Question

ABCD is a square with sides of length 10 units. OCD is an isosceles triangle with base CD. OC cuts AB at point Q and OD cuts AB at point P. The area of trapezoid PQCD is 80 square units. The altitude from O of the triangle OPQ is:

• A. 12
• B. 13
• C. 14
• D. 15

Hint:

Triangle OPQ is similar to triangle OCD because PQ (on AB) is parallel to CD; use the ratio of their altitudes to set up an area equation.

Answer 1

The Correct Option is D

Step 1: Use the ratio of areas of similar triangles instead of expanding algebra.
Triangles OPQ and OCD are similar since PQ is parallel to CD. If k is the ratio of their altitudes (altitude of OPQ over altitude of OCD), then the ratio of their areas is $k^2$.

Step 2: Express the altitudes.
Let h be the altitude of triangle OPQ (what we want). Since AB sits exactly 10 units above CD, the altitude of the full triangle OCD is $h + 10$. So $k = \dfrac{h}{h+10}$.

Step 3: Write the trapezoid area as a fraction of the big triangle.
Area of trapezoid PQCD = Area(OCD) minus Area(OPQ) = Area(OCD)$\left(1 - k^2\right)$, and Area(OCD) $= \dfrac{1}{2}(10)(h+10) = 5(h+10)$. So \[ 5(h+10)\left[1 - \left(\frac{h}{h+10}\right)^2\right] = 80 \]
Step 4: Simplify using the difference of squares.
$1 - \left(\dfrac{h}{h+10}\right)^2 = \dfrac{(h+10)^2 - h^2}{(h+10)^2} = \dfrac{20h + 100}{(h+10)^2}$, so the equation becomes \[ \frac{5(20h+100)}{h+10} = 80 \] \[ 5(20h + 100) = 80(h+10) \]
Step 5: Solve the linear equation.
\[ 100h + 500 = 80h + 800 \] \[ 20h = 300 \] \[ h = 15 \]
Final Answer:
\[ \boxed{15} \]