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

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Re: The figure above shows the shape of a flower bed. If arc QR is a semic [#permalink]
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.

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Re: The figure above shows the shape of a flower bed. If arc QR is a semic [#permalink]
Bunuel, can you provide a more detaliled solution to this question? Thank you!
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Re: The figure above shows the shape of a flower bed. If arc QR is a semic [#permalink]
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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 ?
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Re: The figure above shows the shape of a flower bed. If arc QR is a semic [#permalink]
Anki111 wrote:
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 ?

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.
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Re: The figure above shows the shape of a flower bed. If arc QR is a semic [#permalink]
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.
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Re: The figure above shows the shape of a flower bed. If arc QR is a semic [#permalink]
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.

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Re: The figure above shows the shape of a flower bed. If arc QR is a semic [#permalink]
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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.

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 $$= a + b + b + \pi*\frac{a}{2}= 8 + 6 + 6 + \pi*\frac{8}{2} = 20 + 4\pi$$ .