Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 2006-03-07 03:28:45

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

Triangle problems

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!


triangle.jpg

Offline

#2 2006-03-07 03:41:16

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Triangle problems

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.

Offline

#3 2006-03-07 03:42:45

ganesh
Moderator
Registered: 2005-06-28
Posts: 13,154

Re: Triangle problems

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


Character is who you are when no one is looking.

Offline

#4 2006-03-07 03:44:55

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,908

Re: Triangle problems

Good ganesh.
You was just little slower than me, but you gave the entire solution.
Well done! wink


IPBLE:  Increasing Performance By Lowering Expectations.

Offline

#5 2006-03-07 04:12:57

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

Re: Triangle problems

Thank you!

Offline

Board footer

Powered by FluxBB