SOLUTION: The formula for the area of a triangle is �� =1/2��ℎ, where �� represents the base length, and ℎ
represents the height. The p
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Question 641203: The formula for the area of a triangle is �� =1/2��ℎ, where �� represents the base length, and ℎ
represents the height. The perimeter of the triangle shown is 28 inches. Write an equation for the
area �� of this triangle in terms of its base length ��. then its a triangle and the slanted part is 10 inches and the base is just b inches
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
The formula for the area of a triangle is
, where represents the base length,
and represents the height.
That first sentence suggests that this should be an easy problem, for students knowing just a little algebra and geometry.
If the perimeter is 28 inches and one side other than the base measures 10 inches,
the length of the other two sides would be
and
If we there was some information about the angles in that triangle,
it could be an easy problem, and some information about the angles may be in the drawing of the triangle.
Without any additional information, your best bet is using Heron's formula, based on the semiperimeter, s.
Heron's formula says that the area for a triangle with side lengths a, b, and c, and perimeter 2s is
However, that makes no use of .
Maybe the triangle drawing looks sort of like this,
, indicating somehow that is it an isosceles triangle
(maybe with those short cross-lines showing that the slanted sides are congruent,
or with arcs indicating that the base angles are congruent,
or with a 10 next to each slanted side).
In that case, applying Pythagoras to one of the right triangles, we find
If he triangle drawing looks sort of like this,
, which is a right triangle, then there is a typo in the problem,
because a right triangle with a perimeter of 28 inches can NEVER have a hypotenuse measuring 10 inches.
That's too bad, because in a right triangle, the two perpendicular legs can be taken as base and height, and that would make it a really easy problem..
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