Question #191653

A semi-circle of radius 14 cm is formed from a piece of wire. If it is bent into a rectangle whose length is 2 cm more than its width, find the area of the rectangle.


1
Expert's answer
2021-05-17T03:55:02-0400

 perimeter of semi circle=(2+π)r=(2+π)×14=71.9823cm Given, length of rectangle(l) is 2cm more than its width(b).l=2+b perimeter of rectangle=2(l+b) Given, perimeter of semi circle=perimeter of rectangle(2+π)×14=2(l+2+l)(2+π)×14=4(l+1)l+1=(2+π)×144l=(2+π)×1441l=16.9956cm Therefore, width of rectangle is 18.9956cm.Area of the rectangle=lb=16.9956×18.9956=322.8416cm2\text{ perimeter of semi circle}=(2+\pi)r=(2+\pi)×14=71.9823 cm\newline\text{ Given, length of rectangle(l) is 2cm more than its width(b).}\newline l=2+b\newline\text{ perimeter of rectangle}=2(l+b)\newline\text{ Given, perimeter of semi circle=perimeter of rectangle}\newline (2+\pi)×14=2(l+2+l)\newline (2+\pi)×14=4(l+1)\newline l+1=\frac{(2+\pi)×14}{4}\newline l=\frac{(2+\pi)×14}{4}-1\newline l=16.9956 cm\newline\text{ Therefore, width of rectangle is }18.9956 cm.\newline \text{Area of the rectangle}=lb=16.9956×18.9956=322.8416 cm^2


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