Question

rin) A rectangle has the same area as another, whose length is 6 m more and breadth
is 4 m less. It has also the same area as the third, whose length is 8 m more and
breadth 5 m less. Find the length and breadth of the original rectangle.
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Answer 1

Answer:

length of original rectangle= 24 cm

breath of original rectangle=20 cm

Step-by-step explanation:

lb = (l+6) (b-4). = (l+8) ( b-5)

EQ 1 =EQ 2. = EQ 3

EQ 1 and 2

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

lb = lb-4l+ 6b-24

0 = -4l +6b -24

24= -4l +6b

12= 3b -2l. EQ a

EQ 1 and 3

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

lb=lb-5l+8b-40

0 = -5l +8b-40

40= -5l +8b. EQ b

By using elimination method :

12=3b-2l

40=8b-5l

by further calculations

b= 20

substituting the value of b in any equation

l=24

so the length of original rectangle =24cm

Breath of original rectangle = 20cm

Answer 2

❍ Let the length of original rectangle be l and the breadth of original rectangle be b.

$\\ \underline{\bigstar\boldsymbol{According \;to \;the \;given \;Question:}}$

• Length of rectangle is 6m more so it will be l + 6
• Breadth of rectangle is 4m less so it will be b - 4

$\underline{\dag\frak{As \;we \;know \;that:}}$

$\\ \star{\boxed{\pink{\sf{Area_{(rectangle)} = l\times b}}}}$

Applying the values,

$: \implies \sf{l\times b= (l + 6)(b - 4)} \\ \\ : \implies \sf{l \times b = lb - 4l + 6b - 24} \\ \\ \bf{so \: we \: get;} \\ \\ : \implies \sf{0 = 4l + 6b - 24} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \to \sf{eq(1)}$

$\\ \underline{\bigstar\boldsymbol{Again\;according \;to \;the \;given \;Question:}}$

• Area of original rectangle = Area of third rectangle
• Length is 8m more so it will be l + 8.
• Breadth is 5m less so it will be b - 5.

So,The equation will be

$\\ : \implies \sf{l \times b = (l + 8)(b - 5)} \\ \\ : \implies \sf{l \times b = lb - 5l + 8b - 40} \\ \\ \bf{so \: we \: get;}\\ \\ : \implies \sf{0 =5l - 8b + 40} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \to \sf{eq(2)}$

Now,we have simultaneous equations,

Let us solve it,

$\\ \sf{Eq(1) \times 5 = 0} \\ \\ : \implies \sf{(4l + 6b - 24) \times 5} = 0 \\ \\ : \implies \sf{20l + 30b - 120 = 0} \\ \\ \sf{Eq(2) \times 4 = 0} \\ \\ : \implies \sf{(5l - 8b + 40) \times 4 = 0} \\ \\ : \implies \sf{20l - 32b + 160 = 0}$

Subtracting eq(1) and eq(2),we get:

$\\ \: \: \: \sf{20l \: \: \cancel{ + 30b} \: \: \cancel{- 120} = 0} \\ \: \: \: \sf{20l \: \: \cancel{- 32b} \: \: \cancel{ + 160} = 0} \\ - - - - - - - - - -- \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ - 2b + 40 = 0} \\ - - - - - - - - - - - \\ \\ \sf{multiplying \: minus \: and \: plus \: in \: values} \\ \sf{we \: get:} \\ \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: 2b - 40 = 0} \\ \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 2b = 40} \\ \\ \sf{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: b = \frac{ \cancel{40}}{ \cancel{2}}} \\ \\ \sf{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: b = 20 \: m}$

Now,Subsituting the value of b,

$\\ \implies\sf{4l - 6b + 24 = 0} \\ \\ \implies \sf{4l - 6(20) + 24 = 0} \\ \\ \implies \sf{4l - 120 + 24 = 0} \\ \\ \implies \sf{4l - 96 = 0} \\ \\ \implies \sf{4l = 96} \\ \\ \implies \sf{l = \frac{96}{4}} \\ \\ \implies \sf{l = 24 \: m}$

• Therefore,The length of original rectangle is 24 m and the breadth of original rectangle is 20 m.