Triangle A has an area of #60 # and two sides of lengths #12 # and #15 #. Triangle B is similar to triangle A and has a side of length #9 #. What are the maximum and minimum possible areas of triangle B?

Question

Triangle A has an area of #60    # and two sides of lengths #12  # and #15   #.  Triangle B is similar to triangle A and has a side of length #9   #.  What are the maximum and minimum possible areas of triangle B?

Evelyn Arboleda · Accepted Answer

Maximum area 33.75 and Minimum area 21.6

#Delta s A and B # are similar. To get the maximum area of #Delta B#, side 25 of #Delta B# should correspond to side 12 of #Delta A#. Sides are in the ratio 9 : 12 Hence the areas will be in the ratio of #9^2 : 12^2 = 81 : 144# Maximum Area of triangle #B =( 60 * 81) / 144= 33.75# Similarly to get the minimum area, side 15 of #Delta A # will correspond to side 9 of #Delta B#. Sides are in the ratio # 9 : 15# and areas #81 : 225# Minimum area of #Delta B = (60*81)/225= 21.6#

Claire Akin · Answer

undefined The maximum and minimum possible areas of triangle B can be found by considering the ratios of corresponding sides of similar triangles. Since triangle B is similar to triangle A, the ratio of the sides of triangle B to the sides of triangle A is constant.

Let the area of triangle B be $ x $. Since the ratio of the sides of triangle B to triangle A is 9/15 (or 3/5), the ratio of the areas of similar triangles is the square of the ratio of their sides, which is (3/5)^2 = 9/25.

Therefore, the area of triangle B is 9/25 times the area of triangle A:

$$ x = \frac{9}{25} \times 60 $$

$$ x = \frac{540}{25} $$

$$ x = 21.6 $$

So, the maximum possible area of triangle B is 21.6.

For the minimum possible area of triangle B, the ratio of the sides of triangle B to triangle A is 9/12 (or 3/4). Therefore, the area of triangle B is 3/4 times the area of triangle A:

$$ x = \frac{3}{4} \times 60 $$

$$ x = \frac{180}{4} $$

$$ x = 45 $$

So, the minimum possible area of triangle B is 45.