Okay so we need to figure out the width of the swimming pool using the length, and we know the width is 10 ft shorter than twice the width. I believe the easiest way to do this would be to first do 35-10, and then divide it in half. That gives us 12.5. To check our work I'll do the problem 12.5+12.5+10=35.
The width of the swimming pool is 12.5 ft.
First we can find the perimeter of the given rectangle
P=2L+2W (L=length and W=width)
P=2*3+2*4
P=6+8
P=14
So now we want to come up with a different combination of numbers that would give up a perimeter of 14. We know that 14 is divisible by 2 so let's make out width 2. If we plug in 2 as the width and 14 as the perimeters, we can solve for length
14=2L+2*2
14=2L+4
10=2L (subtract 4 from both sides)
L=5
So the dimensions are 5 units and 2 units
Hope this helps!
Answer:
BD = 35
Step-by-step explanation:
Calculate CD in right triangle ABC, then BD in right triangle BCD
Using Pythagoras' identity in both triangles.
The square on the hypotenuse is equal to the sum of the squares on the other two sides.
In Δ ADC
CD² + AD² = AC² , substitute values
CD² + 9² = 15²
CD² + 81 = 225 ( subtract 81 from both sides )
CD = 144 ( take the square root of both sides )
CD = 
= 12
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In Δ BCD
BD² + CD² = BC² , substitute values
BD² + 12² = 37²
BD² + 144 = 1369 ( subtract 144 from both sides )
BD² = 1225 ( take the square root of both sides )
BD = 
= 35
Answer:
The numerical value of the circumference is greater than the numerical value of the area.
