Triangle A has an area of #5 # and two sides of lengths #4 # and #7 #. Triangle B is similar to triangle A and has a side of length #18 #. What are the maximum and minimum possible areas of triangle B?

Question

Triangle A has an area of #5    # and two sides of lengths #4  # and #7   #.  Triangle B is similar to triangle A and has a side of length #18   #.  What are the maximum and minimum possible areas of triangle B?

Savannah Aguilar · Accepted Answer

Maximum possible area of triangle B = 101.25
Minimum possible area of triangle B = 33.0612

#Delta s A and B # are similar. To get the maximum area of #Delta B#, side 18 of #Delta B# should correspond to side 4 of #Delta A#. Sides are in the ratio 18 : 4 Hence the areas will be in the ratio of #18^2 : 4^2 = 324 : 16# Maximum Area of triangle #B =( 5 * 324) / 16= 101.25# Similarly to get the minimum area, side 7 of #Delta A # will correspond to side 18 of #Delta B#. Sides are in the ratio # 18 : 7# and areas #324 : 49# Minimum area of #Delta B = (5*324)/49= 33.0612#

Allison Allen · Answer

undefined Since triangle B is similar to triangle A, the ratio of corresponding sides of triangle B to triangle A is constant. Let's denote this ratio as $k$. Given that one side of triangle B is 18, and the corresponding side in triangle A is 7, we can find $k = \frac{18}{7}$.

Now, for any similar triangles, the ratio of their areas is the square of the ratio of their corresponding sides. Therefore, the area of triangle B compared to triangle A is $k^2 = \left(\frac{18}{7}\right)^2$.

To find the maximum and minimum possible areas of triangle B, we consider the extremes of this ratio. Since the ratio of areas is positive, the minimum area occurs when the ratio of sides is minimized, and the maximum area occurs when the ratio of sides is maximized.

The maximum area of triangle B occurs when the ratio of sides is maximized, i.e., when the side of length 18 in triangle B corresponds to the longest side of triangle A. In this case, $k = \frac{18}{7}$, and the area of triangle B is $\left(\frac{18}{7}\right)^2 \times \text{Area of triangle A}$.

The minimum area of triangle B occurs when the ratio of sides is minimized, i.e., when the side of length 18 in triangle B corresponds to the shortest side of triangle A. In this case, $k = \frac{18}{4}$, and the area of triangle B is $\left(\frac{18}{4}\right)^2 \times \text{Area of triangle A}$.

Therefore, the maximum possible area of triangle B is $\left(\frac{18}{7}\right)^2 \times 5$ and the minimum possible area of triangle B is $\left(\frac{18}{4}\right)^2 \times 5$.