Rakesh Vishwakarma, who teaches math at Genesis Mentors writes a primer on solving geometry problems in the CAT that involve calculating cleverly-etched triangles inside intricate figures.

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A lot of questions in management entrance exams including CAT are from Geometry and that too from the topic involving the areas of shapes.

Many questions are based on finding the area of triangles, circles and quadrilaterals etc. Most of these questions can be solved in various ways. There is always a traditional method to solve such problems but the problem which most of the aspirants encounter in geometry is that they are not able to apply the theory / theorem when it is required.

Here I am discussing an easy approach to solve a certain type of questions based on finding the area of the desired geometrical figure.

Let’s start with the question itself:

Example 1: In the triangle PQR (refer to the figure), S is the midpoint of PQ, ST is perpendicular to PR, UR is perpendicular to PR, and the area of ∆PQR is 36 sq units. Find the area of ∆PTU?

Won’t it be a good idea to try the above problem before going ahead?

If you are done with your effort and got the answer congratulations to you. In case you are finding it difficult to comprehend let me assure you that its not that difficult as it seems. By the end of the article you would be in a position to attempt such questions with ease.

Lets discuss some theorems which are handy for solving such problems.

We all know that,

Area of Triangle = 1/2 (base * height)

Theorem 1: Triangles with same base and between same parallel lines have equal areas.

In the above figure for all three triangles ABC, ABD and ABE, the base i.e. AB and the height is same hence

[Area] ABC = [Area] ABD = [Area] ABE

Theorem 2: If two triangles have the same base, the ratio of the areas is the ratio of the heights.

In the above figure AB is the common base, hence Area of ∆ABC/Area of ∆ABD = h/H

Corollary: If two triangles have the same height, the ratio of the areas is the ratio of the bases.

Area of ∆ACD/Area of ∆CDB = a/b

Theorem 3: In any Trapezium ABCD (refer figure), AB || CD, O is the intersection of diagonals AD and BC then,

Area of ∆AOC = ∆BOD

Proof:

Since AB||CD, from theorem 1, we conclude that

Area of ∆ACD = Area of ∆BCD

Or Area of ∆AOC + Area of ∆COD = Area of ∆COD + Area of ∆BOD

or Area of ∆ AOC = Area of ∆ BOD {since triangle COD is common}

Theorem 4: In any quadrilateral ABCD, with AC and BD as diagonals intersecting at O.

Let,

Area of ∆ AOB = W

Area of ∆ BOC = X

Area of ∆ COD = Y

Area of ∆ AOD = Z then,

W.Y = X.Z

i.e.

Area of ∆AOB x Area of ∆COD = Area of ∆BOC x Area of ∆AOD

Note: The above relation can easily be proven with the help of theorem 2 as discussed above. I am leaving the proof for you all. Try it.

Now let’s visit the question which I asked in the beginning.

Question: In the triangle PQR (refer to the figure), S is the midpoint of PQ, ST is perpendicular to PR, UR is perpendicular to PR, and the area of ∆PQR is 36 sq. Units. Find the area of ∆PTU?

Solution:

Observe that ST||UR and STRU is a trapezium. Join SR, the figure looks like fig 1.1.

Fig 1.1

Fig 1.2

From theorem 3, we find that area of ∆SOU = area of ∆TOR,

Also area of ∆PTU = area of quad PTOS + area of ∆SOU

Or area of ∆PTU = area of quad PTOS + area of ∆TOR

Or area of ∆PTU = area of ∆PRS

Since S is the midpoint of triangle PQR, and RS is the median then,

Area of triangle PRS = 1/2 * Area of ∆PQR =1/2*36 = 18 sq. units.

Example 2: In the adjacent figure ABCD is a square with side 15 cm. DEFG is another square with G lying on AD, find the area of shaded region?

Solution:

Let FC intersect AD at H.

Join FD, we find FD||AC, also DFAC is a trapezium.

Applying theorem 3 we get Area of ∆FHA = Area of ∆CHD

Therefore, Area of ∆AFC = Area of ∆AHC + Area of ∆DHC = Area of ∆ADC = 1/2 * area of square ABCD = 1/2 * 15² = 112.5 cm²

Alternate approach:

Since the dimension of Square DEFG is not given hence we can conclude that the required area is independent of square DEFG. Without loss of generality we can take the sides of square DEFG = 0, thereby points F, E and G will coincide with point D and the Area of ACF = Area of ACD = ½ * area of square ABCD = 112.5 cm².

Example 3: In a triangle ABC with AB = 14 cm, D and E are points on BC and AC respectively such that BE and AD intersect at point F and the area of ∆ BFD = area of ∆ AFE. Also BD:DC = 2:5. Find the length of DE?

Solution:

Refer to the diagram,

Joining DE we observe that in a quadrilateral ABDE, since area of ∆ BFD = area of ∆AFE from theorem 3, we conclude that DE is parallel to AB. Therefore by similar triangle concept, ∆CDE is similar to ∆CBA with corresponding sides in the ratio:

BC: DC = (2+5): 5 = 7:5,

Therefore, DE/AB = 5/7

Hence the length of DE = 5/7 * 14 = 10cm.

Example 4: As shown in the figure, ABCD is a parallelogram AC and BE intersects at F. Area of ∆ CEF = 6 cm2 and ∆ BCF = 9 cm2. Find the area of quadrilateral ADEF?

Solution:

Join AE. ABCE is a trapezium therefore from theorem 3, Area of ∆BCF = ∆AFE = 9 cm²

Applying theorem 4 in ABCE we get,

Area of ∆ AFB x Area of ∆ EFC = Area of ∆ AFE x Area of ∆ BFC

Or Area of ∆ AFB x 6 = 9 x 9

Or Area of ∆ AFB = 27/2 cm²

Area of ∆ ABC = Area of ∆ ADC = 9 + 27/2 = 22.5 cm²

Therefore Area of ADEF = Area of ∆ ADC – Area of ∆ EFC

Area of ADEF = 22.5 – 6 = 16.5 cm²

Practice Problems

1. ABCD is a quadrilateral. The diagonals of ABCD intersect at the point P. The area of the triangles APD and BPC are 27 and 12, respectively. If the areas of the triangles APB and CPD are equal then the area of triangle APB is (XAT 2008)

a. 12
b. 18
c. 15
d. 16

2. In the given figure the area of ΔABC is 24 cm², D is the midpoint of side BC and E is the midpoint of AD. Then the area of ΔABE is:

a. 4 cm²
b. 8 cm²
c. 6 cm²
d. 12 cm²

3. In a parallelogram PQRS, X is any point in its interior. If area of PQRS = 36 cm², then area of ΔPXS + area of ΔRXQ is :

a. 9 cm²
b. 12 cm²
c. 18 cm²
d. 27 cm²

4. In the adjacent figure, the triangle PQR is divided into 6 parts, the area if four parts is indicated. Find the area of triangle PQR?

The author Rakesh Vishwakarma is a founder partner at Genesis Mentors Pune, where he heads content development for the Quantitative Ability section and also teaches the subject. Visit their website, PaGaLGuY page and Facebook page.

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