Triangle A has an area of #9 # and two sides of lengths #3 # and #9 #. Triangle B is similar to triangle A and has a side with a length of #7 #. What are the maximum and minimum possible areas of triangle B?

1 Answer
Jul 19, 2016

Maximum possible Area of B: #10 8/9# sq.units

Minimum possible Area of B: #0.7524# sq.units (approximately)

Explanation:

If we use the side of A with length #9# as the base
then the height of A relative to this base is #2#
(since the area of A is given as #9# and #"Area"_triangle=1/2xx"base"xx"height"#)

Note that there are two possibilities for #triangleA#:
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The longest "unknown" side of #triangleA# is obviously given by Case 2 where this length is the longest side possible.

In Case 2
#color(white)("XXX")#the length of the "extension" of the side with length #9# is
#color(white)("XXXXXX")sqrt(3^2-2^2)=sqrt(5)#
#color(white)("XXX")#and the "extended length" of the base is
#color(white)("XXXXXX")9+sqrt(5)#
#color(white)("XXX")#So the length of the "unknown" side is
#color(white)("XXXXXX")sqrt(2^2+(9+sqrt(5))^2)#
#color(white)("XXXXXXXX")=sqrt(90+18sqrt(5))#
#color(white)("XXXXXXXX")=3sqrt(10+2sqrt(5))#

The Area of a geometric figure varies as the square of its linear dimensions.

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The maximum area of #triangleB# will occur when #B#'s side of length #7# corresponds to the shortest side of #triangleA# (namely #3#)

#("Area of "triangleB)/("Area of "triangleA)=7^2/3^2#
and since #"Area of "triangleA =2#
#rArr "Area of "triangleB = (7^2)/(3^2)xx2=98/9=10 8/9#

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

The minimum area of #triangleb# will occur when #B#'s side of length #7# corresponds to the longest possible side of #triangleA# (namely #3sqrt(10+2sqrt(5))# as shown above).

#("Area of "triangleB)/("Area of "triangleA)=7^2/((3sqrt(10+2sqrt(5)))^2)#

and since #"Area of "triangleA =2#

#rArr "Area of "triangleB = (7^2)/((3sqrt(10+2sqrt(5)))^2)xx2=98/(90+19sqrt(5))~~0.7524#