1. What must be the dimensions of a rectangle to have an area 125m^2 and a perimeter of 60m?
2. In a right angle the hypotenuse is 15cm. if the second side is twice the length of the third side, calculate the exact length pf the shortest side.
3. THe height h metres of a cricket ball after being struck by a batsman is given by the equation h = 1 + x - x^2/40 where x metres is the horizontal distance travelled by the ball from the bat. How far would the ball travel before it hits the ground?
#1 You need to solve
2(x+y) = 60
xy = 125
so, since y=30-x,
x(30-x) = 125
x = 5 or 25
#2 It sounds like you want x where
x^2 + (2x)^2 = 15^2
#3 it hits the ground when its height is zero, so you need to solve
-x^2/40+x+1 = 0
making it a bit more comfortable by clearing fractions and having a positive leading coefficient, solve
x^2 - 40x - 40 = 0
1. To find the dimensions of a rectangle with a given area and perimeter, we can use the following formulas:
Area of a rectangle = length × width
Perimeter of a rectangle = 2(length + width)
Given that the area is 125m^2 and the perimeter is 60m, we can set up the following equations:
125 = length × width [Equation 1]
60 = 2(length + width) [Equation 2]
To solve this system of equations, we can use substitution or elimination method. Let's solve it using the elimination method:
From Equation 2, we can simplify it to:
30 = length + width [Equation 3]
Now we have two equations (Equation 1 and Equation 3) with two variables (length and width).
From Equation 3, we can rewrite it as:
length = 30 - width
Substituting this value into Equation 1, we get:
125 = (30 - width) × width
Expanding this equation, we have:
125 = 30w - w^2
Rearranging it in the standard quadratic form, we get:
w^2 - 30w + 125 = 0
Now, we can solve this quadratic equation for the value of width. Factoring or using the quadratic formula, we obtain:
(w - 5)(w - 25) = 0
So, w = 5 or w = 25
If w = 5, substituting it back into Equation 3, we get:
length = 30 - 5 = 25
If w = 25, substituting it back into Equation 3, we also get:
length = 30 - 25 = 5
Therefore, the dimensions of the rectangle can be either 5m by 25m or 25m by 5m.
2. In a right angle triangle, we can use the Pythagorean theorem to find the lengths of the sides. The theorem states that the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
Let's call the second side x and the third side (shortest side) y. We can set up the following equation:
x^2 + y^2 = hypotenuse^2
Given that the hypotenuse is 15cm, we can substitute this value into the equation:
x^2 + y^2 = 15^2
x^2 + y^2 = 225
Given that the second side (x) is twice the length of the third side (y), we can write:
x = 2y
Substituting this value into the equation, we get:
(2y)^2 + y^2 = 225
4y^2 + y^2 = 225
5y^2 = 225
y^2 = 225/5
y^2 = 45
Taking the square root of both sides, we find:
y = sqrt(45)
y ≈ 6.708 cm
Therefore, the exact length of the shortest side is approximately 6.708 cm.
3. To find the distance that the ball would travel before it hits the ground, we need to find the horizontal distance traveled by the ball when its height is 0.
The height equation is given as:
h = 1 + x - x^2/40
Setting h equal to 0, we can solve for x:
0 = 1 + x - x^2/40
Multiplying both sides by 40 to eliminate the fraction, we get:
0 = 40 + 40x - x^2
Rearranging it in the standard quadratic form, we get:
x^2 - 40x - 40 = 0
Now, we can solve this quadratic equation for the value of x. Factoring or using the quadratic formula, we obtain:
(x - 20)(x + 2) = 0
So, x = 20 or x = -2
Since the distance cannot be negative, the ball would travel 20 meters before it hits the ground.
1. To find the dimensions of the rectangle, we need to use the formulas for area and perimeter.
Let's assume the length of the rectangle is L metres and the width is W metres.
The formula for the area of a rectangle is A = L * W, where A is the area.
The formula for the perimeter of a rectangle is P = 2L + 2W, where P is the perimeter.
Given that the area is 125m² and the perimeter is 60m, we can set up the following equations:
A = L * W = 125 -------- Equation 1
P = 2L + 2W = 60 --------- Equation 2
To solve the equations simultaneously, we can use substitution or elimination.
Using substitution:
From Equation 2, we can isolate L:
2L = 60 - 2W
L = 30 - W
Substitute this value of L into Equation 1:
(30 - W) * W = 125
Expand and rearrange the equation:
30W - W² = 125
W² - 30W + 125 = 0
This is a quadratic equation that can be factored:
(W - 5)(W - 25) = 0
So, W = 5 or W = 25
Substitute these values of W back into Equation 2 to find the corresponding values of L:
When W = 5, L = 30 - 5 = 25
When W = 25, L = 30 - 25 = 5
Therefore, the dimensions of the rectangle can be 5m x 25m or 25m x 5m.
2. In a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides, known as the Pythagorean theorem.
Let's assume the length of the shortest side is x cm.
According to the problem, the second side is twice the length of the third side, so the third side would be x/2 cm.
Using the Pythagorean theorem:
(x/2)^2 + x^2 = 15^2
x^2/4 + x^2 = 225
5x^2/4 = 225
5x^2 = 900
x^2 = 900/5
x^2 = 180
x = √180
Now, simplify the square root:
x = √(36 * 5)
x = √36 * √5
x = 6√5 cm
So, the exact length of the shortest side is 6√5 cm.
3. To find how far the cricket ball would travel before it hits the ground, we need to determine the horizontal distance x at which the height h of the ball is equal to zero.
Given the equation h = 1 + x - x^2/40, we set h equal to zero and solve for x:
0 = 1 + x - x^2/40
Multiply both sides by 40 to get rid of the fraction:
40 * 0 = 40 * (1 + x - x^2/40)
0 = 40 + 40x - x^2
Rearrange the equation:
x^2 - 40x - 40 = 0
This is a quadratic equation that can be factored:
(x - 20)(x + 2) = 0
So, x = 20 or x = -2
Since distance cannot be negative in this context, the ball would travel 20 meters before it hits the ground.