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Isoceles triangle

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Isoceles triangle

by jai123oct » Sat Apr 14, 2012 12:07 pm
Hey Guys need help !!! i gave a kaplan mock test today and encountered a problem with triangle :

the problem says :

Area of isosceles triangle is given by equation : 2x^2 + 2x + 1/2 ; what would be the perimeter
and then the options in the answer set goes by as i remember factors of x such as
1) (2x+1)(2x+2) and so on the next 3 choices

But the real thing is how to solve this i could not get how to approach to this problem

any suggestion !!!

thanks

jai.
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by ADN1988 » Sat Apr 14, 2012 5:00 pm
Hi,

Can you give us the answer choices please?

Is there any info about the angles?

If it is a right triangle:

Area = 2x^2 + 2x + 1/2

(sxs)/2 = 2x^2 + 2x + 1/2

s^2 = 4x^2 + 4x + 1

s^2 = (2x+1)^2

s = 2x+1

After this you have 2 sides, just get the third and you have the perimeter.
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by seal4913 » Sat Apr 14, 2012 5:09 pm
jai123oct wrote:Hey Guys need help !!! i gave a kaplan mock test today and encountered a problem with triangle :

the problem says :

Area of isosceles triangle is given by equation : 2x^2 + 2x + 1/2 ; what would be the perimeter
and then the options in the answer set goes by as i remember factors of x such as
1) (2x+1)(2x+2) and so on the next 3 choices

But the real thing is how to solve this i could not get how to approach to this problem

any suggestion !!!

thanks

jai.
Area = .5 * b * h; 2A = b*h; area is 2x^2 + 2x +.5 so time that by 2 and it's 4x^2 + 4x + 1 so is equal to (2x^2 + 1)^2.

So base and height is equal to that use a^2 + b^2 = c^2 to hypth side of the triangle. Which is [(2x +1)/2]^2 + (2x + 1)^2 = c^2. You get [(sqrt of 5)/2] * 2x + 1. So perm. is 2 * [(sqrt of 5)/2] + 2x + 1 which is equal to [(sqrt5)/2 + 1)(2x+1)
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by jai123oct » Sun Apr 15, 2012 10:37 am
hey seal thanks ;
but the question says its an isosceles triangle ; no info about angles given ; it just goes on with a quadratic equation as mentioned below .

sorry i dont remember the answer choices ; but i liked your approach

thanks

jai.
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by seal4913 » Sun Apr 15, 2012 11:49 am
You don't need angles.

You got the base and the height from the area.

So the height divides the triangle is two and that's where you use pyth. them. So you use the height squared + .5 the base squared and you get the sides of the triangle. So you have all the sides now.

The answer from the a^2 + b^2 = c^2 gives you to the equal sides. And you have the base. Add them all together and you got your answer
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by ADN1988 » Sun Apr 15, 2012 12:31 pm
If you divide the triangle in two, the height isn't the same that you got before. So how do you get the new height?
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by seal4913 » Sun Apr 15, 2012 12:43 pm
ADN1988 wrote:If you divide the triangle in two, the height isn't the same that you got before. So how do you get the new height?
I don't know how you think you height changes.

Draw a triangle. This is your triangle. You know the height and the base is (2x^2 + 1)^2.


So having the height and the base you know the height divides the triangle equally into two. So the height is (2x^2 + 1)^2 and the base is half of that. Use pythm to get the outter side. So now you have the two outter sides and the base and that means you can get your answer.
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by ADN1988 » Sun Apr 15, 2012 1:06 pm
To do that, aren't you supposing that the height is equal to the base?
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by seal4913 » Sun Apr 15, 2012 3:10 pm
ADN1988 wrote:To do that, aren't you supposing that the height is equal to the base?
Not assuming...

base x height = 2x^2 + 1 * 2x^2 + 1 so both the base and height are the same.
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by ADN1988 » Mon Apr 16, 2012 2:03 am
The area is 2x^2 + 2x + 1/2

If h = 2 and b = 2x^2 + 2x + 1/2 -> area = [2 * (2x^2 + 2x + 1/2)]/2 = 2x^2 + 2x + 1/2

In this case the height and the base can be different and the area is the same as given.

I don't really understand how the height and the base must be equal.

It is true if the triangle is 45 - 45 - 90, so the height and the base are the equal sides.
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by seal4913 » Mon Apr 16, 2012 3:02 am
ADN1988 wrote:The area is 2x^2 + 2x + 1/2

If h = 2 and b = 2x^2 + 2x + 1/2 -> area = [2 * (2x^2 + 2x + 1/2)]/2 = 2x^2 + 2x + 1/2

In this case the height and the base can be different and the area is the same as given.

I don't really understand how the height and the base must be equal.

It is true if the triangle is 45 - 45 - 90, so the height and the base are the equal sides.
Only way to get a 30-60-90 is if the triangle is an eqaualateral
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by ADN1988 » Mon Apr 16, 2012 4:51 am
seal4913 wrote:Only way to get a 30-60-90 is if the triangle is an eqaualateral
That doesn't mean that an isosceles triangle is always 45-45-90.

A triangle that has 2 sides = 7 inch and another = 6 inch is isosceles and it's not 45-45-90.

(6/2)^2 + Height^2 = 7^2

Height^2 = 40

Height = 40^(1/2) that is different from 7.

If you split this triangle in two, the sides of one of the new triangles are 7, 3 and 40^(1/2), which is neither isosceles nor 45-45-90.

A isosceles triangle can have a height different from the base.
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by cindy79 » Sun Apr 22, 2012 11:39 pm
2x^2+2x+1/2= (1/2)*（4x+1)^2 = (1/2)sh=area
which means the s=h and thus the triangle must be a Isosceles triangle.
then, you can find the answer.
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