Question
The area of a square field is $5184 cm^2$. A rectangular field, whose length is twice its breadth has its perimeter equal to the perimeter of the square field. Find the area of the rectangular field.
✓
Answer
First, we have to find the perimeter of the square.
The area of the square is $r^2$, where $r$ is the side of the square.
Then, we have the equation as follows,
$r^2=5184=(2 \times 2) \times(2 \times 2) \times(2 \times 2) \times(3 \times 3) \times(3 \times 3)$
Taking the square root, we get $r =2 \times 2 \times 2 \times 3 \times 3=72$
Hence the perimeter of the square is $4 \times r=288 m$
Now let $L$ be the length of the rectangular field.
Let $W$ be the width of the rectangular field.
The perimeter is equal to the perimeter of square. Hence, we have,
$2(L+W)=288$
Moreover, since the length is twice the width,
$L=2 \times W$
Substituting this in the previous equation, we get,
$2 \times(2 \times W+W)=288$
$3 \times W=144$
$W=48$
To find $L$,
$L=2 \times W=2 \times 48=96$
Area of the rectangular field $= L \times W =96 \times 48=4608 m^2$
The area of the square is $r^2$, where $r$ is the side of the square.
Then, we have the equation as follows,
$r^2=5184=(2 \times 2) \times(2 \times 2) \times(2 \times 2) \times(3 \times 3) \times(3 \times 3)$
Taking the square root, we get $r =2 \times 2 \times 2 \times 3 \times 3=72$
Hence the perimeter of the square is $4 \times r=288 m$
Now let $L$ be the length of the rectangular field.
Let $W$ be the width of the rectangular field.
The perimeter is equal to the perimeter of square. Hence, we have,
$2(L+W)=288$
Moreover, since the length is twice the width,
$L=2 \times W$
Substituting this in the previous equation, we get,
$2 \times(2 \times W+W)=288$
$3 \times W=144$
$W=48$
To find $L$,
$L=2 \times W=2 \times 48=96$
Area of the rectangular field $= L \times W =96 \times 48=4608 m^2$
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