Triangle A has an area of #24 # and two sides of lengths #8 # and #12 #. Triangle B is similar to triangle A and has a side of length #5 #. What are the maximum and minimum possible areas of triangle B?

Question

Triangle A has an area of #24    # and two sides of lengths #8  # and #12   #.  Triangle B is similar to triangle A and has a side of length #5   #.  What are the maximum and minimum possible areas of triangle B?

Avery Allen · Accepted Answer

#A_s = 4.16#
#A_L = 14.36# Using either given side as a base, the third side can be calculated. #A = 1/2 xx b xx h# ; #24 = 1/2 xx 8 xx h# #h = 6# We use this as part of a right triangle to find the third side. #C = 6.46# NOW we know that the SMALLEST similar triangle "B" must have the largest side as 5, and the LARGEST similar triangle will have the smallest side with a length of 5. The corresponding ratios are: #12:8:6.46# (original) #5:3.33:2.69# (smallest) #9.29:6.19:5# (largest) Using Herron's Rule directly saves some further angle computation. #s = (a + b+ c)/2# #A = sqrt(((s(s-a)(s-b)(s-c))# #5:3.33:2.69# (smallest) #s = (5 + 3.33 + 2.69)/2 = 5.51# #A = sqrt(((5.51(5.51-5)(5.51-3.33)(5.51-1.74))# #A_s = 4.16# #9.29:6.19:5# (largest) #s = (9.29 + 6.19+ 5)/2 = 10.24# #A = sqrt(((10.24(10.24-9.29)(10.24-6.19)(10.24-5))# #A_L = 14.36#

Allison Allen · Answer

undefined Since triangles A and B are similar, their corresponding sides are in proportion. Let's denote the ratio of the corresponding sides of triangles A and B as $ k $.

Given that triangle A has sides of lengths 8 and 12, its area is 24. We can use Heron's formula to find the length of the third side:
$$ s = \frac{8 + 12 + x}{2} $$
$$ A = \sqrt{s(s-8)(s-12)(s-x)} = 24 $$
Where $ x $ is the length of the third side.

By solving this equation, we find $ x = 4 $ or $ x = 16 $. However, since the third side must be less than the sum of the other two sides and greater than their difference, $ x $ must be 4.

Now, we can find the ratio $ k $:
$$ k = \frac{5}{4} $$

The maximum possible area of triangle B occurs when its side length is the maximum length it can be in proportion to triangle A, which is when $ k = \frac{5}{4} $. So, the maximum area is $ 24 \times \left(\frac{5}{4}\right)^2 = 37.5 $.

The minimum possible area of triangle B occurs when its side length is the minimum length it can be in proportion to triangle A, which is when $ k = \frac{5}{16} $. So, the minimum area is $ 24 \times \left(\frac{5}{16}\right)^2 = 2.34375 $.