i need help in solving these 2 problems...please help..thank you...
Your home is built on a square lot. To add more space to your yard, you purchase an additional 4 feet along the side of the property. The area of the lot is now 9600 square feet. What are the dimentions of the new lot??
The Garys have a triangular pennant of area 420in.^2 flying from the flagpole in their yard. The height of the triangle is 10 in. than 5 times the base of the triangle. What are the dimensions of the pennant??
For the first problem:
x=length of side before the addition
x+4=the length of the longer side after the the additional 4 feet is purchased.
The new area of the property is:
x(x+4)=9600
x^2 + 4x =9600
x^2 + 4x - 9600=0
which is:
(x-96)(x+100)=0
Ignore the negative solution for x since a negative number would not make much sense for a lot length.
x=96 and x+4=100
x * ( x + 4 ) = 9600
x^2 + 4 x - 9600 = 0
solve quadratic
area = (1/2) b h
420 = (1/2) b (5b+10)
840 = 5 b^2 + 10 b
b^2 + 2 b - 168 = 0
solve quadratic
Sure, I'd be happy to help you solve these problems! Let's take them one at a time.
Problem 1: Finding the dimensions of the new lot
To solve this problem, we can use the information given to set up an equation.
Let's assume the length of the original square lot is x feet. Since you purchased an additional 4 feet along one side, the new length will be x + 4 feet.
Now, we know that the area of the lot is 9600 square feet. The area of a square is calculated by multiplying its length by its width.
So, we can set up the following equation:
(x + 4) * x = 9600
To find the dimensions of the new lot, we need to solve this equation for x. Here's how you can do it:
1. Expand the equation: x^2 + 4x = 9600
2. Rearrange the equation into a quadratic form: x^2 + 4x - 9600 = 0
3. Factorize the quadratic equation: (x + 104)(x - 100) = 0
4. Set each factor equal to zero and solve for x:
x + 104 = 0 or x - 100 = 0
x = -104 or x = 100
Since the length cannot be negative, we disregard the negative solution.
Therefore, the length of the original lot is 100 feet. Adding 4 feet, the new length becomes 104 feet.
Hence, the dimensions of the new lot are 104 feet by 100 feet.
Problem 2: Finding the dimensions of the pennant
To solve this problem, we need to represent the given information in terms of variables.
Let's assume the base of the triangle is represented by b inches.
According to the problem, the height of the triangle is 10 inches more than 5 times the base. We can write this as 5b + 10.
The area of a triangle can be calculated using the formula: (1/2) * base * height.
So, we have the following equation:
(1/2) * b * (5b + 10) = 420 in^2
To find the dimensions of the pennant, we need to solve this equation for b. Here's how you can do it:
1. Expand and simplify the equation: (5/2) * b^2 + 5b - 210 = 0
2. Multiply the entire equation by 2 to get rid of the fraction: 5b^2 + 10b - 420 = 0
3. Factorize the quadratic equation: (b - 10)(5b + 42) = 0
4. Set each factor equal to zero and solve for b:
b - 10 = 0 or 5b + 42 = 0
b = 10 or b = -42/5
Since the base cannot be negative, we disregard the negative solution.
Therefore, the base of the triangle is 10 inches. Plugging this value into our equation for the height, we get:
Height = 5b + 10 = 5(10) + 10 = 50 + 10 = 60 inches
Hence, the dimensions of the pennant are a base of 10 inches and a height of 60 inches.