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02 Jul 2019, 08:00
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A
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Difficulty:
15%
(low)
Question Stats:
79% (01:18) correct
21%
(01:29)
wrong
based on 483
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What is the area of triangle ABC above, with side lengths x, y, and z?
(1) \((x+y)^2- (x-y)^2=80\)
(2) \((x-y)= 1\)
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02 Jul 2019, 08:20
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the question is asking about the value of \(\frac{x*y}{2}\)
from statement (1), 4xy = 80, xy = 20, \(\frac{x*y}{2}\) = 10 --> sufficient
from statement (2), x is taller than y by 1 unit, but no constrains about the values of each --> insufficient
A
from statement (1), 4xy = 80, xy = 20, \(\frac{x*y}{2}\) = 10 --> sufficient
from statement (2), x is taller than y by 1 unit, but no constrains about the values of each --> insufficient
A
General Discussion
Updated on: 02 Jul 2019, 08:53
Originally posted by BelalHossain046 on 02 Jul 2019, 08:18.
Last edited by BelalHossain046 on 02 Jul 2019, 08:53, edited 1 time in total.
Last edited by BelalHossain046 on 02 Jul 2019, 08:53, edited 1 time in total.
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Question: (1/2) xy=?
Here, (1/2) is fixed numerical term.
So, if we find the value of xy, we can can easily find the value of (1/2) xy i.e. AREA of the triangle.
SO, WE SIMPLY NEED the value of xy.
Thus, question becomes xy=?
Statement 1:
(x+y)^2- (x-y)^2=80
=> x^2+2xy+y^2-x^2+2xy-y^2=80
=> 4xy=80
=> xy=20
SUFFICIENT
Statement 2: x-y= 1
Case 1: if x=2 & y=1, then xy=2,
Case 2: if x=5 & y=4, then xy=20,
Not SUFFICIENT.
Answer:A
Here, (1/2) is fixed numerical term.
So, if we find the value of xy, we can can easily find the value of (1/2) xy i.e. AREA of the triangle.
SO, WE SIMPLY NEED the value of xy.
Thus, question becomes xy=?
Statement 1:
(x+y)^2- (x-y)^2=80
=> x^2+2xy+y^2-x^2+2xy-y^2=80
=> 4xy=80
=> xy=20
SUFFICIENT
Statement 2: x-y= 1
Case 1: if x=2 & y=1, then xy=2,
Case 2: if x=5 & y=4, then xy=20,
Not SUFFICIENT.
Answer:A
