SOLUTION: Hello I have done 7 out of 10 problems these are giving me some problems. 1 A right tringle has sides of 20cm and 15 cm. What is the length of the hypothenus. Thanks 6 Simp

Algebra ->  Customizable Word Problem Solvers  -> Age -> SOLUTION: Hello I have done 7 out of 10 problems these are giving me some problems. 1 A right tringle has sides of 20cm and 15 cm. What is the length of the hypothenus. Thanks 6 Simp      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 65188: Hello I have done 7 out of 10 problems these are giving me some problems.
1 A right tringle has sides of 20cm and 15 cm. What is the length of the hypothenus. Thanks
6 Simply 2(sqrt12)/4(sqrt27). Thanks
10 Find the distance between (4,0 and (-9,0)
Thanks for your help and have a happy Holiday.

Found 2 solutions by checkley71, Earlsdon:
Answer by checkley71(8403) About Me  (Show Source):
You can put this solution on YOUR website!
a^2+b^2=c^2 c being the hypotenuse
20^20=15*15=c^2
400+225=c^2
c^2=625
c=25
--------------------------------
2sqrt12/4sqrt27 factor sqrt 12=sqrt4*3 & sqrt 27=sqrt9*3
2*2sqrt3/4*3sqrt3
4sqrt3/12sqrt3 factor out the 4 & the sqrt 3 leaves you with
1/3
------------------------------------
distance between (4,0)& (-9,0)is
4-(-9)=4+9=13 is the distance between thwese two points
------------------------------------------
HAVE A MERRY CHRISTMAS & A HAPPY NEW YEAR

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
11. Find the length of the hypotenuse if the two legs are 20cm and 15cm.
Remember that in a right triangle, the lengths of the sides are in proprtion to: 3, 4, and 5, and the hypotenuse is always the longest side.
So:
15 = 3*5
20 = 4*5
x = 5*5 = 25
The hypotenuse is 25cm.
6. Simplify:
2sqrt%2812%29%2F4sqrt%2827%29 Cancel 2 in the top and bottom.
sqrt%2812%29%2F2sqrt%2827%29 Simplify the square roots.
sqrt%283%2A4%29%2F2sqrt%283%2A9%29
2sqrt%283%29%2F3%2A2sqrt%283%29 Cancel where appropriate.
1%2F3
10. The distance between two points, (x1, y1) and (x2, y2) is given by:
d+=+sqrt%28%28x2-x1%29%5E2+%2B+%28y2-y1%29%5E2%29 Applying this to your two points:(4, 0) and (-9, 0)
d+=+sqrt%28%28-9-4%29%5E2+%2B+%280-0%29%5E2%29
d+=+sqrt%28%28-13%29%5E2%29
d+=+sqrt%28169%29
d+=+13