# Constructing an isosceles triangle with base 12 ft.

Assume you have 60 feet of fencing or border material. Calculate the area of an isosceles triangle with a base of 12 ft.

I'm a little confused by this because if the base is 12 then obviously the remaining two sides will be 24 ft each.

When I construct an isosceles triangle, it is not possible to have the side lengths be 24 while also having the base angles be 45 degrees each. So how would I calculate the area of this triangle? What would the lengths be of the two congruent sides?

• Why can't the two sides be $24$? The problem doesn't appear to specify $45^{\circ}$ base angles anywhere. It's not a right triangle, for sure.
– lulu
May 12, 2023 at 16:44
• The two base angles do not have to be 45 degress in an isosceles triangle. They just have to be equal to each other. May 12, 2023 at 16:44
• If the trouble comes from the computation of the area, like all triangle it's half the base times the height. And if you cut an isoceles triangle along its height, you get two right triangles. So there should be no problem. (even in the more general case there are formulas, see en.wikipedia.org/wiki/Area_of_a_triangle) May 12, 2023 at 16:53
• @lulu oh yes thank you. but if i don't know the angles then how should I find the "height" of the triangle? because when i draw the altitude, it splits it into two right triangles but i do not know the other angles May 12, 2023 at 16:58
• You don't need the angles. Just drop the height to the base and use the Pythagorean Theorem.
– lulu
May 12, 2023 at 17:03

If the base angles of angles of an isosceles triangle are $$45^\circ$$ then it is a isosceles right angled triangle.

For the area, use heron's formula $$\Delta=\sqrt{s(s-a)(s-b)(s-c)}$$ where $$a,b,c$$ are sides of the triangle and $$s=\frac{a+b+c}{2}$$

• Or use the Pythagorean theorem to find the height (since the bisected base and the other sides are known to form two congruent right triangles). May 12, 2023 at 17:00

Yes, if the base is 12ft then the other two sides will be 24ft. But the angles opposite to the equal sides may not nessescarily be 45 degree. When the question says 'base' it does not imply that the triangle is a right angled triangle having base 12 ft and a hypotenuse and perpendicular of 24 ft. That is simply not possible. In an isosceles triangle that has exactly two equal sides, the equal sides are called legs and the third side is called the base.

Moving on to the methods which can be used to calculate the area, there are two that I can think of.

The first is that you can draw a perpendicular from the vertex common to the equal sides to the base. This perpendicular will bisect the base and will divide the triangle into two smaller triangles of equal area. You can calculate the length of the perpendicular using Pythogoras Theorem and thereafter calculate the area of the triangle.

Alternatively you can simply use Heron's formula to calculate the area.

• ah yes thanks. When i did the calculation I got that the area is $6\sqrt{540}$ is this correct? Also this is somewhat of a side note but if I were to draw this on graph paper and I calculated 1 box to be 0.25 in. Then I would simply need to take my ruler and draw 6 inches from each side of the base to the vertex right? May 12, 2023 at 21:24
• The area is correct. But I don't quite understand the second question. You are taking 6 inches or 6 boxes? Either way you are not getting 24 as the length of the sides. May 12, 2023 at 21:55
• sorry for the lack of clarity. I mean to say if I set it up so that 1 box = 1 foot (on a piece of graph paper). And I measure 1 box to be 0.25 inches. then for me to measure one of the sides that is 24 feet, I would need to take my ruler and start from one end of the base and draw 6 inches. That way those six inches would represent 24 feet. May 13, 2023 at 3:41
• @user130306 Yes. Ideally you would measure 6 inches on the compass and draw two arcs (by taking the end points of the base as center) such that they cut each other at a point which will be the vertex. Then you just join the vertices. May 13, 2023 at 9:21