The figure above shows the shape of a flower bed. If arc QR is a semic

Question

https://gmatclub.com/forum/download/file.php?id=21272 The figure above shows the shape of a flower bed. If arc QR is a semicircle and PQRS is a rectangle with QR > RS, what is the perimeter of the flower bed ? (1) The perimeter of rectangle PQRS is 28 feet. (2) Each diagonal of rectangle PQRS is 10 feet long. flower bed.png

MA · Accepted Answer

Given that QR>RS,
Suppose side QR=Length (L), and side RS = Breath (B)
From I, 
2(L+B)=28
L+B=14 ..................... (i)

from II,
L^2+B^2=H^2
L^2+B^2=10^2

add 2LBin both sides 

L^2+2LB+B^2=100+2LB
(L+B)^2=100+2LB ........... (ii)

putting valu of L+B in eq ii
14^2=100 +2LB
LB=48 
L=48/B .............................(iii)

Putting value of L in eq (i)
48/B+B=14
B^2-14B+48 = 0
B=6 and 8, 
A=6 and 8
but from question L>B,
L=8 and B=6.
Hence , OA is C.

GMATT73 · Answer

We can not assume this is a 6:8:10 triangle. There are infinite numbers of triangles with a hypotenuse of 10. B is insufficient.

A and B together confirms that it must be a 6:8:10. Longer side 8 is the diameter of the semicircle, which we can use to solve the length of the arc QR.

Answer must be C.

AKG1593 · Answer

Option C.
S1:l+b=14.
Insufficient as miltiple values are possible for length and breadth.
S2:l^2+b^2=100
Again insuff

Combining the two
We can find out l*b=48
And l+b=14
To get single value of l & b.After that we can find out the perimeter of the whole figure.

Posted from my mobile device

AntonioGalindo · Answer

Bunuel , can you provide a more detaliled solution to this question? Thank you!

Anki111 · Answer

in a rectangle, why can't we use 45-45-90 rule as the corners are 90 degree each to find the remaining two sides if hypotenuse is given? 
but then the the two sides will come equal which is not possible for a rectangle. anyone ?

Bunuel · Answer

A diagonal divides a rectangle into two 45-45-90 triangles only if the rectangle is a square. If a rectangle is NOT a square, then the diagonal can create any angle with an adjacent side, from 0 to 90, not inclusive. In this question we are told that, QR > RS, so PQRS is NOT a square so we don't get 45-45-90 triangles.

Bunuel · Answer

Can you please tell me which part of the  solution here  is not clear? I'll try to elaborate.

MM91 · Answer

If you can please show all the passages, not clear yet for me the whole process.

Many thanks in advance!

Bunuel · Answer

https://gmatclub.com/forum/download/file.php?id=21272

The figure above shows the shape of a flower bed. If arc QR is a semicircle and PQRS is a rectangle with QR > RS, what is the perimeter of the flower bed ?

Let the length of the rectangle be  a  and the width be  b . Then the perimeter will be PS + RS + QR + (length of arc QR)  = a + b + b + \pi*\frac{a}{2} . As you can see to get the perimeter of the figure we only need the width and the length of the rectangle. Notice that the length of the rectangle, will also be the diameter of the semicircle.

(1) The perimeter of rectangle PQRS is 28 feet.

2(a + b) = 28;
a + b = 14. 
Not sufficient, to get the values of  a  and  b .

(2) Each diagonal of rectangle PQRS is 10 feet long.

a^2 + b^2 = 10^2. 
Also, not sufficient, to get the values of  a  and  b .

(1)+(2) Squaring the equation from statement (1) gives us a^2 + 2ab + b^2 = 14^2. We can then substitute a^2 + b^2 = 10^2 into this equation to get 2ab + 10^2 = 14^2. Simplifying, we get ab = 48.

Since b = 14 - a from statement (1), we can substitute this expression into ab = 48 to get a(14 - a) = 48. This quadratic equation can be solved to find that a = 6 or a = 8. Since QR > RS, we know that a > b, so a must equal 8 and b must equal 6.

Therefore, we can calculate the perimeter of the flower bed as  = a + b + b + \pi*\frac{a}{2}= 8 + 6 + 6 + \pi*\frac{8}{2} = 20 + 4\pi  .

Answer: C.

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The figure above shows the shape of a flower bed. If arc QR is a semic

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