Question 950551: I have to solve a problem. My problem is of an equilateral triangle. The sides equal 7ft and the base is 6ft. I have use pythagorean theorem and find the area.
Found 2 solutions by richard1234, addingup:
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! That is not an equilateral triangle. It is an isosceles triangle.
Draw an altitude from the vertex to the base of length 6 feet, and use the Pythagorean theorem to find the height.
Answer by addingup(3677)  (Show Source):
You can put this solution on YOUR website! The equilateral triangle has three identical sides. What you describe, both sides the same and the base a different length, is an isosceles triangle. To find the area, draw a straight vertical line from the vertex above to the base below. Now you have two right angle triangles, each with a base 6/2= 3ft and a hypotenuse that's 7ft Once we find the length of the leg we just drew right down the middle we will have the height of the triangle and we'll be able to calculate the area:
Let's call the hypotenuse h, the base (the short leg) b, and the long leg a
Pythagoras says: h = √(b^2+a^2) Now, we know h= 7 and b= 2, we need to find a:
With a little algebra:
h^2 = b^2+a^2
h^2 - b^2= a^2
7^2 - 3^2 = a^2
49 - 9 = a^2
40 = a^2 Take the square root, both sides, to get rid of the exponent
√40 = √a^2
a = 6.32
The area of the triangle is base x height. Since we split the triangle in 2, we'll find the area of one and multiply x 2:
3 x 6.32 = 18.96 (the answer may be 19, since we had quite a bit of rounding decimals)
Area of your isosceles triangle: 18.96 x 2 = 37.92 (or 38)
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