Question

Triangle A has an area of `12` and two sides of lengths `3` and `8`. Triangle B is similar to triangle A and has a side of length `15`. What are the maximum and minimum possible areas of triangle B?

Answer 1

Maximum possible area of triangle B is `300` sq.unit
Minimum possible area of triangle B is `36.99` sq.unit

Explanation:

Area of triangle `A` is `a_A=12`

Included angle between sides `x=8 and z=3` is

`(x*z*sin Y)/2=a_A or (8*3*sin Y)/2=12 :. sin Y =1`

`:. /_Y= sin^-1(1)=90^0` Therefore, Included angle between

sides `x=8 and z=3` is `90^0`

Side `y=sqrt(8^2+3^2)= sqrt 73`. For maximum area in triangle

`B` Side `z_1=15` corresponds to lowest side `z=3`

Then `x_1=15/3*8=40 and y_1= 15/3*sqrt 73= 5 sqrt 73`

Maximum possible area will be `(x_1*z_1)/2=(40*15)/2=300`

sq. unit. For minimum area in triangle `B` Side `y_1=15`

corresponds biggest side `y=sqrt 73`

Then `x_1=15/sqrt73*8=120/sqrt73` and

`z_1= 15/sqrt73*3= 45/ sqrt 73`. Minimum possible area will be

`(x_1*z_1)/2=1/2*(120/sqrt73* 45/ sqrt 73)= (60*45)/73`

`~~ 36.99 (2 dp)` sq.unit [Ans]