Question

26. The width of a rectangle is 64 inches. The length of the rectangle is twice its width What is the perimeter of the rectangle? A 20 inches B. 40 inches c. 30% inches D. 885 inches​

Answer 1

W = 64
L = 64*2 = 128

P = (64 * 2) + (128 * 2) = 384

Answer 2

$\large \orange{ \frak{Given :}}$

• Width of rectangle = 64 in.

• Length of the rectangle is twice its width.

$\large \orange{ \frak{To \: find:}}$

• Perimeter of rectangle

$\large \orange{ \frak{Solution:}}$

We know :-

• Length = twice of width
• Length = 2 × width
• Length = 2 × 64
• Length = $\small \bf 128$

$\large \orange{ \frak{Digram:}}$

$\tt 64in.\begin{gathered} \small \large\boxed{\begin{array}{cc} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \\ \end{array}}\end{gathered} \\ \: \: \: \: \sf128in.$

Formula of perimeter of rectangle:-

$\bigstar \boxed{\rm{}perimeter \: of \: rectangle = 2(l + w)}$

where :-

l = length of rectangle

w = width of rectangle

So :-

$\dashrightarrow\sf{}perimeter \: of \: rectangle = 2(l + w)$

$\dashrightarrow\sf{}perimeter \: of \: rectangle = 2(64 + 128)$

$\dashrightarrow\sf{}perimeter \: of \: rectangle = 2(64) + 2(128)$

$\dashrightarrow\sf{}perimeter \: of \: rectangle = 2 \times 64+ 2(128)$

$\dashrightarrow\sf{}perimeter \: of \: rectangle =128+ 2(128)$

$\dashrightarrow\sf{}perimeter \: of \: rectangle =128+ 2 \times 128$

$\dashrightarrow\sf{}perimeter \: of \: rectangle =128+ 256$

$\dashrightarrow\bf{}perimeter \: of \: rectangle =384 \: in$