I am just having a hard time visualizing this problem to solve it.
The perimeters of 2 similar triangles are 12 ft and 45 ft. The area of the smaller triangle is 64ft^2. What is the area of the larger triangle?
You may want to review what your textbook says about similar figures. You should find something like this:
If two figures are similar, then the ratios of any pair of corresponding LENGTHS are equal (and since perimeter is simply the sum of the lengths of all the sides of a figure, the ratio of the perimeters is the same as the ratio of any pair of corresponding sides.) If the ratio of a pair of corresponding sides is a/b, then the ratio of the perimeters would ALSO be a/b.
If two figures are similar, then the ratio of the AREAS is the SQUARE of the ratio of any pair of corresponding sides. (so, if the ratio of a pair of corresponding sides was a/b, then the ratio of the areas would be (a/b)
2.)
If two three-dimensional figures are similar, then the ratio of their volumes is the CUBE of the ratio of a pair of corresponding sides. (so, if the ratio of a pair of corresponding sides is a/b, then the ratio of the volumes would be (a/b)
3)
Your problem says that two triangles are similar, and their perimeters are 12 ft and 45 ft respectively. So we know that the ratio of the perimeters, and thus the ratio of any pair of corresponding sides, is 12/45 (or, if you reduce that, 4/15).
The ratio of the areas, then could be expressed this way:
area of small triangle / area of large triangle = (4/15)
2
Area of small triangle given to be 64 ft
2.
Let x = area of large triangle, and substitute into the proportion above:
64/x = (4/15)
2
And solve that for x.....
