Question

Triangle A has an area of `18` and two sides of lengths `8` and `12`. Triangle B is similar to triangle A and has a side with a length of `9`. What are the maximum and minimum possible areas of triangle B?

Answer 1

Maximum area of `Delta` B 729/32 & Minimum area of `Delta` B 81/8

Explanation:

If sides are 9:12, areas will be in their square.
Area of B `=(9/12)^2*18=(81*18)/144=` 81/8

If the sides are 9:8,
Area of B `=(9/8)^2*18=(81*18)/64=` 729/32

Aliter :
For similar triangles, ratio of corresponding sides are equal.

Area of triangle A =18 and one base is 12.
Hence height of `Delta` A `= 18/((1/2)12)=3`
If `Delta` B side value 9 corresponds to `Delta` A side 12, then the height of `Delta` B will be `=(9/12)*3=9/4`

Area of `Delta` B `=(9*9)/(2*4)=` 81/8

Area of `Delta` A = 18 and base is 8.
Hence height of `Delta` A `=18/((1/2)(8))=9/2`
I`Delta` B side value 9 corresponds to `Delta` A side 8, then
the height of `Delta` B `=(9/8)*(9/2)=81/16`

Area of `Delta` B `=((9*81)/(2*16))=`729/32

`:.` Maximum area 729/32 & Minimum area 81/8

Answer 2

Minimum possible area 81/8
Maximum possible area 729/32

Explanation:

Alternate Method :

Sides ratio 9/12=3/4.Areas ratio will be `(3/4)^2`
`:.` Min. possible area `= 18*(3^2/4^2)=18*(9/16)=81/8`

Sides ratio = 9/8.
`:.` Max. possible area `=18*(9^2/8^2)=729/32`