Ratio of Area of Similar Triangles

Chapter 6 Class 10 Triangles
Serial order wise

### Transcript

Question 5 D, E and F are respectively the mid-points of sides AB, BC and CA of ΞABC. Find the ratio of the areas of ΞDEF and ΞABC. Given: Ξ ABC & D,E,F mid-points of AB,BC & CA respectively To find: (ππ βπ·πΈπΉ)/(ππ βπ΄π΅πΆ) Note: Since we need to find ratio of area of ΞDEF and ΞABC. 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 ΞABC , 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 ΞEDF and ΞABC β  DFE =β ABC β  EDF= β  ACB By using AA similarity criterion Ξ DEF βΌ 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΄ (ππ βπ·πΈπΉ)/(ππ βπ΄π΅πΆ) = π·πΈ2/π΄πΆ2 (ππ βπ·πΈπΉ)/(ππ βπ΄π΅πΆ) = πΉπΆ2/π΄πΆ2 (π΄πππ ππ βπ·πΈπΉ)/(π΄πππ ππ βπ΄π΅πΆ)=( π΄πΆ/2 )^2/(π΄πΆ)2 (π΄πππ ππ βπ·πΈπΉ)/(π΄πππ ππ βπ΄π΅πΆ)=((π΄πΆ)2/4)/(π΄πΆ)2 (π΄πππ ππ βπ·πΈπΉ)/(π΄πππ ππ βπ΄π΅πΆ)=(1/4)/1 Hence , (π΄πππ ππ β π·πΈπΉ)/(π΄πππ ππ β π΄π΅πΆ)=1/4