A wire when bent in the form of a square encloses an area of 484 $m^{2}$ . Find the largest area enclosed by the same wire when bent to form:
(i) an equilateral triangle.
(ii) a rectangle of breadth 16 m.

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Solution

area when as square $= 484 m^{2}$
$a^{2} = 484 m^{2}$
$a = \sqrt{484} = 22 m$
length of the wire $= 4 a = 4 \times 22 = 88 m$
(i) equilateral triangle
Refer image,
$3 P =$ length of wire $= 88 m$
$P = \frac{88}{3} m$
area $= \frac{\sqrt{3}}{4} p^{2} = \frac{\sqrt{3}}{4} = {(\frac{88}{3})}^{2}$
$= \frac{1936 \sqrt{3}}{9}$
$a r e a = 215.11 \sqrt{3} m^{2}$

(ii) rectangle of breadth $16 m$
Refer image 2.
$2 (l + b) = 88$
$l + b = 44$
$l = 44 - 16 = 28$
$A r e a = l b = 28 \times 16 = 448 m^{2}$

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A wire when bent in the form of a square encloses an area of 484 m2. Find the largest area enclosed by the same wire when bent to form:(i) an equilateral triangle.(ii) a rectangle of breadth 16 m.

area when as square =484m2a2=484m2a=√484=22mlength of the wire =4a=4×22=88m(i) equilateral triangle Refer image,3P= length of wire =88mP=883marea =√34p2=√34=(883)2=1936√39area=215.11√3m2(ii) rectangle of breadth 16mRefer image 2.2(l+b)=88l+b=44l=44−16=28Area=lb=28×16=448m2

A wire when bent in the form of a square encloses an area of 484 $m^{2}$ . Find the largest area enclosed by the same wire when bent to form:
(i) an equilateral triangle.
(ii) a rectangle of breadth 16 m.