The areas of two similar triangles are 121 cm2 and 64 cm2 respectively. If the median of the first triangle is 12.1 cm, find the corresponding median of the other.
Solution
We have,
ΔABC ~ ΔPQR
Area (ΔABC) = 121 cm2,
Area (ΔPQR) = 64 cm2
AD = 12.1 cm
And AD and PS are the medians
By area of similar triangle theorem
`("Area"(triangle))/("Area"(trianglePQR))="AB"^2/"PQ"^2`
`rArr121/64="AB"^2/"PQ"^2`
`rArr11/8="AB"/"PQ"`
Since, ΔABC ~ ΔPQR
Then, `"AB"/"PQ"="BC"/"QR"` [Corresponding parts of similar Δ are proportional]
`rArr"AB"/"PQ"=(2BD)/(2QS)` [AD and PS are medians]
`rArr"AB"/"PQ"="BD"/"QS"` .......(ii)
In ΔABD and ΔPQS
∠B = ∠Q [ΔABC ~ ΔPQS]
`"AB"/"PQ"="BD"/"QS"` [From (ii)]
Then, ΔABD ~ ΔPQS [By SAS similarity]
`therefore"AB"/"PQ"="AD"/"PS"` .....(iii)[Corresponding parts of similar Δ are proportional]
Compare (i) and (iii)
`11/8="AD"/"PS"`
`rArr11/8=12.1/"PS"`
`rArr"PS"=(8xx12.1)/11=8.8` cm
