Question

wire of length 20 cm is to be Bent to form a rectangle of : (1) maximum and (2) minimum area ​

Answer 1

length of wire = perimeter of wire = 20 cm

$20 = 2(l + b) \\ l + b = \frac{20}{2} \\ l + b = 10$

• We have to find maximum area. So, take l=b . Then
• $l + b = 10 \\ l + l = 10 \\ 2l = 10 \\ l = \frac{10}{2} \\ l = 5 \: cm \\ l = b = 5 \: cm$

Maximum area of ∆

$= l \times b \\ = 5 \times 5 \\ = 25 \: cm {}^{2}$

• Now , we have to find minimum area . So , take l=0 . Then

$l + b = 10 \\ 0 + b = 10 \\ b = 10 \: cm$

Minimum area of ∆

$= l \times b \\ = 0 \times 10 \\ = 0 \: cm {}^{2}$

Maximum area of ∆ is 25 cm² & minimum area of ∆ is 0 cm² .

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