Let
x---------> the length of the rectangle
y--------> the width of the rectangle
we know that
A=84 ft²
[area of rectangle]=x*y-----> 84=x*y-----> equation 1
x=y-5------> equation 2
substitute 2 in 1
84=[y-5]*y-----> 84=y²-5y---------> y²-5y-84=0
using a graph tool-----> to resolve the second order equation
see the attached figure
the solution is
y=12
x=y-5-----> x=12-5-----> x=7
the answer isthe length of the rectangle is 7 ftthe width of the rectangle is 12 ft
Answer:
It is C
Step-by-step explanation:
The bottom of the ladder is 10.5 feet away from the wall
Step-by-step explanation:
The given scenario forms a right triangle.
Where
The length of ladder will be the hypotenuse
The wall on which the window is situated will ebt he perpendicular and
The distance between the foot of ladder and the wall will be the base
So,
Hypotenuse = H = 20 foot
Perpendicular = P = 17 feet
Base = B = ?
Using the Pythagoras theorem
![H^2=P^2+B^2\\(20)^2=(17)^2+B^2\\400=289+B^2\\400-289=B^2\\B^2=111\\Taking\ square\ root\ on\ both\ sides\\\sqrt{B^2}=\sqrt{111}\\B=10.53\\Rounding\ off\ to\ the\ nearest\ tenth\\B=10.5](https://tex.z-dn.net/?f=H%5E2%3DP%5E2%2BB%5E2%5C%5C%2820%29%5E2%3D%2817%29%5E2%2BB%5E2%5C%5C400%3D289%2BB%5E2%5C%5C400-289%3DB%5E2%5C%5CB%5E2%3D111%5C%5CTaking%5C%20square%5C%20root%5C%20on%5C%20both%5C%20sides%5C%5C%5Csqrt%7BB%5E2%7D%3D%5Csqrt%7B111%7D%5C%5CB%3D10.53%5C%5CRounding%5C%20off%5C%20to%5C%20the%5C%20nearest%5C%20tenth%5C%5CB%3D10.5)
The bottom of the ladder is 10.5 feet away from the wall
Keywords: Triangle, Pythagoras Theorem
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Answer:
14 in
Step-by-step explanation: