How to solve intriguing SSC CGL level Geometry problem in a few steps 4

Key information discovery and decision making skills are generally judged

how to solve intriguing ssc cgl geometry problem in a few steps 4 top

Sometimes in the breakneck hurry while struggling with the questions in a cutting edge leading selection test you might meet a problem that doesn't seem right somehow. You figure out the only way to arrive at the solution but out of sheer curiosity you check back the solution value with the given values to verify the sanctity of the problem definition.

In a leading competitive test though you would never have any time to even think of verifying solution results let alone actually verify. The most basic reason of this observation is the fact that, in no leading competitive exam you are expected to finish answering all the questions.

Nevertheless, thankfully one of our students actually verified a solution result of one the problems in the article, SSC CGL level Solution Set 21 Geometry 3, and found that the values given in the particular question apparently are incorrect with respect to the solution value.

This is a rare occasion and the problem itself is deceptive. It required a detailed treatment leading to a general abstract conceptual result.

The problem

We repeat the problem and its solution that we had presented in SSC CGL level Solution Set 21 Geometry 3.

Problem 1.

In a right triangle $\triangle ABC$, the $\angle A = 90^0$ and $AD$ is perpendicular to $BC$. If areas of the triangles $\triangle ABC = 40cm^2$ and $\triangle ACD = 10cm^2$ with $AC = 9cm$, the length of $BC$ is,

  1. 4cm
  2. 12cm
  3. 18cm
  4. 6cm

Solution 1.

The following figure describes the problem,

SSC CGL Solution Set 21 geometry3 q1

In two right-angled triangles $\triangle ABC$ and $\triangle ACD$, apart from the equal right angles, the angle $\angle C$ is common. So the two triangles are similar to each other. The equal ratios of corresponding sides are (first triangle taken as the $\triangle ACD$),


Ratio of their areas is,


In this ratio from equal side ratios because of triangle similarity,

$\displaystyle\frac{CD_1}{AC_2} = \displaystyle\frac{AC_1}{BC_2} = \displaystyle\frac{AC}{BC}$, and

$\displaystyle\frac{AD_1}{AB_2} = \displaystyle\frac{AC_1}{BC_2} = \displaystyle\frac{AC}{BC}$.

So, the ratio of areas = $\displaystyle\frac{1}{4} = \frac{AC^2}{BC^2}$,

Or, $BC^2 = 4\times{9^2}$,

Or, $BC = 2\times{9} = 18$cm.

Answer: c: 18cm.

The Discrepancy

Now out of curiosity you decide to check the correctness of the given values of tiangle areas. You can do so now easily as you have the two values of two sides of the main right triangle,

$AB^2 = 18^2 - 9^2=243$,

Or, $AB=15.59$.

So the area of $\triangle ABC = \frac{1}{2}AC\times{AB} \approxeq{70}$ sq cm, whereas the given value is 40 sq cm. This is a major discrepancy.

Why did it happen? Is our method of solving the problem wrong?

Is the solution result wrong?

Or, the question itself is wrong?

Further analysis and key information discovery

Intending to verify now the last given value, that is, the area of the smaller $\triangle ADC$ our first step is to find the value of AD,

$\text{Area 70 sq cm } = \frac{1}{2}AD\times{BC}$


Or, $AD = \frac{70}{9} \approxeq{7.78}$ cm.

And in right $\triangle ACD$,

$CD^2 = 9^2 - (7.78)^2 = 81 - 60.5 = 20.5$,

Or, $CD \approxeq{4.5}$ cm

First important relation

Under these conditions, length of CD will then be one-fourth of the length of BC ($4.5\times{4}=18$).

Carrying on with our calculation,

$\text{Area of triangle ACD } = \frac{1}{2}AD\times{CD}$. Without even calculating the area value now, as CD is one-fourth of BC, we can say,

  Area of $\triangle ACD$ is one-fourth of the area of $\triangle ABC$.

This is the second important key information in this problem.

So merging the two pieces of key information, we can finally say,

In the figure as above, whatever be the individual values of lengths of the sides of the triangles, if the length of BC is double the length of AC, the area of $\triangle ACD$ will be one-fourth the area of $\triangle ABC$ and vice versa.

Here the ratios are the determining or key elements and not the individual values of length of sides or area of triangles.

Interestingly, as this condition is satisfied in the given problem with area of $\triangle ACD$ as one-fourth the area of $\triangle ABC$, the value of length BC automatically turned out to be double the length of AC, that is, 18 cm, though in actuality the individual area values were wrong. This happens because here the ratio of areas cleanly and directly determines the ratio of the length of two sides, and not the individual values of the areas.

This relationship between ratio of sides to ratio of areas of triangles in this specific right triangle formation, along with further results, constitute a rich geometry concept. As always, this rich concept also is based firmly on one of the more basic concepts, specifically the concept of similar triangle properties in this case.

We refrain from naming this rich concept, as it is a bit too specific, being tied to a particular geometric formation only.

Deductive reasoning that was followed in the original solution

"On quick examination, it is clear that with area of $\triangle ABC$ and height AC given the base AB can be calculated. That would be easy. And from these two values itself we would be able to arrive at the hypotenuse value by using Pythagoras theorem. But that path involves squaring and taking square root of decimal numbers. Not only will the calculation be messy, the answer will not also be an integer, where all choice values are integers. The solution value can't match exactly with any of the choices as well as the calculation would take more than desired time. This is not a very desirable path, we decide.

So we further examine the configuration to find the two right triangles $\triangle ACD$ and $\triangle ABC$ similar with the common side AC and common angle $\angle C$. Knowing that in similar triangles, the ratio of corresponding sides are equal, we sense easy and quick solution along this path. This is the conceptual analytical path and correspondingly a clean and fast one."

End note

Still the question remains, can the question be considered wrong?

Though we rely more on thinking strategies and analytical deductive reasoning, nevertheless we stick to precision as far as practicable. If we were to form this particular question we would have used nearly correct values of the areas of triangles but we would certainly have kept the ratio at one-fourth, because this relationship is to be used in the important concept of triangle similarity for the elegant solution.

More importantly,

In all such tests worldwide, mathematics is perhaps used as a vehicle for judging the core competence in efficient problem solving, reasoning and decision making skills, not merely the mathematical abilities.

We would end the session with a general form of this problem in which we have used abstraction heavily.

The general problem

In a right $\triangle ABC$ with right angle at A, a perpendicular AD is dropped on the hypotenuse BC. If the height of $\triangle ABC$, $AC=x$ and ratio of areas of triangles $\triangle ACD$ and $\triangle ABC$ is $p : q$, the length of BC is,

  1. $\displaystyle\sqrt{\frac{q}{p}}x$
  2. $\displaystyle\sqrt{\frac{p}{q}}x$
  3. $\displaystyle\frac{\sqrt{q}}{\sqrt{p}x}$
  4. $\displaystyle\frac{\sqrt{p}}{\sqrt{q}x}$

Hints to solution

The following proportionate to the scale figure is applicable for area ratio of 1 : 4.

how to solve intriguing ssc cgl geometry problem in a few steps 4

With the inside out view of the concepts on this problem aspects you should find it easy to arrive at the solution of the general form of the problem quickly now.