10.8,1){\line(0,2){0.2}}\put(10.8,1.2){\line(2,0){0.2}}\end{picture}$

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✩ Property of rectangle : Each Angle of Rectangle forms 90°, and so Form two Right Angle Triangle Inside it.

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☢By Pythagoras theorem in ∆ BCD :

: $\sf\implies ( hypotenuse )^2 = (perpendicular )^2+ (base)^2$

:$\sf\implies 41^2 = 40^2 + n^2$

:$\sf\implies 1681 = 1600 + n^2$

:$\sf\implies 1681 - 1600 = n^2$

:$\sf\implies n^2 = 81$

:$\sf\implies n = \sqrt{81}$

:$\sf\implies n = 9$

$\sf{\underline{\boxed{\green{\large{\bold{ n = breadth = 9cm }}}}}}$

__________________________

$\sf{\underline{\boxed{\blue{\large{\bold{ perimeter = 2 ( l + b )}}}}}}$

:$\sf\implies perimeter = 2 ( 40 + 9 )cm$

:$\sf\implies perimeter = 2 \times 49 cm$

:$\sf\implies perimeter = 98cm$

$\sf{\underline{\boxed{\green{\large{\bold{ \dag perimeter = 98cm }}}}}}$