The altitude of a triangle is 2cm longer than its base. The area of the triangle is 60 cm*2. What are the dimensions of the triangle?
Did you try an if so how?
Area= 1/2 x b x h h = b +2
= 1/2 x b x b +2
60= 1/2 x b x b + 2
120=b x b + 2
118= 2b
59=b
h=59 + 2 =61
To solve this problem, we can use the formula for the area of a triangle, which is given by:
Area = (1/2) * base * altitude
Let's assign variables to the dimensions of the triangle. Let's say the base of the triangle is 'b' cm, and the altitude is 'h' cm.
Given that the altitude of the triangle is 2 cm longer than its base, we can write the equation:
h = b + 2
Now, we can substitute the given area into the formula:
60 = (1/2) * b * h
Substituting h with b + 2:
60 = (1/2) * b * (b + 2)
Now, we can simplify the equation by multiplying both sides by 2 to eliminate the fraction:
120 = b * (b + 2)
Expanding the equation:
120 = b^2 + 2b
Rearranging the equation:
b^2 + 2b - 120 = 0
To solve this quadratic equation, we can factor it or use the quadratic formula. Let's factor it:
(b + 12)(b - 10) = 0
This gives us two possible values for the base:
b = -12 or b = 10
Since the dimensions of a triangle cannot be negative, we discard -12 as a solution. Therefore, the base of the triangle is 10 cm.
Now, we can find the altitude by substituting the base into the equation h = b + 2:
h = 10 + 2
h = 12 cm
So, the dimensions of the triangle are:
Base = 10 cm
Altitude = 12 cm