Triangle A has an area of #5 # and two sides of lengths #9 # and #3 #. 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?

Question

Triangle A has an area of #5    # and two sides of lengths #9  # and #3   #.  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?

Mason Adams · Accepted Answer

#45# & #5#

There are two possible cases as follows

Case 1: Let side #9# of triangle B be the side corresponding to the small side #3# of triangle A then the ratio of areas #\Delta_A# & #\Delta_B# of similar triangles A & B respectively will be equal to the square of ratio of corresponding sides #3# & #9# of both similar triangles hence we have

#\frac{\Delta_A}{\Delta_B}=(3/9)^2#

#\frac{5}{\Delta_B}=1/9\quad (\because \Delta_A=5)#

#\Delta_B=45#

Case 2: Let side #9# of triangle B be the side corresponding to the greater side #9# of triangle A then the ratio of areas #\Delta_A# & #\Delta_B# of similar triangles A & B respectively will be equal to the square of ratio of corresponding sides #9# & #9# of both similar triangles hence we have

#\frac{\Delta_A}{\Delta_B}=(9/9)^2#

#\frac{5}{\Delta_B}=1\quad (\because \Delta_A=5)#

#\Delta_B=5#

Hence, maximum possible area of triangle B is #45# & minimum area is #5#

Noah Atkinson · Answer

undefined The maximum possible area of triangle B occurs when it is similar to triangle A with its side length equal to 9. Therefore, the maximum possible area of triangle B is 5.

The minimum possible area of triangle B occurs when it is similar to triangle A with its side length equal to 3. Therefore, the minimum possible area of triangle B is $\frac{1}{9}$ times the area of triangle A, which equals $\frac{1}{9} \times 5 = \frac{5}{9}$.