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# A triangle has side lengths of a, b, and c centimeters. Does each angl

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A triangle has side lengths of a, b, and c centimeters. Does each angl  [#permalink]

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19 Jul 2017, 14:04
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A triangle has side lengths of a, b, and c centimeters. Does each angle in the triangle measure less than 90 degrees?

(1) The 3 semicircles whose diameters are the sides of the triangle have areas that are equal to 3 $$cm^2$$, 4 $$cm^2$$, and 6 $$cm^2$$, respectively.

(2) c < a + b < c + 2
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A triangle has side lengths of a, b, and c centimeters. Does each angl  [#permalink]

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08 Jan 2018, 09:35
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A triangle has side lengths of a, b, and c centimeters. Does each angle in the triangle measure less than 90 degrees?

For one of the angles to be 90º, it has to happen that two squared sides are equal to the third squared side. Thus, if we can get the values of the 3 sides the question can be responded.

(1) The 3 semicircles whose diameters are the sides of the triangle have areas that are equal to 3 cm2cm2, 4 cm2cm2, and 6 cm2cm2, respectively.

In this case we get the area of the semicircles formed with each side as a diameter.

If we randomly take any of the triangle´s sides, just to try, we get that $$a^2 · pi · \frac{1}{2} = 3$$.

From there we can get the value of "a" knowing that $$pi = \frac{22}{7}$$ and the same for the other two sides, thus allowing us to determine whether or not two squared sides are equal to the third squared side (what is not necessary to calculate).

SUFF

(2) c < a + b < c + 2

The best way to prove this is to pick numbers.

First, we pick number that will make $$a^2+b^2 =c^2$$.

$$2 < 1 + 2 < 4$$ causes that $$1^2+2^2=5$$ which is different from $$2^2$$ so NO angle of 90º with this example.

Secondly, we pick numbers that will make $$a^2+b^2 =c^2$$.

We try Pithagorean triples first, but after trying 3, 4, 5 we realize that there is no way to fulfill the condition using integers. Thus, we try a Pithagorean triple with a non-integer as one side of the triangle, the 30-60-90 ratio:

$$2 < 1 + \sqrt{3} < 4$$ which we´ll inevitable have an angle of 90º since $$a^2+b^2 =c^2$$ is fulfilled.

It should be noticed that it would also work out using the Right Isosceles ratio (1, 1, $$\sqrt{2}$$).

INSUFF

AC: A

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##### General Discussion
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A triangle has side lengths of a, b, and c centimeters. Does each angl  [#permalink]

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19 Jul 2017, 15:20
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moonwalking wrote:
A triangle has side lengths of a, b, and c centimeters. Does each angle in the triangle measure less than 90 degrees?

(1) The 3 semicircles whose diameters are the sides of the triangle have areas that are equal to 3 $$cm^2$$, 4 $$cm^2$$, and 6 $$cm^2$$, respectively.

(2) c < a + b < c + 2

It should be A

Statement 1: Sufficient

With area of circles available, we can calculate diameter by using formula $$π*r^2$$ where $$r = \frac{d}{2}$$. With that we know the length of 3 sides, and subsequently the ratio of angles.

Statement 2: Insufficient

c < a + b < c + 2

Suppose c is small at 0.2, a is 0.6 and b is 0.7. a + b is still less than 2 but angle opposite the hypotenuse b is greater than 90.
Now suppose c is 0.4, a and b are both 0.5. They still satisfy the equation in 2 but the angles are below 90 degrees.

Image attached for illustration but it may not be a perfect proportional fit of the values listed. I would appreciate anyone providing a mathematical explanation for this.
Attachments

diagram4.jpg [ 15.08 KiB | Viewed 7868 times ]

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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl  [#permalink]

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19 Jul 2017, 15:56
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jedit wrote:
I would appreciate anyone providing a mathematical explanation for this.

