Question

The area of a rectangle gets reduced by 67 square meters, when its length is increased by 3m and the breadth is decreased by 4m. If the length is reduced by 1m and breadth is increased by 4m, the area is increased by 89 square meters, Find the length of the rectangle.

Answer 1

✴$\mathcal{\large{\fbox{\pink{Easy\:Method\:To\:Solve}}}}$✴

$\mathfrak{\huge{\underline{\underline{\red{Question\:?}}}}}$

✰ ᴛʜᴇ ᴀʀᴇᴀ ᴏғ ᴀ ʀᴇᴄᴛᴀɴɢʟᴇ ɢᴇᴛs ʀᴇᴅᴜᴄᴇᴅ ʙʏ 67 sǫᴜᴀʀᴇ ᴍᴇᴛᴇʀs, ᴡʜᴇɴ ɪᴛs ʟᴇɴɢᴛʜ ɪs ɪɴᴄʀᴇᴀsᴇᴅ ʙʏ 3ᴍ ᴀɴᴅ ᴛʜᴇ ʙʀᴇᴀᴅᴛʜ ɪs ᴅᴇᴄʀᴇᴀsᴇᴅ ʙʏ 4ᴍ. ɪғ ᴛʜᴇ ʟᴇɴɢᴛʜ ɪs ʀᴇᴅᴜᴄᴇᴅ ʙʏ 1ᴍ ᴀɴᴅ ʙʀᴇᴀᴅᴛʜ ɪs ɪɴᴄʀᴇᴀsᴇᴅ ʙʏ 4ᴍ, ᴛʜᴇ ᴀʀᴇᴀ ɪs ɪɴᴄʀᴇᴀsᴇᴅ ʙʏ 89 sǫᴜᴀʀᴇ ᴍᴇᴛᴇʀs, ғɪɴᴅ ᴛʜᴇ ʟᴇɴɢᴛʜ ᴏғ ᴛʜᴇ ʀᴇᴄᴛᴀɴɢʟᴇ.

$\mathcal{\huge{\fbox{\green{AnSwEr:-}}}}$

➱ тнє łєηgтн σƒ тнє ʀєcтαηgłє ıs 28 м.

$\mathcal{\huge{\fbox{\purple{Solution:-}}}}$

✴ ʟᴇᴛ ᴛʜᴇ ʟᴇɴɢᴛʜ ᴀɴᴅ ᴛʜᴇ ʙʀᴇᴀᴅᴛʜ ᴏғ ᴛʜᴇ ʀᴇᴄᴛᴀɴɢʟᴇ ʙᴇ x ᴍ ᴀɴᴅ ʏ ᴍ, ʀᴇsᴘᴇᴄᴛɪᴠᴇʟʏ.

ᴀᴄᴄᴏʀᴅɪɴɢ ᴛᴏ ᴛʜᴇ ǫᴜᴇsᴛɪᴏɴ,

ᴄᴀsᴇ 1 :-

✯ ʜᴇʀᴇ, ʟᴇɴɢᴛʜ ɪs ɪɴᴄʀᴇᴀsᴇᴅ ʙʏ 3ᴍ ᴀɴᴅ ᴛʜᴇ ʙʀᴇᴀᴅᴛʜ ɪs ᴅᴇᴄʀᴇᴀsᴇᴅ ʙʏ 4ᴍ .

⟿ xʏ- (x +3) (ʏ - 4) = 67

⟿ xʏ - xʏ + 4x - 3ʏ + 12 = 67

⟿ 4x - 3ʏ = 55...........(1)

ᴄᴀsᴇ 2 :-

✯ ʜᴇʀᴇ, ʟᴇɴɢᴛʜ ɪs ʀᴇᴅᴜᴄᴇᴅ ʙʏ 1ᴍ ᴀɴᴅ ʙʀᴇᴀᴅᴛʜ ɪs ɪɴᴄʀᴇᴀsᴇᴅ ʙʏ 4ᴍ .

