what is the length of the hypotenuse of a right triangle whose legs are 5 and the square root of two
1 answer:
To begin with, you need to use Pythagorean Theorem (a^2+b^2=c^2)
5 becomes 25
Solve for the square root first which is 1.414213562 and this becomes 2
So you add this two numbers and it will give you 27
Square root 27 and its 5.196152423
You can check it using this equation
5^2+square root of 2^2=5.196152423^2
The length of the hypotenuse is 5.1961524227066
~Revilla03
5 becomes 25
Solve for the square root first which is 1.414213562 and this becomes 2
So you add this two numbers and it will give you 27
Square root 27 and its 5.196152423
You can check it using this equation
5^2+square root of 2^2=5.196152423^2
The length of the hypotenuse is 5.1961524227066
~Revilla03
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(5,24)
- The two areas are the same.
- To find the area, we multiply the side lengths. Y=area
Rectangle 1: side lengths 4 and (x+1)
y=4(x+1)= 4x+4
Rectangle 2: side lengths 3 and (2x-2)
y=3(2x-2)= 6x-6
- Since the two areas are same, we can conclude that
4x+4=6x-6
- Now, simplify.
-2x=-10, x=5
- Since x is 5, we can plug it into the equations to find y.
Option 1 with rectangle 1: y=4(5)+4, y=24
Option 2 with rectangle 2: y=6(5)-5, y=24
I graphed the linear equation on desmos.
