“Fresh sale of individual modules has been closed. Visit anaprep.com for information on new products and write to us for special offers”


Geometry - Solving Questions with Diagrams

We have discussed how drawing extreme diagrams can help solve Geometry questions. Today we will see how to solve another Geometry question by making diagrams. The diagram can help you understand exactly what it is that you need to do; doing it will be quite straightforward.


Question: If 10, 12 and ‘x’ are sides of an acute angled triangle, how many integer values of ‘x’ are possible?

(A) 7
(B) 12
(C) 9
(D) 13
(E) 11


Solution: The question is very interesting. It asks you for an acute triangle i.e. a triangle with all angles less than 90 degrees. It’s a little hard to wrap your head around it, isn’t it? We know that the third side of a triangle can take many values. Right from a little more than the difference of the other two sides to a little less than the sum of the other two sides (Since we know that the sum of any two sides of a triangle is always greater than the third side). So x can be anything from a little more than 2 to a little less than 22. But how do we find out the values for which all the angles will be less than 90?


We want no obtuse or right angles. An obtuse angled triangle has one angle more than 90. So the thought here is that before one of the angles reaches 90, find out all the values that x can take.


Look at the figure given above. The value of x in the first figure is very small – slightly more than 2 – minimum required to make a triangle. There is an obtuse angle in that triangle. We keep making x bigger and bigger and the angle keeps becoming smaller till it reaches 90 (Fig III). We use Pythagorean theorem to get the value of x in that case:

x = sqrt(12^2 – 10^2)
x = sqrt(44) which is 6.something
x should be greater than 6.something because the angle cannot be 90.


We further keep increasing x and all the angles are acute now. We reach Fig V where we hit another right triangle. We use Pythagorean theorem again to get the value of x (the hypotenuse) in this case:

x = sqrt(12^2 + 10^2)
x = sqrt(244) which is 15.something
x should be less than 15.something so that the angle is not 90.

Further on, in Fig VI, we obtain an obtuse angle again.

We only need integral values of x so values that x can take range from 7 to 15 which is 9 values.


Answer (C).


Note: We made two angles 90 and found the values of x in between those two angles. The third angle cannot be 90 because that will make 10 the hypotenuse but hypotenuse is always the greatest side.


Leave a Comment

(Login required to leave a comment.)