Question

A rectangle has the same area as another, whose length is 6 m more and breadth
is 4m less . It has also the same area as the third, whose length 8m more and
breadth 5 m less. Find the length and breadth of the original rectangle.​

Answer 1

Let the length of the rectangle be 'l' and breadth be 'b'.

A rectangle has the same area as another whose length is 6 m more and breadth is 4m less.

We know that,

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

Length is 6 m more => (l + 6) m

Breadth is 4 m less => (b - 4) m

Now, substitute these values in above formula.

=> $\sf{l\:\times\:b\:=\:(l\:+\:6)(b\:-\:4)}$

=> $\sf{lb\:=\:lb\:-\:4l\:+\:6b\:-\:24}$

=> $\sf{0\:=\:-\:4l\:+\:6b\:-\:24}$

=> $\sf{4l\:-\:6b\:=\:-\:24}$

=> $\sf{2l\:-\:3b\:=\:-\:12}$ ---- [1]

Now again a condition is given that, if length of rectangle is 8m more and breadth is 5 m less.

Length is 8 m more => (l + 8) m

Breadth is 5 m less => (b - 5) m

According to question,

=> $\sf{lb\:=\:(l\:+\:8)(b\:-\:5)}$

=> $\sf{lb\:=\:lb\:-\:5l\:+\:8b\:-\:40}$

=> $\sf{5l\:-\:8b\:=\:-\:40}$ ---- [2]

Now, multiple eq [1] with 5 and [2] with 2

On multiplying we get,

=> $\sf{10l\:-\:15b\:=\:-\:60}$ ---- [3]

=> $\sf{10l\:-\:16b\:=\:-\:80}$ ---- [4]

Solve eq [3] & [4] by elimination method

$\sf{10l\:-\:16b\:=\:-\:80}$

$\sf{10l\:-\:15b\:=\:-\:60}$ (Change the signs)

$\rule{150}1$

$\sf{00l\:-\:01b\:=\:-20}$

$\rule{150}1$

=> $\sf{b\:=\:20\:m}$

Now substitute the value of b in eq [1]

=> $\sf{2l\:-\:3(20)\:=\:-\:12}$

=> $\sf{2l\:-\:60\:=\:-\:12}$

=> $\sf{2l\:=\:48}$

=> $\sf{l\:=\:24\:m}$

Therefore,

Length of original rectangle is 24 m.

Breadth of original rectangle is 20 m.

$\rule{100}1$ [ANSWER]

Verification:-

From above calculations we have -

• length of rectangle = l = 24 m
• breadth of rectangle = b = 20 m

Also we have the equation (1) is 2l - 3b = - 12

Substitute the value of l and b in above equation

=> 2(24) - 3(20) = - 12

=> 48 - 60 = - 12

=> - 12 = - 12

Answer 2

Solution:-

Let the length of the rectangle be l .

Let the breadth of the rectangle be b.

As we know that, the rectangle has the same area as another length is 6m more and 4m less.

As we know that,

[ Area of the rectangle = Length × Breadth ]

Length = (l + 6)m

Breadth = (b - 4)m

Putting the given values in the above formula.

=> lb = (l + 6)(b - 4)

=> lb = lb - 4l + 6b - 24

=> 0 = - 4l + 6b - 24

=> 4l - 6b = -24

=> 2l - 3b = -12 (Equation 1)

Another case,

If the length is 8m more and breadth is 5m less.

Length = (l + 8)m

Breadth = (b - 5)m

Putting the given values in the above formula.

=> lb = (l + 8)(b - 5)

=> lb = lb - 5l + 8b - 40

=> 5l - 8b = -40 (Equation 2)

Now, multiplying Equation 1 with 5 and Equation 2 with 2

On multiplying, we get new equations are:

10l - 15b = -60 (Equation 3)

10l - 16b = -80 (Equation 4)

Solve Equation 3 and 4 by elimination method.

10l - 15b = -60 (Equation 3)

10l - 16b = -80 (Equation 4)

Change the sign

=> 0l - b = - 20

=> - b = - 20

(-) ve sign cancelled both LHS and RHS

.°. b = 20

Now, put the value of b in equation 1

=> 2l - 3(20) = -12

=> 2l - 60 = -12

=> 2l = 48

.°. l = 24 m

Therefore,

Original Length = 24m

Original Breadth = 20m