The base of an Isosceles triangle is half as long as the two equal sides. Write the area of the triangle as a function of the length of the base.
the answer is:
A=a^2times the square root of 15
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4
How did they get this answer? Thanks!
The area of a triangle is half its base times its vertical height. Call the base a. Now: you know that the other two sides are twice as long as that, so they're each 2a. What you need to find out now is the vertical height - so think of your isosceles triangle as two right-angled triangles back-to-back. You've got the base of each one (a/2), and you've got the hypoteneuse (2a). Can you now work out the height?
After you have the height, though, what do you do?
Yes,what do you do at that point? then you have the height as 4, but where does the 15 come in?
The length of the base of an isososeles triangle is one fourth the length of one of its legs. If the perimeter of the triangle is 62in, what is the lengthof the base?
What is the formula to work this?
How did you find the height for the isosceles triangle?
To find the area of an isosceles triangle, you need the length of the base and the length of the congruent (equal) sides. Let's call the length of the base "b" and the length of each congruent side "a".
Given that the base is half as long as the two equal sides, we can express this relationship as:
b = (1/2) * a
To find the area of a triangle, you can use the formula A = (1/2) * base * height. In this case, the base of the triangle is "b" and the height can be found using the Pythagorean theorem.
In an isosceles triangle, the height can be calculated by using the Pythagorean theorem on a right-angled triangle created by drawing a line from the top vertex (opposite the base) to the midpoint of the base. This line is perpendicular to the base and creates two right-angled triangles that are congruent.
In the right-angled triangle, one of the equal sides is the height (h) of the isosceles triangle, and the base is half as long as the two equal sides (a). So we can express the Pythagorean theorem as:
(a/2)^2 + h^2 = a^2
Simplifying this:
(a^2/4) + h^2 = a^2
h^2 = a^2 - (a^2/4)
h^2 = 3a^2/4
h = sqrt(3a^2/4)
h = (a * sqrt(3))/2
Now that we have the height, we can substitute it into the area formula:
A = (1/2) * b * h
A = (1/2) * b * [(a * sqrt(3))/2]
A = (1/2) * b * (a * sqrt(3))/2
Substituting b = (1/2) * a:
A = (1/2) * (1/2) * a * (a * sqrt(3))/2
A = a^2 * sqrt(3)/4
To simplify further, we can multiply the numerator by sqrt(3) and divide the whole expression by 4:
A = a^2 * (sqrt(3)/4)
So, the area of the isosceles triangle, expressed as a function of the length of the base (b, or in our case, (1/2) * a), is:
A = (a^2 * sqrt(3))/4
The answer you provided is in the same form, but with simplified multiplication:
A = (a^2 * sqrt(15))/4
I hope this explanation helps! Let me know if you have any more questions.