Asked by mulenga
A piece of wire 400cm long is bent to form the perimeter of a right langted triangle whose hypotenuse is170mm find the length of the shortor side.
Answers
Answered by
Bosnian
Mark the shorter pages with a and b and mark the hypotenuse with c.
The perimeter is:
P = a + b + c = 400 cm
a + b + 170 = 400
Subtract 170 from both sides
a + b = 230
The length of the hypotenuse is:
c = √ (a² + b²)
170 = √ (a² + b²)
Square both sides
28900 = a² + b²
Now you need to solve the system:
a + b = 230
a² + b² = 28900
From the first equation:
a = 230 - b
Put this in the second equation.
( 230 - b )² + b² = 28900
When you compose the equation it will be:
2 b² - 460 b + 24000 = 0
Divide both sides by 2
b² - 230 b + 12000 = 0
The solutions are:
b = 80
and
b = 150
when b = 80 cm
a = 230 - b = 230 - 80 = 150 cm
when b = 150 cm
a = 230 - b = 230 - 150 = 80 cm
So the shorter sides are 80 cm and 150 cm
The perimeter is:
P = a + b + c = 400 cm
a + b + 170 = 400
Subtract 170 from both sides
a + b = 230
The length of the hypotenuse is:
c = √ (a² + b²)
170 = √ (a² + b²)
Square both sides
28900 = a² + b²
Now you need to solve the system:
a + b = 230
a² + b² = 28900
From the first equation:
a = 230 - b
Put this in the second equation.
( 230 - b )² + b² = 28900
When you compose the equation it will be:
2 b² - 460 b + 24000 = 0
Divide both sides by 2
b² - 230 b + 12000 = 0
The solutions are:
b = 80
and
b = 150
when b = 80 cm
a = 230 - b = 230 - 80 = 150 cm
when b = 150 cm
a = 230 - b = 230 - 150 = 80 cm
So the shorter sides are 80 cm and 150 cm
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