⟿ (x-1) (ʏ + 4) - xʏ = 89

⟿ xʏ + 4x - ʏ - 4 - xʏ = 89

⟿ 4x - ʏ = 93.............(2)

✭sᴜʙᴛʀᴀᴄᴛɪɴɢ (ɪ) ᴀɴᴅ (ɪɪ), ᴡᴇ ɢᴇᴛ :-

4x - 3ʏ = 55...........(1)

4x - ʏ = 93.............(2)

________________

2ʏ = 38

_______________

⟿ 2ʏ = 38

➥ ʏ= 19

✭ ᴏɴ sᴜʙsᴛɪᴛᴜᴛɪɴɢ ʏ = 19 ɪɴ (ɪɪ), ᴡᴇ ʜᴀᴠᴇ

⟿ 4x - 19 =93

⟿ 4x = 93 + 19

⟿ 4x = 112

⟿ x = 112/4

➥ x = 28

✵ ʜᴇɴᴄᴇ, ᴛʜᴇ ʟᴇɴɢᴛʜ(ʟ) 28ᴍ ᴀɴᴅ ᴛʜᴇ ʙʀᴇᴀᴅᴛʜ (ʙ) = 19ᴍ.

______________________________________

Answer 2

Given :-

The area of a rectangle gets reduced by 67 m² when it's length is increased by 3 meters and the breadth is decreased by 4 meters .

If the length is reduced by 1 meter and breadth is increased by 4 meters , the area is increased by 89 m²

Required to find :-

• Length of the rectangle ?

Solution :-

Let ,

The length of the rectangle be x

Breadth of the rectangle be y

Area of the rectangle ( A ) = length x breadth

=> ( x ) ( y )

=> xy

Now,

Consider the 1st condition ;

According to which ,

The area of a rectangle gets reduced by 67 m² when it's length is increased by 3 meters and the breadth is decreased by 4 meters .

So,

This implies ;

Length of the rectangle = x + 3 meters

Breadth of the rectangle = y - 4 meters

Area of the rectangle =

( x + 3 ) ( y - 4 ) = A - 67 m²

Here A denotes area of the rectangle ;

x ( y - 4 ) + 3 ( y - 4 ) = A - 67

xy - 4x +3y - 12 = A - 67

xy - 4x + 3y = A - 67 + 12

xy - 4x + 3y = A - 55

Since,

A = xy

xy - 4x + 3y = xy - 55

xy - 4x + 3y - xy + 55 = 0

xy , - xy get's cancelled

- 4x + 3y + 55 = 00

- 4x + 3y = - 55

Take minus as common ;

- ( 4x - 3y ) = - ( 55 )

minus get's cancelled

4x - 3y = 55 $\longrightarrow{\tt{\bf{ Equation - 1 }}}$

consider this as equation 1

Now,

Consider the 2nd condition ;

If the length is reduced by 1 meter and breadth is increased by 4 meters , the area is increased by 89 m²

So,

This implies ;

Length of the rectangle = x - 1 meters

Breadth of the rectangle = y + 4 meters

Area of the rectangle =

( x - 1 ) ( y + 4 ) = A + 89 m²

x ( y + 4 ) - 1 ( y + 4 ) = A + 89

xy + 4x - y - 4 = A + 89

xy + 4x - y = A + 89 + 4

xy + 4x - y = A + 93

Since,

A = xy

xy + 4x - y = xy + 93

xy + 4x - y - xy = 93

4x - y = 93 $\longrightarrow{\tt{\bf{ Equation - 2 }}}$

consider this as equation 2

Now,

Let's solve these 2 equations simultaneously ;

Method used - Elimination method

Subtract equation 1 from equation 2

$\tt{4x - \: \: \: y = 93} \\ \tt{ 4x - 3y = 55} \\ \underline{ ( - )( + ) \: \: \: ( - ) \: \: \: \: } \\. \tt\underline{ \: \: \: \: \: \: \: \: \: \: \: 2y = 38 \: \: } \\ \\ \tt \implies y = \frac{38}{2} \\ \\ \implies \tt y = 19$

Now,

Substitute the value of y in equation 2

4x - y = 93

4x - 19 = 93

4x = 93 + 19

4x = 112

x = 112/4

x = 28

Hence,

• Value of x = 28 meters

• Value of y = 19 meters

Therefore ,

The length of the rectangle ( x ) = 28 meters