Question

Triangle A has an area of `9` and two sides of lengths `4` and `7`. Triangle B is similar to triangle A and has a side with a length of `16`. What are the maximum and minimum possible areas of triangle B?

Answer 1

`color(red)("The maximum possible area of B will be 144")`

`color(red)("and the minimum possible area of B will be 47")`

Explanation:

Given
`"Area Triangle A"=9 " and two sides 4 and 7"`
If the angle between sides 4 & 9 be a then

`"Area"=9=1/2*4*7*sina`
`=>a=sin^-1(9/14)~~40^@`

Now if length of the third side be x then

`x^2=4^2+7^2-2*4*7cos40^@`

`x=sqrt(4^2+7^2-2*4*7cos40^@)~~4.7`

So for triangle A

The smallest side has length 4 and largest side has length 7

Now we know that the ratio of areas of two similar triangles is the square of the ratio of their corresponding sides.

`Delta_B/Delta_A=("Length of one side of B"/"Length of Corresponding side of A")^2`

When the side of length 16 of triangle corresponds to the length 4 of triangle A then

`Delta_B/Delta_A=(16/4)^2`

`=>Delta_B/9=(4)^2=16=>Delta_B=9xx16=144`

Again when the side of length 16 of triangle B corresponds to the length 7 of triangle A then

`Delta_B/Delta_A=(16/7)^2`

`=>Delta_B/9=256/49=16=>Delta_B=9xx256/49=47`

`color(red)("So the maximum possible area of B will be 144")`

`color(red)("and the minimum possible area of B will be 47")`