SOLUTION: what is the area of a triangle with side that are 9 units, 17 units and 10 units?

Algebra ->  Surface-area -> SOLUTION: what is the area of a triangle with side that are 9 units, 17 units and 10 units?       Log On


   



Question 707114: what is the area of a triangle with side that are 9 units, 17 units and 10 units?

Found 2 solutions by lynnlo, KMST:
Answer by lynnlo(4176) About Me  (Show Source):
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I say that the area is highlight%2836%29 and here is the triangle:
The small triangle, with sides measuring 6, 8, and 10 units is a right triangle, because the squares of the lengths of the shorter sides add up to the square of the length of the longer side:
6%5E2%2B8%5E2=36%2B64=100=10%5E2
A large triangle with sides measuring 15, 8, and 17 units is also a right triangle, because the squares of the lengths of the shorter sides add up to the square of the length of the longer side:
15%5E2%2B8%5E2=225%2B64=189=17%5E2
Our triangle with sides measuring 9, 10, and 17 units is the large right triangle minus the small right triangle, and its area is the difference between the areas of the two right triangles.
Area of the large right triangle=%281%2F2%29%2815%2A8%29=60 square units
Area of the small right triangle=%281%2F2%29%286%2A8%29=24 square units
Area of the triangle in the problem=60-24=36 square units

When you only have the lengths of the sides, unless it is a right triangle, the problem is tricky. If you have not been taught enough, you need a combination of luck, ingenuity and triangle construction to figure out what I show in the drawing an explanation above. To construct a triangle with sides measuring 9, 10, and 17 cm, you would draw one of the sides to exact measure, and the use a compass to mark arcs centered at the ends of the first side, with radii matching the other two side lengths. Where the arcs meet is the third vertex of the triangle. Such a construction would help you figure out what I showed above.

If you have been taught about Heron's formula involving the semi-perimeter, you could use that.

If you were taught the law of cosines, you could use it to find the cosine (and then sine) of one angle, and then use the formula for area involving the lengths of two sides and the sine of the angle in between.

USING LAW OF COSINES:
a=9 units = measure of shortest side
b=10 units = measure of medium length side
c=10 units = measure of longest side
C=measure of the largest angle, opposite longest.
The law of cosines says that
c%5E2=a%5E2%2Bb%5E2-2ab%2Acos%28C%29 so 17%5E2=9%5E2%2B10%5E2-2%2A9%2A10%2Acos%28C%29 --> 289=81%2B100-180%2Acos%28C%29
Solving for cos%28C%29 :
cos%28C%29=%2881%2B100-289%29%2F180=%28-108%29%2F180=-3%2F5
sin%28C%29=sqrt%281-%28cos%28C%29%29%5E2%29 so sin%28C%29=sqrt%281-%28-3%2F5%29%5E2%29=sqrt%281-9%2F25%29=sqrt%2816%2F25%29=4%2F5
The area of a triangle can be calculated from the lengths of 2 sides, a and b, and the angle between them, C, as
area=%281%2F2%29ab%2Asin%28C%29 so area=%281%2F2%29%2A9%2A10%2A%284%2F5%29=highlight%2836%29

USING HERON'S FORMULA:
The semi-perimeter, s, is defined as half of the perimeter.
If the side lengths are a, b, and c, then s=%28a%2Bb%2Bc%29%2F2,
and the area of the triangle is
area=sqrt%28s%28s-a%29%28s-b%29%28s-c%29%29
s=%289%2B10%2B17%29%2F2=36%2F2=18 and