Waste less time on Facebook — follow Brilliant.
×

anyone can solve this?

If the length of a rectangle increased by 12 and the width decreased by 8, the area is unchanged, the area is also unchanged if the original length is increased by 5 and the original width decreased by 4. Find the original dimensions of the rectangle.

Note by Haitham Wafa
4 years ago

No vote yet
5 votes

Comments

Sort by:

Top Newest

Let the length of the rectangle be l units, and its breadth be b units. Then the initial area of the rectangle will be lb square units. The length of the rectangle is increased by 12 units, so its latter length is (l+12) units. The width of the rectangle is decreased by 8 units, so its latter width is (b-8) units. Then the area of the rectangle will be (l+12)(b-8) square units. But the question states that this is equal to the previous area of the rectangle, which is lb square units.

Thus, equating...

(l+12)(b-8)= lb Or, l(b-8) + 12(b-8)= lb Or, lb - 8l + 12b -96= lb Or, 12b - 8l - 96= 0 (Note that the term lb appears both in the L.H.S and the R.H.S. So it can be cancelled out. Re-arranging the terms gives this) ............(1)

Again, the original length is increased by 5 units, so latter length= (l+5) units. The original width is deceased by 4 units, so latter width= (b-4) units. So, latter area= (l+5)(b-4) square units. According to the question, this should be equal to the previous area, which is lb square units. Thus, equating...

(l+5)(b-4)= lb Or, l(b-4) + 5(b-4)= lb Or, lb - 4l + 5b - 20 = lb Or, 5b - 4l - 20= 0 (Note that the term lb is present both in the L.H.S and in the R.H.S, so it can be cancelled out. Re-arranging the terms gives this) ........(2)

Now we have two simultaneous linear equations, which have to be solved for l and b. My approach is through elimination process.

Do 5(eq1) - 12(eq)2 to get the following.

5(12b - 8l - 96) - 12(5b - 4l - 20)= 50 - 120 Or, 512b - 58l - 596 - 125b + 124l + 1220 = 0 Or, 60b - 40l - 480 - 60b + 48l + 240 = 0 [multiplying] Or, (60b - 60b) + (48l - 40l) + (240-480)= 0 [arrangini] Or, 0 + 8l - 240= 0 [Note that 60b-60b= 0] Or, 8l= 240 [transposing 240] Or, l= 240/8 = 30

Now substituting this value of l in eq(2) gives the following.

5b - 4*30 - 20= 0 Or, 5b - 120 - 20= 0 Or, 5b - 140= 0 Or, 5b= 140 Or, b= 140/5= 28

So we get l= 30 and b= 28. Thus the length of the rectangle is 30 units, while its width is 28 units. Sreejato Bhattacharya · 4 years ago

Log in to reply

a rect with l= 30,b=28 satisfies the given conditions. Bhargav Das · 4 years ago

Log in to reply

@Bhargav Das Thank YOU GUYS but do i get to know the fully-worked solution like the steps and stuff?? Haitham Wafa · 4 years ago

Log in to reply

@Haitham Wafa full stuff is let the length is l

and the breadth is b

then according the first condition

we have ,

   (l+12) * (b-8) = l*b   ...1

and according to second condition

we have,

  (l+5) * (b-4) = l*b    ....2

now we have 2 equations you can solve by either method substitution , cross multiplication , elimination Mayank Kaushik · 4 years ago

Log in to reply

28*30 are the dimensions Why this question is here ?? Mayank Kaushik · 4 years ago

Log in to reply

let area a= lb = (l+12)(b-8)=lb+12b-8l-96. u get the eqn "8l-12b+96=0" similarly, lb=(l+5)(b-4)=lb-4l+5b-20 new eqn : "4l-5b+20=0" solving these two eqns u will get l=30 and b=28 Chinmay Gore · 4 years ago

Log in to reply

@Chinmay Gore When u omitted the lb in the equation and changed the signs of the other terms why is that?? Haitham Wafa · 4 years ago

Log in to reply

lb =(l+12)(b-8) lb=lb+12b-8l+96 12b-8l+96=0 and similarly get other linear equation and solve together Aryan Garg · 4 years ago

Log in to reply

\[\begin{cases} lb = (l+12)(b -8) \\ lb = (l + 5)(b - 4) \end{cases} \]Substitution. Parth Kohli · 4 years ago

Log in to reply

@Parth Kohli Thank YOU GUYS but do i get to know the fully-worked solution like the steps and stuff?? Haitham Wafa · 4 years ago

Log in to reply

the original dimensions of the rectangle are length=28 and width=11 Aravind Kukkila · 4 years ago

Log in to reply

@Aravind Kukkila That's not correct. \( 28 \times 11 = 308 \) but \( (28+12) \times (11-8) \) is \( 120 \). Tan Li Xuan · 4 years ago

Log in to reply

@Tan Li Xuan sorry i think i made a mistake in my calculations.the length is 30 but the width is correct. i made a wrong substitution before. Aravind Kukkila · 4 years ago

Log in to reply

@Aravind Kukkila That's even not correct. 30*11=330 and. (30+12) * (11-8). = 126 Vyom Chaturvedi · 4 years ago

Log in to reply

@Vyom Chaturvedi sorry Aravind Kukkila · 4 years ago

Log in to reply

@Aravind Kukkila Thank YOU GUYS but do i get to know the fully-worked solution like the steps and stuff?? Haitham Wafa · 4 years ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...