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Having a mental block... can you do 1/2 x base x height to find the area of any triangle? Or is it just for a right angled triangle or just for a non-right angled triangle?

2007-12-27 02:09:34 · 14 answers · asked by anna 1 in Science & Mathematics Mathematics

14 answers

1/2 x base x height will work for any triangle, but you have to be careful that you really have the correct value of base and height, meaning, you cannot just multiply any two legs of any triangle together, the height refers specifically to the distance of a line that goes between one leg to an opposite vertex and is perpendicular to that line...sorry, without a diagram it is hard to say this easily, hope this helps

2007-12-27 02:14:16 · answer #1 · answered by kuiperbelt2003 7 · 0 0

That is valid for all triangles, the only problem is to find the height when the triangles are not a right angled triangle, because the height is the longer line starting on the base with a 90 degree angle with respect to the base and arriving on the top angle of the triangle. Sometimes is not easy to find the measure of that line.

2007-12-27 02:22:06 · answer #2 · answered by Escatopholes 7 · 0 0

All triangles, and, as the others said, the height or altitude is not necessarily the same as the length of a side.

Now here is a method that will work for ANY triangle if you know the lengths of all the sides.

1) Determine the perimeter (add the lengths of the sides A, B and C because some people really don't know how to do that) and divide by 2. This is called the semiperimeter, usually abbreviated as S.

2) Multply S times (S-A) times (S-B) times (S-C).

3) Take the square root of that number and that is the area.

It works every time.

2007-12-27 02:21:19 · answer #3 · answered by Tom 6 · 0 0

You draw the "Height" . Then the height divides the base in to two segments forming two Right angled triangles. The area of Left right angled triangle is Left segment multiplied by height divided by two , since this corrospond to the half of the rectangle formed by that segment and the Height. Similiarly on the Right side there is a right angled triangle formed by right segment & height . Its area is also Right segment mutliplied by Height divided by 2.

Thus total area of triangle is SL *H/2 + SR*H/2

taking common H/2 out side the bracket, H/2 { SL+SR}

But Sum ofd two segments SL+SR is nothing but Base, B

Thus area= B*H/2

I hope this will remove your mental block , consider height dividing the triangle in to two Right triangles........

2007-12-27 02:48:48 · answer #4 · answered by RAJASEKHAR P 4 · 0 0

'1/2 x base x height' is a formula to find the area of ANY triangle.

2007-12-27 02:17:54 · answer #5 · answered by An ESL Learner 7 · 0 0

It's for all triangles,
but be careful! The height is not the other side.
The area is equal 1/2 x base x the height to this base
Merry X-mas!

2007-12-27 02:13:07 · answer #6 · answered by Orfeas 3 · 2 0

for all triangles, but remember that the base and height are at right angles to each other!

2007-12-27 02:15:26 · answer #7 · answered by lispyoon 2 · 0 0

Hey Tom Area of an triangle equals 1/2 base multiplied by height i.e 1/2bxh 1/2 base x height = 36 cm squared Base =8cm 1/2(8) x h = 36 cm squared do the sum on the left i.e divide 8 by 2... 4 x h = 36cm squared you need to take all the numbers over to one side, to do this divide both sides by 4 4/4 x h = 36/4 On the left hand side 4/4=0 so it disappears leaving only H H = 36/4 Calculate the sum on the right hand side 36/4 to give you 9 Height ( or altitude) = 9 Check answer 1/2 of 8cm x 9cm = 36cm squared

2016-04-11 03:02:38 · answer #8 · answered by Anonymous · 0 0

Yes, you can do it for any triangle.

However, if you can't find the base or height easily, then use another formula:

Area = square root of the quantity s(s-a)(s-b)(s-c), given that a, b, c are the lengths of the sides, and s = (a+b+c)/2

2007-12-27 03:13:03 · answer #9 · answered by dennismeng90 6 · 0 0

it is valid for all triangles which is a right angled or a non-right angled triangle.

2007-12-27 02:34:18 · answer #10 · answered by el 2 · 0 0

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