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The area of a triangle having vertices at the points (k, k), (k + 4s,
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26 Nov 2024, 00:12
26 Nov 2024, 00:12
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Question Stats:
50% (00:37) correct
50% (02:26) wrong based on 6 sessions
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The area of a triangle having vertices at the points (k, k), (k + 4s, k) and (k, k + s) is 32
Quantity A |
Quantity B |
s |
4 |
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Re: The area of a triangle having vertices at the points (k, k), (k + 4s,
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28 Nov 2024, 02:29
28 Nov 2024, 02:29
1
given coordinates
(k,k) , (k+4s) and (k,k+s)
by looking at the coordinates we can understand that the point (k+4s) lies 4s units above or below the point (k,k) and (k,k+s) lies s units to the right or to the left of (k,k)
it is clear from the given coordinates that the given triangle is a right angled triangle with base length of s units and height of 4s units,
we know that the area of a triangle is given by 0.5 * B *H, where B=base length and H=height of the triangle
the given area of the triangle is 32 square units
so therefore 0.5 * s * 4s = 32, we get 2s^2 = 32
solving for s we get s =+4 or s = -4
the relationship cannot be determined, so option D is the right answer
(k,k) , (k+4s) and (k,k+s)
by looking at the coordinates we can understand that the point (k+4s) lies 4s units above or below the point (k,k) and (k,k+s) lies s units to the right or to the left of (k,k)
it is clear from the given coordinates that the given triangle is a right angled triangle with base length of s units and height of 4s units,
we know that the area of a triangle is given by 0.5 * B *H, where B=base length and H=height of the triangle
the given area of the triangle is 32 square units
so therefore 0.5 * s * 4s = 32, we get 2s^2 = 32
solving for s we get s =+4 or s = -4
the relationship cannot be determined, so option D is the right answer
gmatclubot
Re: The area of a triangle having vertices at the points (k, k), (k + 4s, [#permalink]
28 Nov 2024, 02:29
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