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  • It is given that Delta ABC sim Delta EDF such that AB = 5 cm, AC = 7 cm, DF= 15 cm and DE = 12 cm. Find the lengths of the remaining sides of the triangles.

Q: \text{It is given that }  \Delta ABC \sim \Delta EDF such that AB = 5 \; cm, AC = 7 \; cm, DF= 15 \; cm and DE = 12 \; cm.

Find the lengths of the remaining sides of the triangles.

Answers (1)

Answer:

[EF=16.8\; cm, BC= 6.25 \; cm]

Given :

\Delta ABC \sim \Delta EDF

\therefore sides of both triangles are in the same ratio

i.e

\frac{AB}{ED}=\frac{AC}{EF}=\frac{BC}{DF}\; \; \; \; \; \; .....(1)

\text{In } \Delta ABC, AB=5\; cm,AC=7\; cm

\text{In }  \Delta DEF, DF=15\; cm,DE=12\; cm

Put all these values in equation (1)

\frac{5}{12}=\frac{7}{EF}=\frac{BC}{15}

Taking the first and second term

\frac{5}{12}=\frac{7}{EF}

EF=\frac{7 \times 12}{5}=\frac{84}{5}=16.8

Taking the first and the last term

\frac{5}{12}=\frac{BC}{15}

BC=\frac{15 \times 5}{12}=\frac{75}{12}=6.25

Posted by

infoexpert23

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