Norman window

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vladmire: Master | Next Rank: 500 Posts; Posts: 132; Joined: Tue Oct 07, 2008 4:59 pm; Thanked: 4 times

by vladmire » Mon Jan 19, 2009 9:20 am

A norman window consists of a retangle surmounted by a semicircle. Find the area of the norman window. How much wood fram is needed to enclose the window?

L-6
W-4

I got the rectangle part of the proble...
A=LW
A=6*4=24
24+2Pi=30.28

How do I get the semicircle?

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Source: Beat The GMAT — Problem Solving | Mon Jan 19, 2009 9:20 am

logitech: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pmFollowed by: 25 members; Thanked: 237 times; GMAT Score:730

by logitech » Mon Jan 19, 2009 9:32 am

This should help

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LGTCH
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coffee5251: Senior | Next Rank: 100 Posts; Posts: 35; Joined: Wed Oct 29, 2008 5:14 am

Re: Norman window

by coffee5251 » Thu Jan 29, 2009 5:26 pm

vladmire wrote:A norman window consists of a retangle surmounted by a semicircle. Find the area of the norman window. How much wood fram is needed to enclose the window?

L-6
W-4

I got the rectangle part of the proble...
A=LW
A=6*4=24
24+2Pi=30.28

How do I get the semicircle?

Why do you add 2Pi to the area? How do you know what the radius of semicrircle is?

Also can you explain why you would find the hypotenuse of the triangle? How is that related to the window frame?

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umaa: Legendary Member; Posts: 727; Joined: Sun Jun 08, 2008 9:32 pmFollowed by: 1 member; Thanked: 8 times

by umaa » Thu Jan 29, 2009 6:55 pm

A rectangle SURMOUNTED by a semi circle. So, the semicircle is on top of the rectangle.

Area of norman window is, Area of rec + Area of S.circle.

Area of S. circle is, (pi * r^2)/2; r = l/2 = 6/2 = 3

= 6*4 + pi * 9/2

Total area = = 24 + 4.5pi

To find the wood frame needed to enclose the window is to add rectangle's perimeter and the length of the arc.

= peri. of rec + length of arc
= 2 (6+4) + 2*pi*3/2
= 20 + 3*pi

Guys, let me know the OA.

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krisraam: Master | Next Rank: 500 Posts; Posts: 221; Joined: Wed Jan 21, 2009 10:33 amFollowed by: 1 member; Thanked: 12 times

by krisraam » Thu Jan 29, 2009 8:27 pm

I think this is ambiguous.

Answer is either 20 + 3pi or 20 + 2Pi depending on where the semi circle is surmounted. If the semicircle over the length its 20+3pi. if its over the width its 20+ 2Pi.

Thanks
Raama

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