1. The area of a rectangle is 12 and the width is 3/4 the length. What are the dimensions?
2. One leg of a right triangle is 2 1/2 times the other and the area is 20. What are the dimensions?
1.
let x = the length
let (3/4) x = the width
The area of a rectangle = width * height.
area = width*height
12 = x*(3/4)x
12 = (3/4)x^2
12/(3/4) = x^2
16 = x^2
x=4. The answer x=-4 does not make sense in the context of the problem.
2.
let x = leg one
let 2.5x = leg two
The area of a triangle = base*height/2
area = base*height/2
20 = 2.5x*x/2
20 = 2.5x^2/2
20/2.5 = x^2/2
8 = x^2/2
16 = x^2
x=4. Again, the answer x=-4 does not make sense in the context of the problem.
1. Let's assume the length of the rectangle is x.
Given that the width is 3/4 the length, we can write the equation: width = 3/4 * length.
Since the area of the rectangle is 12, we can also write the equation: area = length * width.
Substituting the value of width from the first equation into the second equation, we have:
12 = x * (3/4 * x)
To solve for x, we multiply the terms inside the parentheses:
12 = (3/4) * x^2
Next, we divide both sides of the equation by (3/4):
x^2 = (4/3) * 12
Simplifying the right side of the equation, we have:
x^2 = 16
Taking the square root of both sides, we find:
x = ±4
Since the length cannot be negative, the length of the rectangle is 4.
Substituting this value into the first equation, we can find the width:
width = 3/4 * 4
width = 3
Therefore, the dimensions of the rectangle are length = 4 and width = 3.
2. Let's assume one leg of the right triangle is x.
Given that the other leg is 2 1/2 times the first leg, we can write the equation: leg 2 = 2 1/2 * leg 1.
Since the area of the right triangle is 20, we can also write the equation: area = 1/2 * (leg 1) * (leg 2).
Substituting the value of leg 2 from the first equation into the second equation, we have:
20 = 1/2 * (leg 1) * (2 1/2 * leg 1)
To solve for leg 1, we simplify and multiply the terms inside the parentheses:
20 = 1/2 * (leg 1)^2 * (5/2)
Next, we multiply both sides of the equation by 2/5 to isolate (leg 1)^2:
(leg 1)^2 = 20 * 2/5
Simplifying the right side of the equation, we have:
(leg 1)^2 = 8
Taking the square root of both sides, we find:
leg 1 = ±√8
Since the length cannot be negative, the length of the leg is √8.
Substituting this value into the first equation, we can find the length of the other leg:
leg 2 = 2 1/2 * √8
leg 2 = 2.5 * 2.83
leg 2 ≈ 7.07
Therefore, the dimensions of the right triangle are leg 1 ≈ √8 and leg 2 ≈ 7.07.
To find the dimensions of the rectangle and right triangle, we can use algebraic equations.
1. Rectangle:
Let the length of the rectangle be "l" and the width be "w". According to the given information, we have two conditions:
(i) The area of the rectangle is 12: l * w = 12
(ii) The width is 3/4 the length: w = (3/4) * l
To find the dimensions, we can solve these equations simultaneously.
Substituting the value of w from equation (ii) into equation (i), we get:
l * [(3/4) * l] = 12
(3/4) * l^2 = 12
Multiplying both sides by 4/3:
l^2 = (12 * 4) / 3
l^2 = 48 / 3
l^2 = 16
Taking the square root of both sides:
l = √16
l = 4
Substituting the value of l into equation (ii) to find w:
w = (3/4) * 4
w = 12/4
w = 3
Therefore, the dimensions of the rectangle are length = 4 and width = 3.
2. Right Triangle:
Let one leg of the right triangle be "x" and the other leg be "y". According to the given information, we have two conditions:
(i) The area of the triangle is 20: (1/2) * x * y = 20
(ii) One leg is 2 1/2 (or 5/2) times the other: x = (5/2) * y
Again, we can solve these equations simultaneously to find the dimensions.
Substituting the value of x from equation (ii) into equation (i), we get:
(1/2) * [(5/2) * y] * y = 20
(5/4) * y^2 = 20
Multiplying both sides by 4/5:
y^2 = (20 * 4) / 5
y^2 = 80 / 5
y^2 = 16
Taking the square root of both sides:
y = √16
y = 4
Substituting the value of y into equation (ii) to find x:
x = (5/2) * 4
x = 20 / 2
x = 10
Therefore, the dimensions of the right triangle are one leg = 10 and the other leg = 4.