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What is the area of a triangle (easier solution)

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by ganeshrkamath » Thu Sep 05, 2013 6:39 am
melguy wrote:Please help with the problem. Thanks
x=4_________(1)
y=5_________(2)
y = (-3/4)x + 20_________(3)

Put (1) in (3): y = -3 + 20 = 17
Point A = (4,17)

Put (2) in (3): 5 = (-3/4)x + 20
(3/4)x = 15
x = 20
Point B = (20,5)

Point C = (4,5)
Image
From figure, it is clear that the the triangle is a right angled triangle with the right angle at C.
BC = 20 - 4 = 16
AC = 17 - 5 = 12

Area = 1/2 * 16 * 12 = 96

Choose E

Cheers

PS : This might look like a long solution, but you will realize it's actually small when you work it out.
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by Brent@GMATPrepNow » Thu Sep 05, 2013 6:47 am
Original question:
Image




Let's first sketch the lines x = 4 and y = 5
Image

To find the point where y = (-3/4)x + 20 intersects the line x = 4, replace x with 4 to get: y = (-3/4)4 + 20 = 17
So the point of intersection is (4, 17)

To find the point where y = (-3/4)x + 20 intersects the line y = 5, replace y with 5 to get: 5 = (-3/4)x + 20
When we solve for x, we get x = 20
So the point of intersection is (20, 5)

Add this information to our sketch:
Image

From here, we can determine the length of the right triangle's base and height:
Image

Area = (1/2)(base)(height)
= (1/2)(16)(12)
= 96
= E

Cheers,
Brent
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by melguy » Thu Sep 05, 2013 7:11 am
Thanks both of you for your help!

Brent - you mentioned about posting the image within the question. I am not sure how can i do it?
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by Brent@GMATPrepNow » Thu Sep 05, 2013 7:43 am
melguy wrote:Thanks both of you for your help!

Brent - you mentioned about posting the image within the question. I am not sure how can i do it?
Hey melguy,

Use the "add image" option. This, I believe, makes the image viewable for those logged in and for those not.

So, how many more days until your test?

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by melguy » Thu Sep 05, 2013 7:47 am
Thanks!

My test is on 10th Sep :-)
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by Brent@GMATPrepNow » Thu Sep 05, 2013 7:55 am
I predict great things!

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by melguy » Thu Sep 05, 2013 8:05 am
Thanks a lot Brent :-)
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