Triangle A has an area of #15 # and two sides of lengths #8 # and #7 #. Triangle B is similar to triangle A and has a side with a length of #14 #. What are the maximum and minimum possible areas of triangle B?

2 Answers
Dec 21, 2017

Maximum possible area of triangle B = 60
Minimum possible area of triangle B = 45.9375

Explanation:

#Delta s A and B # are similar.

To get the maximum area of #Delta B#, side 14 of #Delta B# should correspond to side 7 of #Delta A#.

Sides are in the ratio 14 : 7
Hence the areas will be in the ratio of #14^2 : 7^2 = 196 : 49#

Maximum Area of triangle #B =( 15 * 196) / 49= 60#

Similarly to get the minimum area, side 8 of #Delta A # will correspond to side 14 of #Delta B#.
Sides are in the ratio # 14 : 8# and areas #196 : 64#

Minimum area of #Delta B = (15*196)/64= 45.9375#

Dec 21, 2017

Maximum area: #~~159.5# sq. units
Minimum area: #~~14.2# sq. units

Explanation:

If #triangle_A# has sides #a=7#, #b=8#, #c=?# and an area of #A=15#
then #c~~4.3color(white)("XXX")"or"color(white)("XXX")c~~14.4#
(See below for indication of how these values were derived).

Therefore #triangleA# could have a minimum side length of #4.3# (approx)
and a maximum side length of #14.4# (approx.)

For corresponding sides:
#color(white)("XXX")("Area"_B)/("Area"_A)=(("Side"_B)/("Side"_A))^2#
or equivalently

#color(white)("XXX")"Area"_B="Area"_A * (("Side"_B)/("Side"_A))^2#

Notice that the greater the length of the corresponding #"Side"_A#,
the smaller the value of #"Area"_B#
So given #"Area"_A=15#
and #"Side"_B=14#
and the maximum value for a corresponding side is #"Side"_A~~14.4#
the minimum area for #triangleB# is #15 * (14/14.4)^2 ~~14.164#

Similarly, notice that the smalle the length of the corresponding #"Side"_A#,
the greater the value of #"Area"_B#
So given #"Area"_A=15#
and #"Side"_B=14#
and the minimum value for a corresponding side is #"Side"_A~~4.3#
the maximum area for #triangleB# is #15 * (14/4.3)^2 ~~159.546 #

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Determining possible lengths for #c#

Suppose we place #triangleA# on a standard Cartesian plane with the side with length #8# along the positive X-axis from #x=0# to #x=8#
Using this side as a base and given that the Area of #triangleA# is #15#
we see that the vertex opposite this side must be at a height of #y=15/4#

If the side with length #7# has one end at the origin (coterminal there with the side of length 8) then the other end of the side with length #7# must be on the circle #x^2+y^2=7^2#
(Note that the other end of the line of length #7# must be the vertex opposite the side with length #8#)

Substituting, we have
#color(white)("XXX")x^2+(15/4)^2=7^2#

#color(white)("XXX")x^2=559'16#

#color(white)("XXX")x=+-sqrt(559)/4#

Giving possible coordinates: #(-sqrt(559)/4,15/4)# and #(+sqrt(559)/4,15/4)#

We can then use the Pythagorean Theorem to calculate the distance to each of the points from #(8,0)#
giving the possible values shown above (Sorry, details missing but Socratic is already complaining about the length).

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