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**kakenx****Member**- Registered: 2006-03-07
- Posts: 2

Hi! I'm very new with math and I'm having problems with this excercise.. seems simple but I need some help! This here is to find the perimeter and area of the triangle. Thank you!

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Is this triangle rectangular (sorry but I have to ask)?

If it is, there's a theorem, called pythagorean. It will help you.

And why c² is in square inches?

I think that c must be a line.

(I'm asking because c is actually length and c² must be "length" too)

If it is, then simply

c=5in

This pytagorean theorem states that

a²+b²=c²

You know c, know b too, so it won't be hard to find a.

I leave you to continue with the approach.

IPBLE: Increasing Performance By Lowering Expectations.

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 19,175

Hi kakenx!

Welcome to MathsIsFun!

Given c²= 25in², it can be known that c=5 in (taking only the positive square root);

b = 4in.

From the diagram, it can be seen that the triangle is a right-angled triangle.

For any right-angled triangle (a triangle in which one of the angles is 90 degrees), Pythagoras theorem holds good.

That is, the sum of the squares of two sides of a right-angled triangle would be equal to the square of the third side (the longest side, called the hypotenuse).

From the diagram, it can be seen that c is the longest side.

Hence, c²=a²+b²

25 = b²+a²

It is given that b=4

Therefore,

25 = 4²+a²= 16 + a²

25 - 16 = a², 9=a²

Therefore, a=3in (Since the unit is inch).

Perimeter = a+b+c = 3+4+5 = 15 inches

Area of a triangle = 1/2 (base) x (height)

For any right-angled triangle,

area is given by the formula 1/2 (a)(b) where a and b are the two sides other than the hypotenuse.

Therefore, in this problem,

Area = 1/2 x (3) x(4) = 6 in²

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

Good ganesh.

You was just little slower than me, but you gave the entire solution.

Well done!

IPBLE: Increasing Performance By Lowering Expectations.

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**kakenx****Member**- Registered: 2006-03-07
- Posts: 2

Thank you!

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