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?

Question

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?

Violet Aldrich · Accepted Answer

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

#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#

Wyatt Anderson · Answer

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

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).

James Atkins · Answer

undefined Since Triangle B is similar to Triangle A, the ratio of corresponding sides of Triangle B to Triangle A is constant. Let this ratio be $ k $.

Given that Triangle A has an area of 15 and two sides of lengths 8 and 7, its area can be calculated using the formula:

$$ \text{Area of Triangle A} = \frac{1}{2} \times \text{base} \times \text{height} $$

Using the side lengths given, we can find the height of Triangle A:

$$ 15 = \frac{1}{2} \times 8 \times \text{height} $$

$$ \text{height} = \frac{15}{4} $$

Since Triangle B is similar to Triangle A and has a side with a length of 14, the corresponding side in Triangle A is $ 14 \div k $.

The height of Triangle B would be $ \frac{15}{4} \times k $ (since the height is also scaled by $ k $).

Thus, the area of Triangle B can be expressed as:

$$ \text{Area of Triangle B} = \frac{1}{2} \times 14 \times \frac{15}{4} \times k $$

$$ = \frac{105}{2} k $$

To find the maximum and minimum possible areas of Triangle B, we need to consider the possible values of $ k $. Since $ k $ is a scale factor, it must be positive. Also, the length of the corresponding side in Triangle A must be less than or equal to the sum of the other two sides (Triangle Inequality Theorem).

The minimum value of $ k $ occurs when the corresponding side in Triangle A is equal to 7, the shorter side of Triangle A. In this case, $ k = \frac{14}{7} = 2 $, and the minimum area of Triangle B is $ \frac{105}{2} \times 2 = 105 $.

The maximum value of $ k $ occurs when the corresponding side in Triangle A is equal to 15, the height of Triangle A. In this case, $ k = \frac{14}{15} $, and the maximum area of Triangle B is $ \frac{105}{2} \times \frac{14}{15} \approx 49 $.