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Math Expert V
Joined: 02 Sep 2009
Posts: 64174
What is the area of triangle ABC above, with side lengths x,y, and z?  [#permalink]

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3
3 00:00

Difficulty:   15% (low)

Question Stats: 80% (01:15) correct 20% (01:23) wrong based on 367 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 3203 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
##### General Discussion
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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
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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, 07:18.
Last edited by BelalHossain046 on 02 Jul 2019, 07:53, edited 1 time in total.
DS Forum Moderator V
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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
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

Area of right angled triangle = xy/2

Statement 1 gives
(x+y)^2-(x-y)^2 = 80
(x^2+y^2+2xy) - (x^2+y^2-2xy) = 80
4xy = 80
xy/2 = 10 = Area of the triangle
Statement 1 Sufficient.

Statement 2 gives
x-y = 1
It will provide infinite possibilities of x, y & z
Statement 2 Insufficient.

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
Solution:

Question Stem analysis:

Triangle ABC is a right angled triangle making AB represented as x as it's height, and BC represented by y as it's base. Area of the triangle is 1/ X Base X Height Therefore we need the values of x & y to identify he area.

Statement one alone:

(x + y)^2 - (x - y)^2 = 80, when we solve this, we get 4xy = 80 & xy=20 Hence we can find out the area of the triangle which comes to 10
We can safely eliminate choices B,C & E.

Statement 2 alone:

(x - y)= 1
Clearly, this statement is not sufficient to find out the values of x & y, Thus statement two alone is not sufficient.
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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
Question: What is the area of triangle?

Observations:
The given image is a right angled triangle with the sides x,y,z. Consider, base = y, and height = x
Area of Triangle = (1/2)*base*height = (x*y)/2

Statement 1: (x+y)$$^2$$−(x−y)$$^2$$=80. Simplifying this,
x$$^2$$+y$$^2$$+2xy-x$$^2$$-y$$^2$$+2xy = 80
4xy = 80. From this we can easily find (x*y)/2 which is the area of the triangle.
Remember we don't need to find the exact value in DS problems, we just need to have one true answer and we have it.
Therefore Statement 1 is sufficient. ---> AD/BCE

Statement 2: (x−y)=1
In this case, x and y can take more than one single value and still make the statement true. In this case, we don't need to find the value of z even though it is easy because we don't need it.

For example,
when x=4, y=3 -> Area of triangle = 6
when x=5, y=4 -> Area of triangle = 10.

Therefore, Statement 2 is insufficient

The correct answer choice is A
Attachments Question.png [ 5.28 KiB | Viewed 627 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
DS Question (Geometry):
Attachments

File comment: Solution: A IMG_20190703_001913.jpg [ 1.91 MiB | Viewed 512 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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1
Area of a triangle = 1/2 * base * height

For given figure, area = 1/2 * y * x

Option 1:
(x + y)^2 - (x - y)^2 = 80
x^2 + y^2 + 2xy - x^2 - y^2 + 2xy = 80
4xy = 80
xy = 20

Therefore using the value of xy we can find area of given figure.
Hence option is sufficient.

Option 2:
(x - y) = 1
x = y + 1

There can be multiple right angled triangles formed where the base and height have a difference of just 1

From our knowledge of pythagorean triplets we know
3,4,5
20,21,29
are valid pythagorean triplets.

Hence Option 2 is not sufficient.

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Joined: 08 Jan 2018
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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 want to find the value of the area of right triangle ABC = $$\frac{1}{2}*x*y$$
So, knowing the values of both x and y (or) knowing the product of x and y will be sufficient.

(1) $$(x+y)^2 – (x-y)^2 = 80$$
=> $$x^2 + 2xy + y^2 – [x^2 – 2xy + y^2] = 80$$
=> $$x^2 + 2xy + y^2 – x^2 + 2xy – y^2 = 80$$
=> $$4xy = 80$$
=> $$xy = 20$$

At this point, we can say (1) is sufficient.

(2) (x-y) = 1
=> $$x = y + 1$$

Substituting the above value of x in the Pythagoras equation: $$x^2 + y^2 = z^2$$ we get
$$(y+1)^2 + y^2 = z^2$$
=> $$y^2 + 2y + 1 + y^2 = z^2$$
=> $$2y^2 + 2y + 1 = z^2$$

Now we have the following values:
$$y = y$$
$$x = y + 1$$
$$z^2 = 2y^2 + 2y + 1$$

Let, y=2
Therefore, $$x = y+1 = 3$$ ; $$x^2 = 9$$ ; $$y^2 = 4$$
$$z^2$$ = 2(4) + 2(2) + 1 = 13
Therefore, (2 , 3 , $$\sqrt{13}$$) is one possible triangle.

Let, y = 3
Therefore, $$x = 4$$ ; $$z^2 = 25$$
Thus, (3, 4, 5) is another possible triangle which satisfy the condition.
Hence, (2) 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
What is the area of triangle ABC above, with side lengths x, y, and z?

(1) $$(x+y)^2−(x−y)^2=80$$

(2) $$(x−y)=1$$

Question:(!/2)*x*y = ? i.e. x*y = ?

Statement 1: $$(x+y)^2−(x−y)^2=80$$
i.e. x^2 + y^2 + 2xy - x^2 - y^2 + 2xy = 4xy = 80
i.e. x*y = 40 hence area = 40/2 = 20
SUFFICIENT

Statement 2: $$(x−y)=1$$
x and y may be 4 and 3 or 5 and 4 respectively hence inconsistent value of area
NOT SUFFICIENT

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