I'm not sure if this is what you're looking for, but: if the sides of a triangle work in the Pythagorean Theorem, so a^2 + b^2 = c^2, then there is an angle of exactly 90 degrees between sides a and b. We can extend the Pythagorean Theorem. If a^2 + b^2 > c^2, then c is shorter than what you'd find in a right triangle, so the angle between a and b will be less than 90 degrees. And if a^2 + b^2 < c^2, then c is longer than what you'd have in a right triangle, so the angle between a and b is greater than 90 degrees.

You could use that here (going by memory, I think the OG solution does), but I like the approach you've used above - there's no real need to use any algebra.
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A triangle has side lengths of a, b, and c centimeters. Does each angl  [#permalink]

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30 Jan 2018, 00:51
jedit wrote:
moonwalking wrote:
A triangle has side lengths of a, b, and c centimeters. Does each angle in the triangle measure less than 90 degrees?

(1) The 3 semicircles whose diameters are the sides of the triangle have areas that are equal to 3 $$cm^2$$, 4 $$cm^2$$, and 6 $$cm^2$$, respectively.

(2) c < a + b < c + 2

It should be A

Statement 1: Sufficient

With area of circles available, we can calculate diameter by using formula $$π*r^2$$ where $$r = \frac{d}{2}$$. With that we know the length of 3 sides, and subsequently the ratio of angles.

Statement 2: Insufficient

c < a + b < c + 2

Suppose c is small at 0.2, a is 0.6 and b is 0.7. a + b is still less than 2 but angle opposite the hypotenuse b is greater than 90.
Now suppose c is 0.4, a and b are both 0.5. They still satisfy the equation in 2 but the angles are below 90 degrees.

Image attached for illustration but it may not be a perfect proportional fit of the values listed. I would appreciate anyone providing a mathematical explanation for this.

Hi Jedit,

Can you explain in detail the highlighted part. Is it a property? Thanks.
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl  [#permalink]

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20 Mar 2018, 22:24
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl  [#permalink]

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21 Mar 2018, 08:13
3
moonwalking wrote:
A triangle has side lengths of a, b, and c centimeters. Does each angle in the triangle measure less than 90 degrees?

(1) The 3 semicircles whose diameters are the sides of the triangle have areas that are equal to 3 $$cm^2$$, 4 $$cm^2$$, and 6 $$cm^2$$, respectively.

(2) c < a + b < c + 2

The question asks whether each angle is less than 90 degrees. The only thing most of us know about sides and measure of angles is that in a right triangle, a^2 + b^2 = c^2 where c is the side opposite to 90 degrees angle.

What happens when we have an obtuse triangle? Say the legs stay the same. In the obtuse triangle, the angle will increase and with it the side opposite this angle (c) will increase while a and b stay the same. So we can reason that in an obtuse triangle, a^2 + b^2 < c^2. Then in an acute triangle, a^2 + b^2 > c^2
Attachment:

drawingdividedobtuse_angle.png [ 5.8 KiB | Viewed 4500 times ]

(1) The 3 semicircles whose diameters are the sides of the triangle have areas that are equal to 3 $$cm^2$$, 4 $$cm^2$$, and 6 $$cm^2$$, respectively.

Area of a semicircle $$= (\pi/2)*r^2 = (\pi/8)*d^2$$

Each side is d, the diameter. The ratio of the areas gives the ratios of the square of the sides. Ignoring the constant since we can cancel them across the equation, we see that 3 + 4 > 6 so it must be an acute triangle. Sufficient.

(2) c < a + b < c + 2

c < a+b holds for all triangles. I am not sure what to do with c + 2. Let me go back to the right triangle and try one with sides 1, 1 and $$\sqrt{2}$$ (close to the values being discussed).
$$\sqrt{2} < 1 + 1 < \sqrt{2} + 2$$
Now if c is slightly greater than $$\sqrt{2}$$, the inequality will still hold but it will become an obtuse triangle.
Now if c is slightly less than $$\sqrt{2}$$, the inequality will still hold but it will become an acute triangle.
Not sufficient

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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl  [#permalink]

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21 Mar 2018, 08:32
Hi
For statement1: Since we can find the sides length as per info given, can we straight away say that this is sufficient, as it will make a unique triangle, and it is sufficient to answer the question asked.

