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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] New post   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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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink] New post   21 Mar 2018, 08:13
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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



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
drawingdividedobtuse_angle.png [ 5.8 KiB | Viewed 62232 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

Answer (A)
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink] New post   21 Jan 2019, 11:50
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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


Target question: Does each angle in the triangle measure less than 90 degrees?

Given: A triangle has side lengths of a, b, and c centimeters

Statement 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.
This statement illustrates an important rule concerning Data Sufficiency questions: DO NOT DO MORE WORK THAN IS NECESSARY
Since we have the area for each circle, we COULD calculate the diameter of each circle.
So, after some tedious calculations, we WOULD know the lengths of all 3 sides of the triangle.
IF we know the lengths of all 3 sides of the triangle, we COULD draw the triangle, and we COULD then measure each angle with a protractor.
This means we COULD definitely determine whether each angle is less than 90 degrees.
In other words, we COULD answer the target question with certainty.
This means statement 1 is SUFFICIENT

Statement 2: c < a + b < c + 2
There are several values of a, b and c that satisfy statement 2. Here are two:
Case a: a = 1, b = 1 and c = 1. In this case, we have an EQUILATERAL triangle in which all three angles measure 60 degrees. So, the answer to the target question is YES, each angle in the triangle measures less than 90 degrees
Case b: a = 1.5, b = 2 and c = 2.5.
ASIDE: We know that a 3-4-5 triangle is a RIGHT triangle.
If we make each side HALF as long, we get a 1.5-2-2.5 triangle, which is also a RIGHT triangle.
In this case, we have a RIGHT triangle which means one angle EQUALS 90 degrees. So, the answer to the target question is NO, it is NOT the case that each angle in the triangle measures less than 90 degrees
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

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Brent
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A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink] New post   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.
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink] New post   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] New post   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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A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink] New post   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] New post   21 Mar 2018, 08:32
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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.

Please clarify.


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

Answer (A)
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink] New post   21 Mar 2018, 21:36
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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.

Please clarify.



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] New post   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.
Not sufficient" VeritasPrepKarishma. please clarify.
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink] New post   22 Mar 2018, 00:02
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sadikabid27 wrote:
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.
Not sufficient" VeritasPrepKarishma. please clarify.


Attachment:
drawingsmallobtuseangle.png
drawingsmallobtuseangle.png [ 6.04 KiB | Viewed 56419 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 [#permalink] New post   21 Jan 2019, 13:11
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



we know that the sum of the " angles is 180° if one of the angles is not 90° than all the angles are less then 90 °.
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink] New post   21 Jan 2019, 14:41
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Amelias wrote:

we know that the sum of the " angles is 180° if one of the angles is not 90° than all the angles are less then 90 °.


I'm not sure if this is what you meant to write, but it's not true.

How about 100°, 20° and 60°?

Cheers,
Brent
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink] New post   28 Apr 2019, 08:33
I think main clue to this problem is quite simple.

Even if you draw a microscopic right triangle, rule a^2+b^2=c^2 is still appliable for it.
When you change sides of this potential micro-triangle, you will see that equation becomes inequation:
1) a^2+b^2<c^2 for obtuse triangle
2) a^2+b^2>c^2 for acute triangle

But when "c+2" enters the game, it totally ruins everything.
The inequations above still holds true for a very big values of a,b and c
But remain wrong for all small values of a,b and c (in the potential micro-triangle)

Final point: what the questions asks for? "Does each angle in the triangle measure less than 90 degrees?".
So, it does not ask about range of values for a,b and c
So, insufficient
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink] New post   28 Jun 2019, 07:04
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 cm2cm2, 4 cm2cm2, and 6 cm2cm2, respectively.

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

Note: i will solve this question in the simplest way possible without getting into too much details.Remember, this is a data sufficiency question so you don't really need to solve the whole question and waste time to know if the data is sufficient or not, you simply need to check if the data given is enough to solve the problem.

