Math, asked by ppkk21732, 8 months ago

day was Rs 90, find the number of articles
Find the dimensions of a rectangle whose perimeter is 28 meters and whose area is 40
square meters
If the area of the triangle is 48 sq.cm

Answers

Answered by kartik2507
0

Step-by-step explanation:

perimeter of the rectangle = 28 m

area of rectangle = 40 m²

perimeter

2(l + b) = 28

l + b = 28/2

l + b = 14 equ (1)

area = l × b = 40

l = 40/b

substitute l = 40/b in equ (1)

l + b = 14 \\  \frac{40}{b}  + b = 14 \\  \frac{40 +  {b}^{2} }{b}  = 14 \\ 40 +  {b}^{2}  = 14b \\  {b}^{2}   - 14b + 40 = 0 \\  {b}^{2}  - 10b - 4b + 40 = 0 \\ b(b - 10) - 4(b - 10) = 0 \\ (b - 10)(b - 4) = 0 \\ b - 10 = 0 \:  \:  \:  \:  \:  \: b - 4 = 0 \\ b = 10 \:  \:  \:  \:  \:  \: b = 4

as breadth of rectangle is considered smaller we take b = 4

l = 40/b

l = 40/4

l = 10 m

length of rectangle = 10 m

breadth of rectangle = 4 m

Hope you get your answer

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