the altitude of a triangle is 1 cm less than its base, if the area is 105 cm2. what are the lengths of the base and the altitude
base --- x
height = x-1
area of triangle = (1/2) base x height
(1/2)(x)(x+1) = 105
x^2 + x = 210
x^2 + x - 210 = 0
(x+15)(x-14) = 0
x = -15 or x = 14, but x > 0
base is 14 cm, height is 15 cm
check:
area = (1/2)(15)(14) = 105
all is good!
To find the lengths of the base and the altitude of a triangle, we can use the formula for the area of a triangle.
Let's denote the base of the triangle as 'b' and the altitude as 'h'.
Given that the altitude is 1 cm less than the base, we can express it as h = b - 1.
The formula for the area of a triangle is: Area = (1/2) * base * height.
Since we know the area of the triangle is 105 cm^2, we can substitute the given values into the formula:
105 = (1/2) * b * (b - 1)
Now, let's solve this equation step by step to find the lengths of the base and the altitude.
Step 1: Multiply both sides of the equation by 2 to eliminate the fraction:
2 * 105 = b * (b - 1)
210 = b^2 - b
Step 2: Rearrange the equation to form a quadratic equation:
b^2 - b - 210 = 0
Step 3: Factorize the quadratic equation or use the quadratic formula to find the values of 'b'.
Since -210 has factors of -14 and 15 that differ by 1, we can rewrite the equation as:
(b - 15)(b + 14) = 0
Setting each factor equal to zero, we have:
b - 15 = 0 or b + 14 = 0
Solving these equations, we get:
b = 15 or b = -14
Since the length of a triangle's side cannot be negative, we discard the solution b = -14.
Therefore, the length of the base of the triangle is 15 cm.
To find the length of the altitude, we substitute this value back into the expression h = b - 1:
h = 15 - 1
h = 14 cm
So, the lengths of the base and the altitude are 15 cm and 14 cm, respectively.