Triangles-testing reasoning and critical skills

[harsh.champ]

Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer, then how many such triangles exist?

(A)13
(B)21
(C)10
(D)15
(E)14

The OA is C. 10

Don't look at the soln before answering by yourself.

Solution:-
[spoiler]We know that for an obtuse triangle of sides a, b and c (where c is the largest side),

a2 + b2 < c2

We also know that for a triangle, a + b > c

These present us with two limiting cases.

Let 8 cm and 15 cm be the length of shorter sides. The value of the largest side (x) must be greater than sqrt(8^2 + 15^2) which is 17cm.

The possible integer values of x are 18, 19, 20, 21 and 22 cm.

We cannot consider values from 23 onwards because 8 + 15 = 23 and this violates the second condition.

Now, consider the case where 15 cm is the measure of the largest side.

The value of the remaining side (x) must be less than sqrt(15^2 - 8^2) which is 12.69 cm.


The possible integer values are 12, 11, 10, 9 and 8 cm.

We cannot consider values less than 8 because 7 + 8 = 15 and this violates the second condition.

Thus, we have in total 10 possible values for x. Hence, option C.[/spoiler]

Just try it out.

[akahuja143]

Good questions Thanks for sharing

[Stuart@KaplanGMAT]

Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer, then how many such triangles exist?

(A)13
(B)21
(C)10
(D)15
(E)14

The general rule for sides of a triangle is that, for each side:

|diff of other sides| < side < sum of other sides

In this case, we know that:

7 < x < 23

If our triangle didn't have to be obtuse, therefore, there would be 15 possible integers for x (i.e. 8, 9, ... 22).

However, we need an obtuse triangle (just FYI, I've never seen that term actually used in a GMAT question), so we need one of our angles to be greater than 90 degrees, which means that we can modify the Pythagorean theorem to get:

a^2 + b^2 < c^2

in which c is the side opposite the obtuse angle.

So, if 15 is our big side, then:

8^2 + x^2 < 15^2

64 + x^2 < 225

x^2 < 161

x < 12. something

So, keeping with our other range, x could be 8, 9, 10, 11 or 12.

If we make x our large side, we have:

8^2 + 15^2 < x^2

64 + 225 < x^2

289 < x^2

17 < x

So, keeping with our other range, x could be 18, 19, 20, 21 or 22.

That's a total of 10 possible values for x... choose (C).

[sumanr84]

harsh,
Really nice Question..thanks to u.

Stuart,
thanks for providing detailed explanation. It will really help in attacking such Qs in future.

[harsh.champ]

Good questions Thanks for sharing

Anytime akahuja.

Hey sumanr84,

harsh,
Really nice Question..thanks to u.

Stuart,
thanks for providing detailed explanation. It will really help in attacking such Qs in future.

Welcome.

Anyways I had posted this question because it really test your reasoning skills.
Also I thought it was better to give the solution before hand as the solution approach really needs to be ponder upon.
Also I learnt the triangle equality which needs to be remembered.Though it isvery logical that,|diff of other sides| < side < sum of other sides but it doesn't seem to click in my mind.

However,I hope that everyone eill benefit with the question.

[shashank.ism]

Harsh really a good question... Generally people would consider only on case and would leave out other one..