CAT 2015 Quantitative Ability

Question : 13 of 34

Marks: +1, -0

Through point P. lines are drawn parallel to the sides of triangle ABC The areas of the ∆PED, ∆PFG and ∆PHI are 9,16 and 49 sq. cm respectively. Find the area (in sq cm) of triangle ABC

Your Answer:

Validate

Solution:

ΔPED is similar to ΔGFP
Ratio of area = 9:16
Therefore ratio of sides =

√9/16 = 3:4

Hence, P divides GD in the ratio 3 : 4.

ΔAGD = (7/3)^2 × 9 = 49 \text"sq.cm"

[ΔAGD similar to ΔFGP]
So area of AEPF = 49 - [16 + 9] = 24 sq. cm
Similarly area of BFI =

(10/3)^2 × 9 = 100 \text"sq.cm"

Therefore area of BHPG = 100 - (49 + 9) = 42 sq. cm
Similiarty area of PDCI =

(11/7)^2 × 49 - 49 - 16

= 56 sq.cm
Area of triangle ABC = (9 + 16 + 49 + 24 + 42 + 56) = 196 sq. cm

Alternative method:
Ratio of the corresponding sides is 3:4:7 since the areas are in the ratio 9:16:49 and all the triangles are similar. Hence GP + PD = BH+IC = HI. So HI is half of BC. Since triangle 3 is similar to Triangle ABC and HI is thecorresponding Side to BC . and is half of it the area of triangle 3 must be 1/4 the area of triangle ABC So area of triangle ABC = 196 sq cm.

Go to Question:

Prev Question

Next Question