Question

A rectangular Picture Frame has Length 24cm and Area 672cm². Find Breadth and Perimeter of Rectangle Frame.​

Answer 1

Given:,

• Length of Rectangle = 24cm
• Area of Rectangle = 672cm²

Find:

• Breadth of Rectangle
• Perimeter of Rectangle

Solution:

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\linethickness{5mm}\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf D}\put(-0.5,4.2){\bf A}\put(4.2,-0.5){\bf C}\put(4.2,4.2){\bf B} \put(4.8,2){\bf 24cm} \linethickness{1m}\put(4.2,2){\vector(0,1){2}}\put(4.2, 2){\vector(0, - 1){2}}\put( - 0.8,2){\vector(0,1){2}}\put( -0.8,2){\vector(0, - 1){2}}\put( - 2,2){\bf 24cm} \end{picture}$

Here, we know that

$\boxed{ \sf Area \: of \: rectangle = length \times breadth}$

where,

• Area of Rectangle = 672cm²
• Length, l = 24cm

So,

$\dashrightarrow \sf Area \: of \: rectangle = l \times b$

$\dashrightarrow \sf 672 = 24 \times b$

$\dashrightarrow \sf 672 = 24b$

$\dashrightarrow \sf \dfrac{672}{24} = b$

$\dashrightarrow \sf 28cm = b$

$\therefore \sf b = 28cm$

_____________________________

Now, we know that

$\boxed{ \sf Perimeter \: of \: rectangle = 2(length + breadth)}$

where,

• Length, l = 24cm
• Breadth, b = 28cm

So,

$\dashrightarrow \sf Perimeter \: of \: rectangle = 2(l + b)$

$\dashrightarrow \sf Perimeter \: of \: rectangle = 2(24+ 28)$

$\dashrightarrow \sf Perimeter \: of \: rectangle = 2(52)$

$\dashrightarrow \sf Perimeter \: of \: rectangle = 104cm$

$\therefore \sf Perimeter = 104cm$

_____________________________

Hence,

• Breadth of Rectangle = 28cm
• Perimeter of Rectangle = 104cm