Question

The length of a rectangle is twice its width. Find its area, if its perimeter is 7 1/3 cm The area of the rectangle is _sq. cm.

Answer 1

Answer:

The area of rectangle is $2\frac{80}{81}\ cm^2$

Step-by-step explanation:

we know that

The perimeter of a rectangle is equal to

$P=2(L+W)$

we have

$P=7\frac{1}{3}\ cm=7+\frac{1}{3}=\frac{22}{3}\ cm$

so

$\frac{22}{3}=2(L+W)$

$\frac{11}{3}=(L+W)$ ----> equation A

Remember that

The length of a rectangle is twice its width.

so

$L=2W$ ----> equation B

substitute equation B in equation A

$\frac{11}{3}=(2W+W)$

solve for W

$W=\frac{11}{9}\ cm$

Find the value of L

$L=2(\frac{11}{9})$

$L=\frac{22}{9}\ cm$

Find the area of rectangle

$A=LW$

substitute the values

$A=(\frac{22}{9})(\frac{11}{9})$

$A=\frac{242}{81}\ cm^2$

Convert to mixed number

$\frac{242}{81}\ cm^2=\frac{162}{81}+\frac{80}{81}=2\frac{80}{81}\ cm^2$