# CAT Geometry: An Alternate Approach to Find Area

CAT Geometry: An Alternate Approach to Find Area- Many questions in CAT and other management entrance exams are from Geometry, particularly from the topic called AREA. These questions involve finding the area of triangles, circles, and quadrilaterals, among others. While there is a traditional method to solve such problems, many aspirants struggle to apply the relevant theory and theorems when required.

In this article, we will discuss an easy approach to solving a certain CAT Geometry and An alternate approach to find an area type of area-based geometry questions. We will start by examining a specific example:

Example 1: Find the area of ∆PTU

In the triangle PQR (refer to the figure), S is the midpoint of PQ, ST \$\bot\$ PR, UR \$\bot\$ PR, and the area of ∆PQR is 36 sq. Units.

Before moving forward, take a moment to attempt solving this problem yourself.

Now, let’s discuss some theorems that are useful for solving area-based geometry problems:

Theorem 1: Triangles with the same base and between parallel lines of the same distance have equal areas.
Theorem 2: If two triangles have the same base, the ratio of their areas is the ratio of their heights.
Theorem 3: The area of triangle AOC in any Trapezium ABCD (refer to the figure) where AB || CD and O is the intersection of diagonals AD and BC is equal to the area of triangle BOD.

Theorem 4: In any quadrilateral ABCD, with AC and BD as diagonals intersecting at O, if the areas of triangles AOB, BOC, COD, and AOD are denoted as W, X, Y, and Z respectively, then WY = XZ.

Now, let’s revisit the example we discussed earlier:

A solution to Example 1:

From the given information, we can observe that ST || UR and STRU is a trapezium. By applying Theorem 3, we find that the area of ∆SOU is equal to the area of ∆TOR.

Furthermore, the area of ∆PTU can be calculated as the sum of the area of quad PTOS and the area of ∆SOU, or alternatively, as the area of ∆PTU = area of ∆PR. Since S is the midpoint of triangle PQR and RS is the median, we can conclude that the area of triangle PRS is half the area of triangle PQR, which is equal to 18 sq. Units.

Example 2: Find the area of the shaded region

In the adjacent figure, ABCD is a square with a side of 15 cm. DEFG is another square with G lying on AD.

Solution to Example 2:

By joining FC and FD, we can observe that FD || AC. Applying Theorem 3, we find that the area of ∆FHA is equal to the area of ∆CHD. Therefore, the area of ∆AFC can be calculated as the sum of the area of ∆AHC and the area of ∆DHC, which is equal to the area of ∆ADC. By applying the formula for the area of a square, we can calculate the area of square ABCD as 15^2 = 225 sq. cm. Finally, we divide this by 2 to find the area of square ABCD is 112.5 sq. cm.

In the next section, we will discuss a few more examples to further illustrate the application of these theorems.

Example 3: Find the length of DE

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 is equal to the area of ∆AFE. Also, BD:DC = 2:5.

Solution to Example 3:

By analyzing the given information and observing the diagram, we can deduce that DE is parallel to AB. Using similar triangles, we can establish that ∆CDE is similar to ∆CBA, with corresponding sides in the ratio BC:DC = 7:5. Therefore, the length of DE can be calculated as (5/7) * 14 = 10 cm.

As shown in the figure, ABCD is a parallelogram and AC and BE intersect at F. The area of ∆CEF is 6 cm² and ∆BCF is 9 cm².

Solution to Example 4:

By applying Theorem 3, we can establish that the area of ∆BCF is equal to the area of ∆AFE, which is equal to 9 cm². Using Theorem 4, we can calculate the area of ∆AFB as (6 * 9) / 9 = 6 cm². By noting that triangles ABC and ADC have the same area, we can conclude that their combined area is 9 + 27/2 = 22.5 cm². Finally, we find the area of quadrilateral ADEF by subtracting the area of ∆EFC from the area of ∆ADC.

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