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[The Preacher]

Think of an equilateral triangle in which the length of a side is s, the perimeter (distance around the triangle) is p, and the area is A. Find the missing parts.

s=4
p=12


I took the triangle and made two right triangles, giving me a leg of 2 and a hypotenuse of 12.

Leg (l) = 2
Hypotenuse (h) =12

I took the Pythagorean Theorem and found out that 140 is the square of my other leg. I've tried factoring 140 but I can't seem to find the right answer. Can anyone give me a push in the right direction? Many thanks.

 

[stapel]

Find the missing parts.

What "parts" are "missing"?

Note: If the perimeter is "p" and the side length is "s", then p = 3s, since the triangle is equilateral (equal-sided).

Eliz.

 

[The Preacher]

Whoops, double post.

 

[The Preacher]

Find the missing parts.

What "parts" are "missing"?

Note: If the perimeter is "p" and the side length is "s", then p = 3s, since the triangle is equilateral (equal-sided).

Eliz.

My bad. Usually I give a picture with this stuff, and don't have to explain it. I found the side lengths before I asked for help here. They equal 4. The only problem I'm having is finding the area, because I'm having trouble factoring 140 to find a satisfactory answer. It's supposed to be in the radical form.

Sorry for being too vague.

God bless,
-Zac

EDIT: I've decided it might be a good thing to show some of my steps. I've tried getting the greatest factor of 140, which is 70. So... 140/70=2. Wouldn't that simplify to like... 7sqrt(2) or something? That's not the answer though... this stuff is confusing to me, and I apologize.

 

[wjm11]

Think of an equilateral triangle in which the length of a side is s, the perimeter (distance around the triangle) is p, and the area is A. Find the missing parts.

s=4
p=12


I took the triangle and made two right triangles, giving me a leg of 2 and a hypotenuse of 12.

Leg (l) = 2
Hypotenuse (h) =12

Hello, Preacher,

If you have an equilateral triangle with s = 4, 4 is the hypotenuse when you split the triangle into two right triangles. Note also that the right triangle thus created is a 30-60-90 triangle with sides in the ratio 1:2:sqrt3 (which can also be determined/verified by the Pythagorean Th.). The altitude of your right triangle is 2sqrt3. Can you take it from here?

 

[The Preacher]

Think of an equilateral triangle in which the length of a side is s, the perimeter (distance around the triangle) is p, and the area is A. Find the missing parts.

s=4
p=12


I took the triangle and made two right triangles, giving me a leg of 2 and a hypotenuse of 12.

Leg (l) = 2
Hypotenuse (h) =12

Hello, Preacher,

If you have an equilateral triangle with s = 4, 4 is the hypotenuse when you split the triangle into two right triangles. Note also that the right triangle thus created is a 30-60-90 triangle with sides in the ratio 1:2:sqrt3 (which can also be determined/verified by the Pythagorean Th.). The altitude of your right triangle is 2sqrt3. Can you take it from here?

Oh snap. Yes, thank you. For some reason I kept mistaking the perimeter for the hypotenuse. *rolls eyes* That would be why I had so much trouble. I had no more trouble with this problem after you cleared that up. Thank you, wjm11. =]