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## DS Section 16, #17

This topic has 3 member replies
jaheer Junior | Next Rank: 30 Posts
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15 Feb 2007
Posted:
11 messages

#### DS Section 16, #17

Mon Feb 19, 2007 6:38 am
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
17. 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.

I've attached the picture. It is a rectangle PQRS with Semi-circle on the top.

Jaheer

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Vasudha Senior | Next Rank: 100 Posts
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Sun Mar 11, 2007 7:07 pm
Hi

I dont see the attached image ..

jayhawk2001 Community Manager
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Sun Mar 11, 2007 8:20 pm
I can't find the image but given that we have the semicircle on top of
the rectange, lets assume

PS = QR = l
RS = QP = w

We should find l and w to solve this.

1 - insufficient. All we know is 2(l + w) = 28

2 - insufficient. All we know is l^2 + w^2 = 100. Cannot solve for l or w.

Using 1 and 2, we have 2 equations with 2 variables. So, a solution
should exist and hence is sufficient.

Sub l = 14 - w

196 + w^2 - 28w + w^2 = 100
w^2 - 14w + 48 = 0
w = 8 or 6

We are given that l > w and so we know what l is. Perimeter of the
flower-bed can hence be computed.

Vasudha Senior | Next Rank: 100 Posts
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Sun Mar 11, 2007 8:50 pm
Thank you, jayhawk2001.

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