Question

It is given that `DeltaABC~DeltaEDF` 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.

Answer 1

Given, `DeltaABC~DeltaEDF`, so the corresponding sides of `DeltaABC and DeltaEDF` are in the same ratio.
i.e, `(AB)/(ED)=(AC)/(EF)=(BC)/(DF)`

Also, AB=5 cm,AC=7 cm
DF =15 cm and DE=12 cm
On putting these values in Eq. (i), we get
`5/12=7/(EF)=(BC)/15`
On taking first and second terms, we get
`5/12=7/(EF)`
`rArr EF=(7xx12)/5=16.8 cm` ltbr On taking first and third terms we get
`5/12=(BC)/15`
`rArrBC=(5xx15)/12=6.25 cm`
Hence, length of the remaining sides of the triangle are EF=16.8 cm and BC =625 cm.