Triangle A has an area of #18 # and two sides of lengths #8 # and #12 #. Triangle B is similar to triangle A and has a side with a length of #9 #. What are the maximum and minimum possible areas of triangle B?

2 Answers
Oct 10, 2017

Maximum area of #Delta# B 729/32 & Minimum area of #Delta# B 81/8

Explanation:

If sides are 9:12, areas will be in their square.
Area of B #=(9/12)^2*18=(81*18)/144=# 81/8

If the sides are 9:8,
Area of B #=(9/8)^2*18=(81*18)/64=# 729/32

Aliter :
For similar triangles, ratio of corresponding sides are equal.

Area of triangle A =18 and one base is 12.
Hence height of #Delta# A #= 18/((1/2)12)=3#
If #Delta# B side value 9 corresponds to #Delta# A side 12, then the height of #Delta# B will be #=(9/12)*3=9/4#

Area of #Delta# B #=(9*9)/(2*4)=# 81/8

Area of #Delta# A = 18 and base is 8.
Hence height of #Delta# A #=18/((1/2)(8))=9/2#
I#Delta# B side value 9 corresponds to #Delta# A side 8, then
the height of #Delta# B #=(9/8)*(9/2)=81/16#

Area of #Delta# B #=((9*81)/(2*16))=#729/32

#:.# Maximum area 729/32 & Minimum area 81/8

Oct 13, 2017

Minimum possible area 81/8
Maximum possible area 729/32

Explanation:

Alternate Method :

Sides ratio 9/12=3/4.Areas ratio will be #(3/4)^2#
#:.# Min. possible area # = 18*(3^2/4^2)=18*(9/16)=81/8#

Sides ratio = 9/8.
#:.# Max. possible area #=18*(9^2/8^2)=729/32#