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Math Expert V
Joined: 02 Sep 2009
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What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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Difficulty:   15% (low)

Question Stats: 80% (01:16) correct 20% (01:22) wrong based on 345 sessions

### HideShow timer Statistics What is the area of triangle ABC above, with side lengths x, y, and z?

(1) $$(x+y)^2- (x-y)^2=80$$

(2) $$(x-y)= 1$$ This question was provided by Veritas Prep for the Game of Timers Competition Attachment: Untitled.png [ 5.28 KiB | Viewed 2605 times ]

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Re: What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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2
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
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Manager  B
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Re: What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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2
1) (x+y)²−(x−y)²=80
gives,
x²+y²+2xy-x²-y²+2xy=80
4xy=80
xy=20

Possible values of x,y can be 4,5 or 5,4

Voila!!
A is sufficient.

2. (x-y)=1
could be any value.
Naah... Insufficient.

So,
IMO A.

Posted from my mobile device
Intern  B
Joined: 15 Feb 2019
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What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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1
(X+y)^2 -(x-y)^2 equals 4xy
4XY =80
XY=20

Statement B is good for nothing
Hit Like and give a kudos if u like the solution

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What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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1
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.

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.
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Re: What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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1
Area of triangle = 1/2*x*y

Statement 1=(x+y)^2−(x−y)^2= (x^2+y^2+2xy)- (x^2+y^2-2xy)=4xy=80

4xy=80
1/2*x*y=10
Sufficient

Statment 2 x-y=1
We can't find x*y with the given information
Insufficient

A
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Re: What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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1
Area is dependent on XY , clearly we get XY from 1 using a2-b2 equals (a+b)(a-b), so we get ans from A , while 2 will give a equation with infinite possible ans

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Re: What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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1
I guess A is the answer for above problem.

Area= 1/2 *x*y

So we need xy.

Now simplifying equation 1
we know a^2 - b^2 = (a+b)(a-b)

So (x+y)^2−(x−y)^2=80

(x+y+ x−y)(x+y-x+y) = 80

2x*2y= 80
xy = 20

So statement 1 is sufficient to find the area.

STAT2:
(x−y)=1

we dont' know what is xy.No other information is provided.

Clearly ,STAT2 is insufficient.

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Re: What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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1
St 1- (x+y)^2- (x-y)^2=80
=> (x+y+x-y)(x+y-x+y)=80
=> 2x*2y=80
=.xy=20
Area= 1/2xy------ ------sufficient

St2- (x−y)=1
x=y+1
x can take any value depending o value of y.-----------not sufficient

Ans.A
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Re: What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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1
since triangle, ABC is a right-angled triangle right angle at B

area of triangle ABC = $$\frac{1}{2}$$xy

so we need to find the value of xy

statement (1) $$(x+y)^2 - (x-y)^2$$ = 80
$$x^2+y^2$$+2xy - $$x^2-y^2$$+2xy = 80
4xy =80
xy = $$\frac{80}{4}$$
so from here we can find the area of the triangle so SUFFICIENT

statement (2) (x−y) = 1
from this statement, we cannot get the definite values of x and y
hence we cannot get the difinite value of xy so this statement is INSUFFICIENT

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Re: What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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1
A is sufficient to answer considering the expansion we get xy = 20 and area is 1/2*xy and triplets in B option not sufficient to answer the question

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Re: What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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1
Area of triangle ABC = $$\frac{xy}{2}$$

1) Considering (x+y) = a and (x-y)=b, we have
a^2-b^2=80 => (a-b) (a+b)=80
=> xy = 20
Area of ABC = 10. Sufficient

2) Given, (x-y)=1
For x=5 and y=4, area of ABC=10
For x=6 and y=5, area of ABC=15

Not sufficient.

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GMAT 1: 640 Q45 V35 GMAT 2: 660 Q48 V33 Re: What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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1
i) Sufficient as when the equation is expanded gives 4xy=80; xy=20 and area of triangle is 1/2*leg1*leg2=1/2xy=10
ii)Insufficient as neither xy or individual values of x and y can be found from x-y=1

IMO A
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Re: What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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1
We have to find the area of the given triangle. So, indirectly we have to see whether we can find the value of XY from the given information.

(1) Solving the given equation will give us the value of XY. So one can easily find an area of the given triangle.

So, 1 can independently answer the question.

(2) Values of X, Y & Z can be fractions or integers. (We don't have enough information about them in the question)

Therefore (X, Y, Z) pair can be anything like (3, 4, 5) / (4, 5, 6.4) and so on.

So, 2 can not answer the question independently.

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GMAT 1: 760 Q50 V42 GRE 1: Q169 V168 Re: What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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1
My answer is (A). This is a relatively simple question.

In order to know the area of the right triangle, we just need to know the product of x and y.

From (a), we can tell 4xy = 80. That is sufficient.
From (b), x = y + 1. There are indefinite number of possibilities. Not sufficient.

So, we should choose (A).
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What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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1

Area = xy/2, so if we can find xy, we can determine the area.

Statement 1:
(X+y)^2 - (x-y)^2 = 80
x^2 + y^2 +2xy -x^2 - y^2 +2xy = 80
4xy = 80 => xy = 20
Therefore we can get value of xy and hence area.

So statement 1 is sufficient.

Statement 2: x-y=1.
We cannot determine xy using this.
So statement 2 is not sufficient.

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Originally posted by prashanths on 02 Jul 2019, 08:28.
Last edited by prashanths on 02 Jul 2019, 22:23, edited 2 times in total.
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Re: What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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1
IMO A

By the image and sign of the angle, we can find it's a right triangle.
So the area is 1/2*base*height = 1/2*y*x

St1: (x+y)^2−(x−y)^2=80 => 4xy = 80 [by simplifying (a+b)^2 and (a-b)^2 formula)
xy = 20, 1/2*x*y = 10
Sufficient

St2: (x−y)=1 => x=1+y => Area = 1/2*y*(1+y) => Not sufficient
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What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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#1
(X+y)^2 - (x-y)^2 = 80
x^2 + y^2 +2xy -x^2 - y^2 +2xy = 80
4xy = 80 => xy = 20
sufficient
#2
(x-y)=1
again x & y can be any integer value
insufficient

IMO A
What is the area of triangle ABC above, with side lengths x, y, and z?

(1) (x+y)2−(x−y)2=80(x+y)2−(x−y)2=80

(2) (x−y)=1

Originally posted by Archit3110 on 02 Jul 2019, 08:30.
Last edited by Archit3110 on 03 Jul 2019, 08:06, edited 1 time in total.
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Re: What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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What is the area of triangle ABC above, with side lengths x, y, and z?
means (1/2)xy =?

(1) (x+y)^2−(x−y)^2=80 ==> correct : 4xy = 80 => (1/2)xy =10

(2) (x−y)=1 ==> can't say (1/2)xy =?
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Re: What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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Area of triangle = 1/2*base*height = 1/2*y*x

(1) (x+y)^2−(x−y)^2=80
--> x^2 + y^2 + 2xy - (x^2 + y^2 - 2xy) = 80
--> x^2 + y^2 + 2xy - x^2 - y^2 + 2xy = 80
--> 4xy = 80
--> xy = 20

Area = 1/2*xy = 1/2*20 = 10

Sufficient

(2) (x−y)=1
--> x = y + 1

Area = 1/2*(y + 1)*y
--> Many values are possible

Insufficient

IMO Option A

Pls Hit Kudos if you like the solution Re: What is the area of triangle ABC above, with side lengths x,y, and z?   [#permalink] 02 Jul 2019, 08:32

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