math help please?

Question

I've read the lesson and sill don't understand these at ALL
http://i118.photobucket.com/albums/o117/Hannahrae2011/untitled2-3.jpg

http://i118.photobucket.com/albums/o117/Hannahrae2011/untitled-16.jpg

Answer 1

The relationship you need is the Pythagorean Theorem, which says that if you have a right triangle with sides a, b forming the right angle and side h forming the long side completing the triange that

a² + b² + h²

Problem 1: You are given one leg of length 4 and a hypotenuse of length 6, so

4² + b² = 6²
16 + b² = 36
b² = 20
b = sqrt ( 20 ) = sqrt ( 4 * 5 ) = sqrt ( 4 ) * sqrt ( 5 ) = 2 sqrt ( 5 )

On the second you are given the two short sides of lengths 5 and 8, so

5² + 8² = h²
25 + 64 = h²
89 = h²

h = sqrt ( 89 ) which doesn't simplify like the first one did.

Answer 2

1. By Pythagoras`:

6^2 = 4^2 + x^2, where x is the leg you wish to find, so:

36 = 16 + x^2, so

20 = x^2, so

x = rt20, so

x = rt(4x5) so

x = rt4 rt5. so

x = 2 rt5.

2. H^2 = 5^2 + 8^2, so

H^2 = 25 + 64, so

H^2 = 89, so

H = rt(89)

13. If the triangle is right angled, then the square on the longest side is equal to the sum of the squares on the other two sides, so, does:

4^2 = 2^2 + 3^2 ? ? ?.........let`s see:

16 = 4 + 9.

well, 4 + 9 = 13, it doesn`t equal 16, so the triangle is not right angled.

14. E, acute equilateral. Equi lateral means equal sided.

15. B, an acute angle.

Hope this is clear and helpful, Twiggy.

Answer 3

Geometry needs to be done with paper and scissors to see how it works. Cut the areas up. The square areas represent the square amount. Fit the smaller areas on top of the large areas. If they all fit it is a sum.