Problem Solving

In the figure shown above, line segment QR has length 12, and rectangle MPQT is a square. If the area of rectangular region MPRS is 540, what is the area of rectangular region TQRS?

144
216
324
360
396

OA is B Please explain

Area of MPRS = Area of MPQT + Area of TQRS

540 = a*a + 12*a

(where 'a' is the side of the square MPQT or the breadth of rectangle MPRS)

a^2 + 12*a - 540 = 0

(a + 30)*(a - 18) = 0

a = 18 is the only positive root and hence the only possible value.

Area of TQRS = 18*12 = 216 sq. units

(B) is the answer

Thanks Aneesh, however how can you quickly go from a^2 + 12*a - 540 = 0 to (a + 30)*(a - 18) = 0 ? I spent a lot of time on this point.


Thanks.

Well, that largely comes from practice. So, I'd suggest you to solve a lot of quadratic equations.

Anyway, for this problem, let me try to explain my thought-process.

a^2 + 12*a - 540 = 0

I knew that I had to split 12*a into two parts such that they multiply to -540*a^2. Since -540*a^2 is a negative number, one of the two parts should be positive and one should be negative. Then I went on looking for two numbers that multiply to 540 and whose difference is 12.

I tried out these pairs which multiply to 540:

20 and 17, doesn't work because the difference is 3.
30 and 18, Voila! It works! So, the two parts must be 30*a - 12*a.

It might look like a long procedure, but when you've practiced hundreds of quadratic expressions, it is a 5-second thing. So, Practice!

Thanks Aneesh, However, I don't get when you take 20 and 17: in fact, they multiply to 340 instead of 540. Can you please clarify your steps? Thanks.

Correction:
I meant 20 and 27