The area of the trapezoid shown below is 90 square units. find the value of X. The sides are x and 2x and the base is x+6. There are 2 90 degree angles at the base. x and 2x are parallel. the top is like one half of a roof. I will try to add a picture if this doesn’t make sense. I have tried all morning to upload a picture and can’t figure out what I am doing wrong. It does look like the trapezoid on this page
http://www.mathgoodies.com/lessons/vol1/area_trapezoid.html under example one only b1 is X B2 is 2X and h=x+6. I know I have to solve it using the quadric formula or completing the square, that is what the chapter is about.
I hope the picture on the website helps. I tried to find a place to attach this to my original question but I can’t find one.
The Correct Answer and Explanation is:
Thanks for the clear description and the reference to the trapezoid example on Math Goodies! I understand your trapezoid has bases $b_1 = x$ and $b_2 = 2x$, and height $h = x + 6$, with an area of 90 square units. You mentioned two right angles at the base, which matches the perpendicular height in the trapezoid.
Step 1: Write down the area formula for a trapezoid
$$
\text{Area} = \frac{1}{2} (b_1 + b_2) \times h
$$
Plug in what we know:
$$
90 = \frac{1}{2} (x + 2x) \times (x + 6)
$$
Simplify inside the parentheses:
$$
90 = \frac{1}{2} (3x)(x + 6)
$$
Multiply by 2 on both sides to clear the fraction:
$$
180 = 3x (x + 6)
$$
Expand the right side:
$$
180 = 3x^2 + 18x
$$
Step 2: Rearrange the equation
Bring all terms to one side to form a quadratic:
$$
3x^2 + 18x – 180 = 0
$$
Divide the entire equation by 3 to simplify:
$$
x^2 + 6x – 60 = 0
$$
Step 3: Solve the quadratic equation $x^2 + 6x – 60 = 0$
You can solve by the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
$$
Where $a=1, b=6, c=-60$:
Calculate the discriminant:
$$
\Delta = b^2 – 4ac = 6^2 – 4(1)(-60) = 36 + 240 = 276
$$
So,
$$
x = \frac{-6 \pm \sqrt{276}}{2}
$$
Simplify $\sqrt{276}$:
$$
\sqrt{276} = \sqrt{4 \times 69} = 2\sqrt{69}
$$
So,
$$
x = \frac{-6 \pm 2\sqrt{69}}{2} = -3 \pm \sqrt{69}
$$
Step 4: Choose the positive solution
Since $x$ represents lengths, it must be positive:
$$
x = -3 + \sqrt{69}
$$
Approximate $\sqrt{69} \approx 8.31$:
$$
x \approx -3 + 8.31 = 5.31
$$
Final answer:
$$
\boxed{x \approx 5.31}
$$
Explanation:
You started with the trapezoid’s area formula, substituting given values for the bases $x$ and $2x$ and height $x+6$. Multiplying and simplifying yielded a quadratic equation. Dividing by 3 made it easier to handle, and then applying the quadratic formula gave two solutions — one positive and one negative. Because length cannot be negative, the acceptable value for $x$ is approximately 5.31 units.
This value means the shorter base is about 5.31 units, the longer base about 10.62 units, and the height about 11.31 units, satisfying the trapezoid’s area condition of 90 square units.
Quadratic equations often arise in geometry problems involving variables for lengths. Using the quadratic formula is a reliable method when factoring is difficult or impossible. This problem showcases the power of algebraic methods to find geometric measurements precisely.
