A Tough Triangles Question | Geometry

We aim to solve all our Geometry questions without the use of trigonometry because many standardised tests do not require it. This question makes us feel that we need trigonometry to handle it but it is not so. In fact, the question looks familiar at first but presents unforeseen problems later on. While going through this exercise, we will learn a few tips and tricks which will be useful in our mainstream standardised tests questions. Hence, it will add value to our repertoire (especially in elimination techniques).

Let’s go on to the question now.

Question: In triangle ${ABC}$ , if angle ${ABC}$ is 30 degrees, ${AC = 2 \times \sqrt2}$ and ${AB = BC = X}$ , what is the value of ${X}$ ?

(A) ${\sqrt3 - 1}$

(B) ${\sqrt3 + 2}$

(D) ${\frac{(\sqrt3 + 1)}{2}}$

(E) ${2\times (\sqrt3 + 1)}$

Solution:

What we see here is an isosceles triangle with one angle as 30 degrees and other two angles as $\frac{(180 - 30)}{2} = 75$ degrees each.

The side opposite the 30 degrees angle is $2\times \sqrt2$ . One simple observation is that $X$ will be greater than $2\times \sqrt2$ because these sides are opposite to the greater angles (75 degrees).

$2\times \sqrt2$ is a bit less than $2\times1.5$ because $\sqrt2 = 1.414$ . So $2\times \sqrt2$ is a bit less than 3. Note that options (A), (B) and (D) are much smaller than 3 so these cannot be the value of $X$ . We have already improved our chances of getting the correct answer. Now we have to choose out of (B) and (E). So we still need to solve.

Here is what is given: Angle $ABC$ = 30 degrees, $AC = 2\times \sqrt2$ . We need to find the value of $X$ . Now, 30 degree angle reminds us of our 30-60-90 triangle in which we know the ratio of sides. So given one side, we can find the other two.

But the problem is this – if we drop an altitude from $B$ to $AC$ , the angle 30 degrees will be split into half and we will actually get a 15-75-90 triangle. We don’t have a 30-60-90 triangle anymore. What do we do now? Let’s try to maintain the 30 degree angle as it is and try to get the 30-60-90 triangle.

Let’s drop an altitude from $C$ to $AB$ instead and call it $CE$ . Now we have a 30-60-90 triangle. since $BCE$ is a 30-60-90 triangle, its sides are in the ratio $1 : \sqrt3 : 2$ . Side $X$ corresponds to 2 on the ratio so $CE = \frac{X}{2}$ .

Area of triangle ABC = $\frac{1}{2} \times BD \times AC = \frac{1}{2} \times CE \times AB$

$\frac{1}{2} \times BD \times 2 \times \sqrt2 = \frac{1}{2} \times \frac{X}{2}\times X$

$BD = \frac{X^2}{4\times \sqrt2}$

Now $DC = \frac{1}{2}AC = \frac{2 \times \sqrt2}{2} = \sqrt2$

Let’s use the Pythagorean theorem on triangle $BDC$ now:

$BD^2 + DC^2 = BC^2$

${\frac{X^2}{4}\times \sqrt2}^2 + {\sqrt2}^2 = X^2$

$\frac{X^4}{32} + 2 = X^2$

$X^4 - 32 \times X^2 + 64 = 0$

$X^4 - 16X^2 + 8^2 - 16X^2 = 0$

${X^2 - 8}^2 - {4X}^2 = 0$

$(X^2 -8 + 4X) \times (X^2 - 8 - 4X) = 0$

Normally, this would require us to use the quadratic roots formula but let’s not get into that complication. We can just plug in the the two shortlisted options and see if either of the factor is 0. If one of the factors becomes 0, the equation will be satisfied and we would have got a root of the equation.

Since both options have both terms positive, it means the co-efficient corresponding to $B$ in $Ax^2 + Bx + C = 0$ must be negative.