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?

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?

Sadie Adams · Accepted Answer

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

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

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")#