Question

perimeter of a rectangle is 480 if it's length is increased by 10% and breadth is decreased by 20% we get the same parameter find the length and breadth of rectangle. please fast I'll mark u as brainlist please with full explained​

Answer 1

Let the length of the rectangle be L  cm and breadth be B cm

Given :

Perimeter of the rectangle = 480 cm

⇒ 2( L + B ) = 480

⇒ L + B = 480 / 2

⇒ L + B = 240

⇒ L = 240 - B → ( 1 )

According to the question :

If it's length is increased by 10% and breadth is decreased by 20% we get the same perimeter

Length of the newly formed rectangle = increased by 10 % = L + ( 10/100 × L ) = L + ( L/10 ) = 11L/10

Given : P

Breadth of the newly formed rectangle = decreased by 20 % = B - ( 20/100 × B ) = B - ( 2B/10 ) = 8B/10

Given : Perimeter of the newly formed rectangle remains the same

⇒ Perimeter of the newly formed rectangle = 480 cm

⇒ 2( Length + Breadth ) = 480  

⇒ 2( 11L/10 + 8B/10 ) = 480

⇒ ( 11L + 8B ) / 10 = 480/2

⇒ 11L + 8B = 240 × 10

⇒ 11L + 12B = 2400

Substituting Eq( 1 ) in the above equation we get,

⇒ 11( 240 - B ) + 8B = 2400

⇒ 2640 - 11B + 8B = 2400

⇒ 2640 - 3B = 2400

⇒ 2640 - 2400 = 3B

⇒ 240 = 3B

⇒ 240/3 = B

⇒ B = 80

Substituting the value of ' B ' in ( 1 )

⇒ L = 240 - B

⇒ L = 240 - 160

⇒ L = 160

∴ the length of the rectangle is 160 cm and breadth of the rectangle is 80 cm.

Answer 2

$\huge\sf\pink{Answer}$

☞ Length = 160

☞ Breadth = 80

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ Perimeter of a rectangle = 480

✭ The length is increased by 10% and the breadth is decreased by 20% we get the same perimeter

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

☆ Length and breadth of the rectangle?

$\rule{110}1$

$\huge\sf\purple{Steps}$

We know that the perimeter of a rectangle can be found by,

$\underline{\boxed{\green{\sf Perimeter = 2(l+b)}}}$

So as per the given information,

➢ $\sf 2(l+b) = 480$

➢ $\sf l+b = \dfrac{480}{2}$

➢ $\sf l = 240 - b \qquad\frac{\quad}{} eq(1)$

So the Length after the 10% increase,

➼ $\sf L +\bigg \lgroup \dfrac{10}{100} × L\bigg \rgroup$

➼ $\sf L+ \lgroup \dfrac{L}{10}\rgroup$

➼ $\sf \dfrac{11L}{10}$

Similarly new breadth,

➜ $\sf \dfrac{8B}{10}$

So the new perimeter is the same as the old one (480)

➠ $\sf 2 \bigg\lgroup \dfrac{11L}{10} + \dfrac{8B}{10}\bigg\rgroup= 480$

➠ $\sf \dfrac{11L+8B}{10} = \dfrac{480}{2}$

➠ $\sf 11L + 8B = 240 × 10$

➠ $\sf 11L + 12B = 2400$

Substituting in eq(1)

➝ $\sf 11\bigg\lgroup240-B\bigg\rgroup+8B = 2400$

➝ $\sf 2640 - 11B + 8B = 2400$

➝ $\sf 2640 -3B = 2400$

➝ $\sf 2640 -2400 = 3B$

➝ $\sf \dfrac{240}{3} = B$

➝ $\sf\color{aqua}{B = 80}$

Substituting the value of B in eq(1)

➳ $\sf L = 240 - B$

➳ $\sf L = 240 - 160$

➳ $\sf\color{lime}{L = 160}$

$\rule{170}3$