# Let ∆ ABC ~ ∆ DEF and their areas be, respectively, 64 cm^{2} and 121 cm^{2}. If EF = 15.4 cm, find BC

**Solution:**

We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Here it is given that ΔABC ~ ΔDEF

Given, EF = 15.4 cm

Therefore, Area of ΔABC / Area of ΔDEF = (BC)^{2}/(EF)^{2}

64 cm^{2 }/ 121 cm^{2} = (BC)^{2}/(15.4)^{2}

(BC)² = [(15.4)^{2} × 64] / 121

BC = (15.4 × 8) / 11

BC = 11.2 cm

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 6

**Video Solution:**

## Let ∆ ABC ~ ∆ DEF and their areas be, respectively, 64 cm² and 121 cm². If EF = 15.4 cm, find BC

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.4 Question 1

**Summary:**

Let ∆ ABC ~ ∆ DEF and their areas be, respectively, 64 cm^{2} and 121 cm^{2}. If EF = 15.4 cm, then the value of BC is 11.2 cm.

**☛ Related Questions:**

- Diagonals of a trapezium ABCD with AB || DC intersect each other at the point If AB = 2 CD,find the ratio of the areas of triangles AOB and COD.
- In Fig. 6.44, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area( ABC) / area(DBC) = AO/DO
- If the areas of two similar triangles are equal, prove that they are congruent.
- D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ ABC. Find the ratio of the areas of ∆ DEF and ∆ ABC

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