i need help finding the base. this is the equation i got after work it out halfway.
42/h+h+sqroot(42/h)^2+h^2=20 ft
how can i find B & H. How should i go about. Thanks.
Isnt the sqrt of (42/h)^2 just 42/h ?
Then, multipy both sides of the equation by h, that gives you a third degree equation. I have no idea how b relates to h.
Are you certain your work is correct? This is an odd equation. It bothers me that the right side dimension is ft, which means that on the left, 42/h is in feet, h^2 is in feet, and h is in feet, a most unusual occurance.
Here's my worK. im trying to find the dimension of this rectangle.
A = (1/2) bh = 21 sq ft
bh/h = 42/4 sq ft.
a^2+b^2 = C^2
P= b+h+ sqroot b^2+h^2 = 20ft.
plug it in
this is how i got the equation
42/h+h+sqroot(42/h)^2+h^2=20 ft
Here's the question to the problem:"
The perimeter of a rectangle is 20 ft and its area is 21 sq feet. what are the dimension of the rectangle?
1/2 bh=21 is not the formula for a rectangle. And, for a rectangle the perimeter is 2(b+h)
1/2 bh=21 is not the formula for a rectangle. And, for a rectangle the perimeter is 2(b+h)
To find the dimensions of the rectangle, we can solve the system of equations given:
Perimeter: P = b + h + √(b^2 + h^2) = 20 ft
Area: A = b * h = 21 sq ft
Let's start by solving the area equation for one variable, say b. We have:
b = 21 / h
Substituting this into the perimeter equation, we get:
20 = (21 / h) + h + √((21 / h)^2 + h^2)
Now, let's simplify this equation:
1. Simplify the square root term:
√((21 / h)^2 + h^2) = √(441 / h^2 + h^2) = √((441 + h^4) / h^2) = (441 + h^4)^0.5 / h
2. Substitute the square root term back into the equation:
20 = (21 / h) + h + (441 + h^4)^0.5 / h
Multiplying through by h to clear the fraction, we get:
20h = 21 + h^2 + (441 + h^4)^0.5
Now, we have a third-degree equation, which may not have an easy algebraic solution. It may require numerical methods or approximation techniques to find the values of b and h.
If you provide the specific requirements of the problem, I can help you further to find an approximation for the dimensions of the rectangle.
To find the dimensions of the rectangle given the perimeter and area, we can use the formulas for perimeter and area of a rectangle.
Let's start by using the formula for the area of a rectangle, which is A = length * width. In this case, we have A = 21 sq ft.
Next, let's use the formula for the perimeter of a rectangle, which is P = 2(length + width). In this case, we have P = 20 ft.
Now let's solve for the dimensions of the rectangle.
1. From the area equation, we can solve for one variable in terms of the other. Let's solve for length, L, in terms of width, W:
A = L * W
21 = L * W
L = 21 / W
2. Substitute the value of L into the perimeter equation, and solve for W:
P = 2(L + W)
20 = 2(21 / W + W)
10 = 21 / W + W
Multiply through by W to get rid of the denominator:
10W = 21 + W^2
Rearrange the equation:
W^2 + 10W - 21 = 0
3. We now have a quadratic equation. Solve it to find the possible values of W:
Factoring or using the quadratic formula, we find that W = 1 or W = -21. However, since we are dealing with dimensions (which cannot be negative), we discard the negative solution. Therefore, W = 1 ft.
4. Substitute the value of W back into the equation for length to find L:
L = 21 / W
L = 21 / 1
L = 21 ft.
So the dimensions of the rectangle are width = 1 ft and length = 21 ft.