Solve all your doubts with Teachoo Black (new monthly pack available now!)

Area of similar triangles

Theorem 6.6
Important
Deleted for CBSE Board 2023 Exams

Ex 6.4, 9 (MCQ) Deleted for CBSE Board 2023 Exams

Ex 6.4, 1 Important Deleted for CBSE Board 2023 Exams

Ex 6.4, 4 Deleted for CBSE Board 2023 Exams

Ex 6.4, 8 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 6.4, 2 Deleted for CBSE Board 2023 Exams

Ex 6.4, 6 Important Deleted for CBSE Board 2023 Exams

Example 9 Important Deleted for CBSE Board 2023 Exams

Ex 6.4, 7 Important Deleted for CBSE Board 2023 Exams

Ex 6.4, 5 Important Deleted for CBSE Board 2023 Exams You are here

Ex 6.4, 3 Important Deleted for CBSE Board 2023 Exams

Last updated at May 29, 2018 by Teachoo

Ex 6.4, 5 D, E and F are respectively the mid-points of sides AB, BC and CA of ΞABC. Find the ratio of the areas of ΞDEF and ΞABC. Given: Ξ ABC & D,E,F mid-points of AB,BC & CA respectively To find: (ππ βπ·πΈπΉ)/(ππ βπ΄π΅πΆ) Note: Since we need to find ratio of area of ΞDEF and ΞABC. We first need to prove these triangles are similar Solution: We know that line joining mid-points of two sides of a triangle is parallel to the 3rd side In ΞABC , D and F are mid-points of AB and AC resp., β΄ DF β₯ BC So, DF β₯ BE also Similarly, E and F are mid-points of BC and AC resp. EF β₯ AB Hence, EF β₯ DB From (1) & (2) DF β₯ BE & FE β₯ DB Therefore, opposite sides of quadrilateral is parallel DBEF is a parallelogram DBEF is a parallelogram Now we know that , in parallelogram, opposite angle are equal Hence β DFE =β ABC Similarity, we can prove DECF is a parallelogram In a parallelogram, opposite angles are equal Hence, β EDF= β ACB Now , in ΞEDF and ΞABC β DFE =β ABC β EDF= β ACB By using AA similarity criterion Ξ DEF βΌ Ξ ABC We know that if two triangles are similar, the ratio of their area is always equal to the square of the ratio of their corresponding side β΄ (ππ βπ·πΈπΉ)/(ππ βπ΄π΅πΆ) = π·πΈ2/π΄πΆ2 (ππ βπ·πΈπΉ)/(ππ βπ΄π΅πΆ) = πΉπΆ2/π΄πΆ2 (π΄πππ ππ βπ·πΈπΉ)/(π΄πππ ππ βπ΄π΅πΆ)=( π΄πΆ/2 )^2/(π΄πΆ)2 (π΄πππ ππ βπ·πΈπΉ)/(π΄πππ ππ βπ΄π΅πΆ)=((π΄πΆ)2/4)/(π΄πΆ)2 (π΄πππ ππ βπ·πΈπΉ)/(π΄πππ ππ βπ΄π΅πΆ)=(1/4)/1 Hence , (π΄πππ ππ β π·πΈπΉ)/(π΄πππ ππ β π΄π΅πΆ)=1/4