Answer: we have to find whether the triangle abc is right or not. to find that we should check if a^2+b^2=c^2

let us discuss statement 1:

statement 1 gives us the areas of the semi circles who diameter are the sides of the triangle, we all know that area of is A=πr^2

step1: since the area is given and π value is already known as 3.14, we can solve and get r for each circle.

step2: know that we got r for each circle we can multiply it by 2 and get the diameter of each circle noting that according to the given the sides of the triangle are the diameters of the circles.

step 3: now that we got all the sides length we can solve a^2+b^2=c^2 and see that the statement is sufficient

know let us discuss statement 2: c < a + b < c + 2

it doesnt really tell us anything about a^2, b^2 and c^2 so it is not sufficient.

the answer is (A)
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink] New post   06 Nov 2020, 06:54
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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 cm2cm2, 4 cm2cm2, and 6 cm2cm2, respectively.

This statement tells us that 3 semicircle have areas that are equal to 3 cm, 4 cm, and 6cm. With this information, we can conclude with certainty that we will get a yes/no answer. SUFFICIENT.

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

Lets test some values:

a = b = c = 1
\(1 < 2 < 3\); each angle measures less than 90 degrees since all sides are equal

c = \(\sqrt{2}\), a = b = 1
\(\sqrt{2}\) < 2 < \(2 + \sqrt{2}\); we have a right triangle.

INSUFFICIENT.

Answer is A.
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink] New post   09 Apr 2021, 16:39
@
VeritasKarishma 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

Answer (A)


Can someone explain this: 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.

What I understand is we can derive the sides from statement 1. That's clear. Then using the values of the sides we can look to see how the sides are related by squaring the sides

a^2 + b^2 < c^2 (obtuse)
a^2 + b^2 > c^2 (acute)

Is this the takeaway here?
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink] New post   19 May 2021, 06:55
pretty easy if you can remember these. In a triangle :

if \(a^2+b^2=c^2\) (all angles =90)
if \(a^2+b^2>c^2\) (all angles <90)
if \(a^2+b^2<c^2\) (all angles are >90)

although, I did not completely solve the question, but it seems from the first we can get each side of the triangle, that is, a,b,c (each side = diameter of the semi-circle). so, it definitely looks sufficient.

B> just looks funny. I wouldn't know how to proceed on B.
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink] New post   11 Jun 2021, 16:40
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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

Solution:

We need to determine whether each angle of a triangle whose side lengths are a, b and c centimeters measures less than 90 degrees. That is, we need to determine whether the triangle is an acute triangle. Notice that if c is the longest side of the triangle, then the triangle is an acute triangle if and only if a^2 + b^2 > c^2. Therefore, we need to determine whether a^2 + b^2 > c^2.

Statement One Alone:

If we let a < b < c, from statement one, we see that π(a/2)^2 / 2 = 3, π(b/2)^2 / 2 = 4 and π(c/2)^2 / 2 = 6. Let’s simplify the first equation:

π(a/2)^2 = 6

a^2 / 4 = 6 / π

a^2 = 24 / π

Similarly, if we simplify the other two equations, we will have b^2 = 32/π and c^2 = 48/π. Since a^2 + b^2 = 24/π + 32/π = 60/π is greater than c^2 = 48/π, we see that the triangle is indeed an acute triangle. Statement one alone is sufficient.

Statement Two Alone:

Statement two alone is not sufficient. If a = 1 and b = 1.8 and c = 2, then a^2 + b^2 = 1^2 + 1.8^2 = 1 + 3.24 = 4.24 is greater than c^2 = 4. In this case, the triangle is an acute triangle. However, if a = 1, b = 1.7 and c = 2, then a^2 + b^2 = 1^2 + 1.7^2 = 1 + 2.89 = 3.89 is not greater than c^2 = 4. In this case, the triangle is not an acute triangle.

Answer: A
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink] New post   13 Jun 2021, 11:40
(1) Here I thought that we are given the ratio of the sides. And for a given ratio we cant form both an acute and an obtuse triangle. Im not 100 % sure if Im correct here though, but it felt reasonable. For example, side ratios 3:4:5 can only be a right triangle. Would be nice if anyone could confirm this?

(2) This statement was a little easier to work with. If c is 0,001 cm, we should be able to make any type of triangle and still end up with the inequality c < a+b < c+2
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Re: A triangle has side lengths of a, b, and c centimeters. Does each angl [#permalink]
13 Jun 2021, 11:40
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