Answers:
1. 8 m
2. 17 m
3. 7 cm
4. 2 s
Explanations:
1. Let x = length of the base
x + 6 = height of the base
Then, the area of the triangle is given by
(Area) = (1/2)(base)(height)
56 = (1/2)(x)(x + 6)
56 = (1/2)(x² + 6x)
Using the symmetric property of equations, we can interchange both sides of equations so that
(1/2)(x² + 6x) = 56
Multiplying both sides by 2, we have
x² + 6x = 112
The right side should be 0. So, by subtracting both sides by 112, we have
x² + 6x - 112 = 112 - 112
x² + 6x - 112 = 0
By factoring, x² + 6x - 112 = (x - 8)(x + 14). So, the previous equation becomes
(x - 8)(x +14) = 0
So, either
x - 8 = 0 or x + 14 = 0
Thus, x = 8 or x = -14. However, since x represents the length of the base and the length is always positive, it cannot be negative. Hence, x = 8. Therefore, the length of the base is 8 cm.
2. Let x = length of increase in both length and width of the rectangular garden
Then,
14 + x = length of the new rectangular garden
12 + x = width of the new rectangular garden
So,
(Area of the new garden) = (length of the new garden)(width of the new garden)
255 = (14 + x)(12 + x) (1)
Note that
(14 + x)(12 + x) = (x + 14)(x + 12)
= x(x + 14) + 12(x + 14)
= x² + 14x + 12x + 168
= x² + 26x + 168
So, the equation (1) becomes
255 = x² + 26x + 168
By symmetric property of equations, we can interchange the side of the previous equation so that
x² + 26x + 168 = 255
To make the right side becomes 0, we subtract both sides by 255:
x² + 26x + 168 - 255 = 255 - 255
x² + 26x - 87 = 0
To solve the preceding equation, we use the quadratic formula.
First, we let
a = numerical coefficient of x² = 1
Note: if the numerical coefficient is
hidden, it is automatically = 1.
b = numerical coefficient of x = 26
c = constant term = - 87
Then, using the quadratic formula
![x = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} = \frac{-26 \pm \sqrt{26^2 - 4(1)(-87)} }{2(1)} \newline x = \frac{-26 \pm \sqrt{1,024} }{2} \newline \newline x = \frac{-26 \pm 32 }{2}](https://tex.z-dn.net/?f=x%20%3D%20%20%5Cfrac%7B-b%20%5Cpm%20%20%5Csqrt%7Bb%5E2%20-%204ac%7D%20%7D%7B2a%7D%20%3D%20%20%5Cfrac%7B-26%20%5Cpm%20%20%5Csqrt%7B26%5E2%20-%204%281%29%28-87%29%7D%20%7D%7B2%281%29%7D%20%20%0A%5Cnewline%20x%20%3D%20%20%5Cfrac%7B-26%20%5Cpm%20%20%5Csqrt%7B1%2C024%7D%20%7D%7B2%7D%0A%5Cnewline%0A%5Cnewline%20x%20%3D%20%20%5Cfrac%7B-26%20%5Cpm%20%2032%20%7D%7B2%7D)
So,
![x = \frac{-26 + 32 }{2} \text{ or } x = \frac{-26 - 32 }{2} \newline x = \frac{6 }{2} \text{ or } x = \frac{-58 }{2} \newline \boxed{ x = 3 \text{ or } x = -29}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B-26%20%2B%2032%20%7D%7B2%7D%20%5Ctext%7B%20%20or%20%7D%20x%20%3D%20%5Cfrac%7B-26%20-%2032%20%7D%7B2%7D%0A%5Cnewline%20x%20%3D%20%5Cfrac%7B6%20%7D%7B2%7D%20%5Ctext%7B%20%20or%20%7D%20x%20%3D%20%5Cfrac%7B-58%20%7D%7B2%7D%0A%5Cnewline%20%5Cboxed%7B%20x%20%3D%203%20%5Ctext%7B%20%20or%20%7D%20x%20%3D%20-29%7D)
Since x represents the amount of increase, x should be positive.
