`DeltaMNT ~ DeltaQRS`. Length of altitude drawn from point T is 5 and length of altitude drawn from point S is 9. Find the ratio `(A(DeltaMNT))/(A(DeltaQRS))`.

Explanation

Let te altitude of `DeltaMNT` drawn from vertex T be TX and the altitude of `DeltaQSR` from vertex S be SY such that `M-X-N` and `R-Y-Q`.
`TX=6` and `SY=9`
`DeltaMNT~DeltaQRS` ………..(Given)
`:._M~=/_Q` ......(Corresponding angles of similar triangles are congruent) .....1
In `DeltaMXT` and `DeltaQYS`
`/_XMT~=/_YQS` .......[From 1]
`/_MXT~=/_QYS`.(Each measures `90^(@)`)
`:.DeltaMXT~DeltaQYS`...(AA test of similarity)
`:(TM)/(SQ)=(TX)/(SY)`...(Corresponding sides of similar triangles are in proportion)
`:.(TM)/(SQ)=5/9`........2
`DeltaMNT~DeltaQRS`..........(Given)
`(A(DeltaMNT))/(A(DeltaQRS))=(TM^(2))/(SQ^(2))`
..............(Theorem of area of similar triangles)
`:.(A(DeltaMNT))/(A(DeltaQRS))=((TM)/(SQ))^(2)`
`:.(A(DeltaMNT))/(A(DeltaQRS))=(5/9)^(2)`......[From 2]
`:.(A(DeltaMNT))/(A(DeltaQRS))=25/81`
`:.(A(DeltaMNT))/(A(DeltaQRS))=25/81`