We are told that the height of a triangle is 3m less than its base. Therefore, if we let b be the base, the height can be expressed as h=b-3. We are also told that the area is 44 square meters. Let's draw a diagram to illustrate the situation.
The area of a triangle is half the product of its base and its height.
A=1/2bhWe can substitute A= 44 and h= b-3 into this formula and solve for b, the base of the triangle. Let's do it!
Note it's a quadratic equation. To solve this equation we will use the Quadratic Formula.
ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a
The variable in our equation is b. Let's replace it with x not to mistake it with the b in the Quadratic Formula.
b^2-3b-88=0
⇕
x^2-3x-88=0
Next, we need to identify the values of a, b, and c.
x^2-3x-88=0 ⇕ 1x^2+( - 3)x+( - 88)=0
We see that a= 1, b= - 3, and c= - 88. Let's substitute these values into the Quadratic Formula.
Now, we can go back to the previous notation and replace x with b. The solutions of our equation are b= 3 ± 192. Let's separate them into the positive and negative cases.
b=3 ± 19/2
b_1=3+19/2
b_2=3-19/2
b_1=22/2
b_2=- 16/2
b_1=11
b_2=- 8
Since b is the base of a triangle it must be positive. Therefore, the base of the triangle is 11m. The height is 3m less than its base, so we can subtract 3 from 11 to obtain the height.
Height: 11-3=8 m