Hence x = 3.
Therefore, the length of the new garden is given by
14 + x = 14 + 3 = 17 m.
3. The area of the shaded region is given by
(Area of shaded region) = π(outer radius)² - π(inner radius)²
= π(2x)² - π6²
= π(4x² - 36)
Since the area of the shaded region is 160π square centimeters,
π(4x² - 36) = 160π
Dividing both sides by π, we have
4x² - 36 = 160
Note that this equation involves only x² and constants. In these types of equation we get rid of the constant term so that one side of the equation involves only x² so that we can solve the equation by getting the square root of both sides of the equation.
Adding both sides of the equation by 36, we have
4x² - 36 + 36 = 160 + 36
4x² = 196
Then, we divide both sides by 4 so that
x² = 49
Taking the square root of both sides, we have
Note: If we take the square root of both sides, we need to add the plus minus sign
![(\pm)](https://tex.z-dn.net/?f=%28%5Cpm%29)
because equations involving x² always have 2 solutions.
So, x = 7 or x = -7.
But, x cannot be -7 because 2x represents the length of the outer radius and so x should be positive.
Hence x = 7 cm
4. At time t, h(t) represents the height of the object when it hits the ground. When the object hits the ground, its height is 0. So,
h(t) = 0 (1)
Moreover, since
![v_0 = 27](https://tex.z-dn.net/?f=v_0%20%3D%2027)
and
![h_0 = 10](https://tex.z-dn.net/?f=h_0%20%3D%2010)
,
h(t) = -16t² + 27t + 10 (2)
Since the right side of the equations (1) and (2) are both equal to h(t), we can have
-16t² + 27t + 10 = 0
To solve this equation, we'll use the quadratic formula.
Note: If the right side of a quadratic equation is hard to factor into binomials, it is practical to solve the equation by quadratic formula.
First, we let
a = numerical coefficient of t² = -16
b = numerical coefficient of t = 27
c = constant term = 10
Then, using the quadratic formula
![t = \frac{-b \pm \sqrt{b^2 - 4ac} }{2a} = \frac{-27 \pm \sqrt{27^2 - 4(-16)(10)} }{2(-16)} \newline t = \frac{-27 \pm \sqrt{1,369} }{-32} \newline \newline t = \frac{-27 \pm 37 }{32}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B-b%20%5Cpm%20%5Csqrt%7Bb%5E2%20-%204ac%7D%20%7D%7B2a%7D%20%3D%20%5Cfrac%7B-27%20%5Cpm%20%5Csqrt%7B27%5E2%20-%204%28-16%29%2810%29%7D%20%7D%7B2%28-16%29%7D%20%5Cnewline%20t%20%3D%20%5Cfrac%7B-27%20%5Cpm%20%5Csqrt%7B1%2C369%7D%20%7D%7B-32%7D%20%5Cnewline%20%5Cnewline%20t%20%3D%20%5Cfrac%7B-27%20%5Cpm%2037%20%7D%7B32%7D)
So,
![t = \frac{-27 + 37 }{-32} \text{ or } t = \frac{-27 - 37 }{-32} \newline t = \frac{-10}{32} \text{ or } t = \frac{-64 }{-32} \newline \boxed{ t = -0.3125 \text{ or } t = 2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B-27%20%2B%2037%20%7D%7B-32%7D%20%5Ctext%7B%20or%20%7D%20t%20%3D%20%5Cfrac%7B-27%20-%2037%20%7D%7B-32%7D%20%5Cnewline%20t%20%3D%20%5Cfrac%7B-10%7D%7B32%7D%20%20%5Ctext%7B%20or%20%7D%20t%20%3D%20%5Cfrac%7B-64%20%7D%7B-32%7D%20%20%20%5Cnewline%20%5Cboxed%7B%20t%20%3D%20-0.3125%20%5Ctext%7B%20or%20%7D%20t%20%3D%202%7D)
Since t represents the amount of time, t should be positive.
Hence t = 2. Therefore, it takes 2 seconds for the object to hit the ground.