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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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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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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 appears to me that if one wants to make progress in mathematics, one should study the masters and not the pupils. - Niels Henrik Abel.
Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.
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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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Thank you!
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