Question

The length of a rectangle is 20 cm more than its breadth. If the perimeter is 100 cm, find the dimension of the rectangle.
Don't spam otherwise reported!!​

Answer 1

Dimensions of the rectangle:

Breadth (b) = 15 cm

Length (l) = 35 cm

Explanation:

Let breadth of a rectangle = b cm

length (l) = 20 cm more than its breadth

= (b+20)cm

Perimeter of the rectangle (P) = 100 cm [given]

=> 2(length + breadth) = 100cm

=> 2(b + 20 + b)=100 cm

$\rm\implies 2b+20=\frac{100cm}{2} \\\rm\implies 2b = 50cm -20cm \\ \rm\implies 2b = 30cm \\\rm\implies b = \frac{30cm}{2} \\ \rm\implies \bold b = 15cm$

Therefore,

Dimensions of the rectangle:

• Breadth (b) = 15 cm
• Length (l) = b + 20 = 15+20 = 35 cm

Answer 2

$\red \bigstar \sf \: \: { \underline{ \underline {G \: I \: V \: E \: N :}}}$

$\:$

Dimensions of the rectangle:

• Breadth (b) = 15 cm

• Length (l) = 35 cm

$\:$

$\red \bigstar \: \: { \sf{ \underline{ \underline{ Explanation: }}}}$

• Let breadth of a rectangle = b cm

• length (l) = 20 cm more than its breadth

$\sf \: = (b + 20) \: cm$

Perimeter of the rectangle (P) = 100 cm [given]

$\:$

=> 2(length + breadth) = 100cm

$\:$

=> 2(b + 20 + b)=100 cm

$\\ \rm \implies \sf \: 2b + 20 = \dfrac{100 \: cm}{2} \\ \\ \rm \implies \sf \: 2b = 50 - 30 \: \\ \\ \rm \implies \sf \: 2b = 30 \\ \\ \rm \implies \sf \: b = \dfrac{30}{2}$

$\\ \implies { \boxed{ \frak{ \blue{b = 15 \: cm}}}}$

$\: \:$

$\therefore \sf \: dimensions \: of \: the \: rectangle : :$

$\:$

$\implies { \boxed{ \frak{ \green{breadth = 15 \: cm}}}}$

$\:$

$\longmapsto { \boxed{ \frak{ \red{length = \: b + 20 = 15 + 20 }}}}$

$\:$

$\longmapsto \: { \boxed{ \frak{ \red{l = 35 \: cm}}}}$