Triangle A has an area of #84 # and two sides of lengths #18 # and #15 #. 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?

1 Answer
Feb 15, 2016

#16.634# and #2.161#

Explanation:

Let me say something obvious:
- the smallest possible similar triangle has the minimum area and the largest possible similar triangle has the maximum area.

Suppose that in triangle A, the unknown side is #a#.
Suppose that in triangle B, the known side is #d#.

The smallest triangle B occurs when side #d# is proportional to the largest side of triangle A (then the other sides will be smaller than #d#). The largest triangle B occurs when side #d# is proportional to the smallest side of triangle A (then the other sides will be larger than #d#).
(The triangle may also be an isosceles one, in which case there will two big congruent sides or two small congruent sides).

Basically is a matter of knowing the length of side #a#.

In the Heron's formula for the area of the triangle:

#S=sqrt(s(s-a)(s-b)(s-c))#
#s=(a+b+c)/2=(a+18+15)/2=(a+33)/2#
#84=sqrt((a+33)/2*((a+33)/2-a)((a+33)/2-18)((a+33)/2-15)#
#7056=(a+33)/2*(-a+33)/2*(a-3)/2*(a+3)/2#
#112896=(-a^2+1089)(a^2-9)#
#112896=-a^4+9a^2+1089a^2-9801#
#a^4-1098a^2+122697=0#
#-> Delta=1,205,604-490,788=714,816#
#-> sqrt(Delta)=845.468#
#a^2=(1098+-845.468)/2#
#->a_1^2=971.733# => #a_1=31.173#
#->a_2^2=126.666# => #a_2=11.236#

As we can see triangle A can have 2 different shapes, one in which side #a# is the largest one and other in which side #a# is the smallest one.

If two triangles are similar their sides are directly proportional (#s"'"=k*s#) and so are their heights (#h"'"=k*h#), then:
#(S"'")/S=((b"'"*h"'")/cancel(2))/((b*h)/cancel(2))=((k*cancel(b))(k*cancel(h)))/(cancel(b)*cancel(h))=k^2#
#-> S"'"=k^2*S#, where k is the ratio between corresponding sides

For #a=11.236#

#S"'"=(5/11.236)^2*84=16.634# (maximum area)

For #a=31.173#

#S"'"=(5/31.173)^2*84=2.161# (minimum area)