MM91 wrote:
Bunuel wrote:
AntonioGalindo wrote:
Bunuel, can you provide a more detaliled solution to this question? Thank you!
Can you please tell me which part of the
solution here is not clear? I'll try to elaborate.
If you can please show all the passages, not clear yet for me the whole process.
Many thanks in advance!
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.
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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