Area of square = s^2
Area of Rectangle = lw
l = 2s
w+3 = s
solve for w, w=s-3
Now we have l and w.
Plug both into area of rectangle formula so: (2s)(s-3)
Since both areas are equal set both equations equal to each other:
(2s)(s-3)=s^2
Now simplify
2s^2-6s=s^2
s^2-6s=0, Solve for s.
Factor polynomial. s(s-6)=0 , s can be equal to 0 or 6. HOWEVER, you cannot have a side length of 0 therefore the side length has to be 6.
Now plug in s for the length formula for the rectangle:
l = 2s so... l = 2(6) so length of rectangle = 12.
Now plug in s for the width formula for the rectangle:
w+3=s so... w+3=6 so width of rectangle = 3.
Now the dimensions of the rectangle are 12 by 3. 12 being length and 3 width.
To CHECK:
Find area of rectangle:
A=lw so A=3 times 12 so A=36
Find area of square:
We know the side is equal to 6 so
A=s^2 so 6^2 = 36
The areas are equal that verifies the answer of 12 by 3.
This is a quadrilateral, so the sum of interior angle measures, like a rectangle or square ( who’s interior angle measures are4 x 90) —— is 360
So ...
E + 89 + 130 + 90 = 360
E + 309 = 360
- 309 - 309
E = 51 degrees
Answer:
Step-by-step explanation:
Let the equation of the line be 
where, 'm' is its slope and 
is a point on it.
Given:
The equation of a known line is:
A point on the unknown line is:
Both the lines are perpendicular to each other.
Now, the slope of the known line is given by the coefficient of 'x'. Therefore, the slope of the known line is 
When two lines are perpendicular, the product of their slopes is equal to -1.
Therefore,
Therefore, the equation of the unknown line is determined by plugging in all the given values. This gives,
The equation of a line perpendicular to the given line and passing through (4, -6) is .
Answer:
y = 2/3x - 6
Step-by-step explanation:
Use the slope intercept equation, y = mx + b
Plug in the slope and given point, then solve for b
y = mx + b
-4 = 2/3(3) + b
-4 = 2 + b
-6 = b
Plug in the slope and b into the equation
y = 2/3x - 6
So, the equation of the line is y = 2/3x - 6