VeritasPrepKarishma wrote:
moonwalking wrote:
A triangle has side lengths of a, b, and c centimeters. Does each angle in the triangle measure less than 90 degrees?

(1) The 3 semicircles whose diameters are the sides of the triangle have areas that are equal to 3 $$cm^2$$, 4 $$cm^2$$, and 6 $$cm^2$$, respectively.

(2) c < a + b < c + 2

The question asks whether each angle is less than 90 degrees. The only thing most of us know about sides and measure of angles is that in a right triangle, a^2 + b^2 = c^2 where c is the side opposite to 90 degrees angle.

What happens when we have an obtuse triangle? Say the legs stay the same. In the obtuse triangle, the angle will increase and with it the side opposite this angle (c) will increase while a and b stay the same. So we can reason that in an obtuse triangle, a^2 + b^2 < c^2. Then in an acute triangle, a^2 + b^2 > c^2
Attachment:
drawingdividedobtuse_angle.png

(1) The 3 semicircles whose diameters are the sides of the triangle have areas that are equal to 3 $$cm^2$$, 4 $$cm^2$$, and 6 $$cm^2$$, respectively.

Area of a semicircle $$= (\pi/2)*r^2 = (\pi/8)*d^2$$

Each side is d, the diameter. The ratio of the areas gives the ratios of the square of the sides. Ignoring the constant since we can cancel them across the equation, we see that 3 + 4 > 6 so it must be an acute triangle. Sufficient.

(2) c < a + b < c + 2

c < a+b holds for all triangles. I am not sure what to do with c + 2. Let me go back to the right triangle and try one with sides 1, 1 and $$\sqrt{2}$$ (close to the values being discussed).
$$\sqrt{2} < 1 + 1 < \sqrt{2} + 2$$
Now if c is slightly greater than $$\sqrt{2}$$, the inequality will still hold but it will become an obtuse triangle.
Now if c is slightly less than $$\sqrt{2}$$, the inequality will still hold but it will become an acute triangle.
Not sufficient

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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl  [#permalink]

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21 Mar 2018, 21:36
1
gmatbusters wrote:
Hi
For statement1: Since we can find the sides length as per info given, can we straight away say that this is sufficient, as it will make a unique triangle, and it is sufficient to answer the question asked.

Yes, as discussed by jedit in the first comment above, you can get the sides of the triangle which means you can draw a unique triangle and you will know whether it is acute or obtuse. That is all you need to figure out for a Sufficiency question.
My comment above discusses the conceptual approach.
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl  [#permalink]

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21 Mar 2018, 22:18
I did not understand this part
"Now if c is slightly greater than √2, the inequality will still hold but it will become an obtuse triangle.
Now if c is slightly less than √2, the inequality will still hold but it will become an acute triangle.
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl  [#permalink]

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22 Mar 2018, 00:02
1
I did not understand this part
"Now if c is slightly greater than √2, the inequality will still hold but it will become an obtuse triangle.
Now if c is slightly less than √2, the inequality will still hold but it will become an acute triangle.

Attachment:

drawingsmallobtuseangle.png [ 6.04 KiB | Viewed 4351 times ]

Say the right triangle $$1:1:\sqrt{2}$$ is as shown in figure with legs as base and the dotted line and the red line is the hypotenuse.
Say you rotate the dotted line a little to the left to make an angle slightly greater than 90 degrees. The legs are still 1 and 1. But now the green line is slightly greater than $$\sqrt{2}$$. We get an obtuse angle in which
c < a + b < c + 2
Slightly more than $$\sqrt{2}$$ < 1 + 1 < $$\sqrt{2}$$ + 2

You can do the same thing for an acute angle. Hence even if we know that c < a+b < c + 2, we still cannot say whether the triangle is acute or not. This relation could hold for both an acute angle and an obtuse angle. Hence not sufficient.
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl &nbs [#permalink] 22 Mar 2018, 00:02
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