Question

plz solve it .......​

Answer 1

Answer:

perimeter of rectangle =L×b

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D

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a

g

o

n

a

l

)

2

=

(

L

e

n

g

t

h

)

2

+

(

W

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d

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2

Answer 2

$\large\blue{\tt Given :-}$

Length of the rectangle = 40 cm

Diagonal of the rectangle = 50 cm

$\large\orange{\tt To \: Find :-}$

Perimeter of the rectangle

$\large\green{\tt Solution :-}$

We have,

Length of the rectangle = 40 cm

Let,

The breadth of the rectangle be “x”

We know that,

$\boxed{\tt Diagonal \: of \: rectangle = \sqrt{l^{2} + b^{2}}}$

Where,

l = Length

b = breadth

Substituting the given values of leagnth and breadth we get,

$:\implies\tt\sqrt{40^{2} + x^{2}} = 50$

$:\implies\tt 40^{2} + x^{2} = 50^{2}$

$:\implies\tt 1600 + x^{2} = 2500$

$:\implies\tt x^{2} = 2500 - 1600$

$:\implies\tt x^{2} = 900$

$:\implies\tt x = \sqrt{900}$

$:\implies\tt x = 30$

$\therefore\bf Breadth \: of \: the \: rectangle = 30 \: cm$

We need to find the perimeter of the rectangle :-

We also know that,

$\boxed{\tt Perimeter \: of \: rectangle = 2(l + b)}$

Where,

l = Length

b = breadth

Substituting the given values of leagnth and breadth we get,

$= \tt2(40+30)$

$= \tt2 \times 70$

$= \tt140 \: cm$

$\therefore\bf Perimeter \: of \: the \: rectangle = 140 \: cm$

$\large\red{\tt Verification :-}$

We have,

Length of the rectangle = 40 cm

Breadth of the rectangle = 30 cm

Diagonal of the rectangle = 50 cm

Then,

$:\implies\tt\sqrt{40^{2}+30^{2}} = 50$

$:\implies\tt\sqrt{1600+900} = 50$

$:\implies\tt\sqrt{2500} = 50$

$:\implies\tt 50 = 50$

HENCE VERIFIED