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16 Apr 2017, 05:00
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
72% (01:14) correct
28%
(01:30)
wrong
based on 195
sessions
History
Date
Time
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Not Attempted Yet
Given the three points below, what is the area of the triangle formed by connecting them?
(0,3), (-3,-3), (3,-3)
(A) 4.5
(B) 9
(C) 18
(D) 36
(E) 72
(0,3), (-3,-3), (3,-3)
(A) 4.5
(B) 9
(C) 18
(D) 36
(E) 72
Updated on: 31 May 2021, 06:13
Originally posted by BrentGMATPrepNow on 15 Nov 2017, 12:45.
Last edited by BrentGMATPrepNow on 31 May 2021, 06:13, edited 1 time in total.
Last edited by BrentGMATPrepNow on 31 May 2021, 06:13, edited 1 time in total.
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saswata4s
Here's a sketch of the triangle:
From here, we can see that the base has length 6, and the height is 6
Area = (base)(height)/2
= (6)(6)/2
= 36/2
= 18
= C
Cheers,
Brent
General Discussion
16 Apr 2017, 10:50
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With the given co-ordinates find the distance between the two points using \(\sqrt{(x2-x1)^2 + (y2-y1)^2}\)
Distance between points (0,3) & (3,-3) is \(\sqrt{-3^2+-6^2}\)=\(\sqrt{45}\)
Distance between points (0,3) & (-3,-3) is \(\sqrt{3^2+-6^2}\) = \(\sqrt{45}\)
Distance between points (-3,-3) & (3,-3) is\(\sqrt{-6^2 +0}\)= 6
which is an isosceles triangle.
We need to find the height so that we can use the formula \(\frac{1}{2}\)*b*h
Height by Pythagoras theorem, h=\(\sqrt{45-6}\)= \(\sqrt{36}\)= 6
Area =\(\frac{1}{2}\)*6*6
=18
Distance between points (0,3) & (3,-3) is \(\sqrt{-3^2+-6^2}\)=\(\sqrt{45}\)
Distance between points (0,3) & (-3,-3) is \(\sqrt{3^2+-6^2}\) = \(\sqrt{45}\)
Distance between points (-3,-3) & (3,-3) is\(\sqrt{-6^2 +0}\)= 6
which is an isosceles triangle.
We need to find the height so that we can use the formula \(\frac{1}{2}\)*b*h
Height by Pythagoras theorem, h=\(\sqrt{45-6}\)= \(\sqrt{36}\)= 6
Area =\(\frac{1}{2}\)*6*6
=18
