Triangle A has an area of #18 # and two sides of lengths #5 # and #9 #. Triangle B is similar to triangle A and has a side of length #12 #. What are the maximum and minimum possible areas of triangle B?

Question

Triangle A has an area of #18    # and two sides of lengths #5  # and #9   #.  Triangle B is similar to triangle A and has a side of length #12   #.  What are the maximum and minimum possible areas of triangle B?

Allison Allen · Accepted Answer

Maximum area of triangle B = 103.68

Minimum area of triangle B = 32

#Delta s A and B # are similar# To get the maximum area of #Delta B#, side 12 of #Delta B# should correspond to side 5 of #Delta A#. Sides are in the ratio 12 : 5. Hence the areas will be in the ratio of #12^2 : 5^2 = 144 : 25# Maximum Area of triangle #B =( 18 * 144) / 25 = 103.68# Similarly to get the minimum area, side 9 of #Delta A # will correspond to side 12 of #Delta B#. Sides are in the ratio # 12 : 9# and areas #144 : 81# Minimum area of #Delta B = (18*144)/81 = 32#

Ella Armstrong · Answer

undefined Since Triangle B is similar to Triangle A, their corresponding sides are proportional. Therefore, if the side length of Triangle B is 12, and the corresponding side length of Triangle A is 9, we can use the ratio of corresponding sides to find the scale factor.

Scale factor $ k = \frac{12}{9} = \frac{4}{3} $.

The area of similar triangles is proportional to the square of the scale factor. Therefore, the maximum and minimum possible areas of Triangle B can be found by squaring the scale factor and multiplying it by the area of Triangle A.

Maximum possible area:
$ \text{Max area of } B = (\frac{4}{3})^2 \times 18 = \frac{16}{9} \times 18 = 32 $

Minimum possible area:
$ \text{Min area of } B = (\frac{4}{3})^2 \times 18 = \frac{16}{9} \times 18 = 32 $

Therefore, the maximum and minimum possible areas of Triangle B are both 32 square units.