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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 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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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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diagram4.jpg [ 15.08 KiB | Viewed 2854 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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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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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]

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New post 29 Jan 2018, 23:51
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.
A triangle has side lengths of a, b, and c centimeters. Does each angl   [#permalink] 29 Jan 2018, 23:51
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