Question

23. If the length of a rectangle decreased by 5 units and breadth increased by
3 units then its area decreases 9 square units. Again if the length increased
by 3 units and breadth increased by 2 units its area increases 67 square units . find the measure of the rectangle.
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Answer 1

Gívєn :

• If the length of a rectangle decreased by 5 units and breadth increased by 3 units then its area decreases square units. Again if the length increased by 3 units and breadth increased by 2 units its area increases 67 square units . find the measure of the rectangle.

Tσ fínd :

• The measure of the rectangle.

Sσlutíσn :

• Let length and breadth of rectangle be x unit and y unit.              
• Area = xy    

             

According to the question,                  

⇒ (x - 5) (y + 3) = xy - 9                  

⇒ 3x - 5y - 6 = 0 ... (i)                  

⇒ (x + 3) (y + 2) = xy + 67                  

⇒ 2x - 3y – 61 = 0 ... (ii)                  

By cross multiplication, we get                  

⇒ x/305 - (-18) = y/-12 -(-183) = 1/9 - (-10)                  

⇒ x/323 = y/171 = 1/19                  

⇒ x = 17, y = 9        

         

• Length of the rectangle = 17 units.
• Breadth of the rectangle = 9 units.

Answer 2

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✺Answer:

♦️GiveN

• Firstly, Length decreased by 5 units and breadth increased by 3 units, Area decreased by 9 sq. units.

• Again, length increased by 3 units and breadth increased by 2 units, Area increased by 67 sq.units.

♦️To FinD

• Length and Breadth of Rectangle

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✺Explanation of Q.

• The above question deals with two variables that are length and breadth. When we alter the value of length and breadth, The area changes accordingly.
• Hence, According to question, Two equations with two variables(i.e. length and breadth) will be formed. Now we can solve it by using various methods, But Here I will use elimination method for clear and easy solution. You can also use methods of your choice(Substitution, Graphical or Cross multiplication)

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✺Solution

Let the length be l and breadth be b.

Then area of rectangle would be lb.

$\Large{\underline{\underline{\rm{\red{Case-1}}}}}$

Length is reduced by 5 units and breadth is increased by 3 units, the area of rectangle reduced by 9 units.

$\large{\rm{\rightarrow (l-5)(b+3)=lb-9}}\\ \\ \large{\rm{\rightarrow \: \cancel{lb}+3l-5b-15=\cancel{lb}-9}}\\ \\ \large{\rm{\rightarrow 3l-5b-15+9=0}}\\ \\ \large{\rm{\purple{\rightarrow 3l-5b-6=0...........(1)}}}$

$\Large{\rm{\underline{\underline{\red{Case-2}}}}}$

Length is increased by 3 units and breadth is increased by 2 units, the area of rectangle increased by 67 units.

$\large{\rm{\rightarrow \:(l+3)(b+2)=lb+67}}\\ \\ \large{\rm{\rightarrow \cancel{ lb}+2l+3b+6 = \cancel{lb}+67}}\\ \\ \large{\rm{\rightarrow \: 2l+3b+6-67=0}} \\ \\ \large{\rm{\purple{\rightarrow \: 2l+3b-61=0.............(2)}}}$

Now multiplying equation (1) with 2, and (2) with 3

$\large{\rm{\rightarrow \: 6l-10b-12=0}}\\ \\ \large{\rm{\purple{\rightarrow \: 6l-10b=12...........(3)}}}\\ \\ \large{\rm{\rightarrow \: 6l+9b-183=0}} \\ \\ \large{\rm{\purple{\rightarrow \: 6l+9b=183...........(4)}}}$

Subtracting equation 4 from equation 3

$\large{\rm{\rightarrow \: 6l-10b-(6l+9b)=12-183}}\\ \\ \large{\rm{\rightarrow \: \cancel{6l}-10b-\cancel{6l}-9b= -171}}\\ \\ \large{\rm{\rightarrow \: \cancel{-19}b=\cancel{-171}\: \: 9}}\\ \\ \large{\rm{\green{\rightarrow \: b=9 }}}$

Putting this value of b in equation (1)

$\large{\rm{\rightarrow \: 3l-5(9)=6}} \\ \\ \large{\rm{\rightarrow \: 3l-45=6}}\\ \\ \large{\rm{\rightarrow \: 3l=51}} \\ \\ \large{\rm{\green{\rightarrow \: l= \frac{51}{3}=17}}}$

Thus, the required length of recatangle is 17 units and required breadth of rectangle is 9 units.

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✺In the attachment

• A rectangle with length l and breadth b

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