1) The first leg of a right triangle is 6 cm longer than its second leg. If the hypotenuse has a length of 8 cm, find the exact lengths of the two legs.
First Leg:
Second Leg:
2) The length of a rectangle is 10 m longer than its width and the area of the rectangle is 480 m2. Find the exact dimensions of the rectangle.
W =
L =
I have no idea dude
1. 8^2 x^2(x+6)^2
=
2. 480 = (w+10)w
1. x^2 + (x+6)^2 = 8^2
x^2 + x^2 + 12x + 36 - 64 = 0
2x^2 + 12x - 28=0
x^2 + 6x - 14 = 0
x^2 + 6x = 14 , completing the square ....
x^2 + 6x + 9 = 14+9
(x+3)^2 = 23
x+3 = ± √23
x = -3 ± √23
but x has to be positive
so x = √23 - 3
so one leg is √23 - 3 , the other is √23 + 3
#2. solve the equation that kuai gave you
To solve these problems, we can use algebraic equations. Let's start with the first problem.
1) Let's assign variables to represent the lengths of the legs of the right triangle. We can let x represent the length of the second leg. Since the first leg is 6 cm longer, its length can be represented as x + 6.
According to the Pythagorean theorem, the sum of the squares of the lengths of the legs will be equal to the square of the length of the hypotenuse.
Applying this to our problem, we get the following equation:
(x + 6)^2 + x^2 = 8^2
Simplifying the equation, we get:
x^2 + 12x + 36 + x^2 = 64
Combining like terms, we get:
2x^2 + 12x + 36 = 64
Rearranging the equation, we have:
2x^2 + 12x - 28 = 0
We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, let's use factoring:
2(x^2 + 6x - 14) = 0
(x^2 + 6x - 14) = 0
Now we can factor the quadratic equation:
(x - 2)(x + 7) = 0
Setting each factor equal to zero, we get:
x - 2 = 0 or x + 7 = 0
Solving for x, we have:
x = 2 or x = -7
Since lengths cannot be negative, the second leg cannot be -7 cm. Therefore, the second leg has a length of x = 2 cm.
Now, we can find the length of the first leg by substituting this value back into our equation: first leg = x + 6 = 2 + 6 = 8 cm.
So, the lengths of the two legs of the right triangle are:
First Leg: 8 cm
Second Leg: 2 cm
Moving on to the second problem:
2) Let's assign variables to represent the dimensions of the rectangle. We can let W represent the width and L represent the length. According to the problem, the length is 10 m longer than the width, so we can say L = W + 10.
The area of a rectangle is calculated by multiplying the length and width, so we have the equation:
L * W = 480
Substituting the value of L from above, we get:
(W + 10) * W = 480
Expanding the equation, we get:
W^2 + 10W = 480
Rearranging the equation and setting it equal to zero, we have:
W^2 + 10W - 480 = 0
Now we can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. In this case, let's use factoring:
(W + 30)(W - 16) = 0
Setting each factor equal to zero, we get:
W + 30 = 0 or W - 16 = 0
Solving for W, we have:
W = -30 or W = 16
Since dimensions cannot be negative, the width cannot be -30 m. Therefore, the width has a value of W = 16 m.
Now, we can find the length by substituting this value back into our equation: L = W + 10 = 16 + 10 = 26 m.
So, the dimensions of the rectangle are:
Width (W): 16 m
Length (L): 26 m