Question

Three congruent isosceles triangles $DAO$, $AOB$ and $OBC$ have $AD=AO=OB=BC=10$ and $AB=DO=OC=12$. These triangles are arranged to form trapezoid $ABCD$, as shown. Point $P$ is on side $AB$ so that $OP$ is perpendicular to $AB$.

Point $X$ is the midpoint of $AD$ and point $Y$ is the midpoint of $BC$. When $X$ and $Y$ are joined, the trapezoid is divided into two smaller trapezoids. The ratio of the area of trapezoid $ABYX$ to the area of trapezoid $XYCD$ in simplified form is $p:q$. Find $p+q$.

Answer 1

p+q = 12

Explanation:

Let's draw the diagram obtained from the given information. Find attached the diagram.

When X and Y divides the line AD and line BY into half respectively, the diagram splits into two trapezoid.

See attachment for diagram.

Ratio of area ABYX to XYCD = p:q

p+q = ?

Area of trapezoid = ½(base 1 + base 2)height

For trapezoid ABYX:

Base 1 = AB = 12, Base 2 = XY

Using Pythagoras theorem to find height of ∆AOB

AO² = AP² + PO²

PO² = (10²-6²) = (100-36)

PO = √64 = 8

height of trapezoid ABYX =h = ½ PO

= 8÷2 = 4

base of triangles in trapezoid ABYX:

base² = 5² -4² = (25-16)

base = √9 = 3

XY = 12+3+3 = 18

Area of trapezoid ABYX = ½(AB + XY)height

= ½(12+18)×4

= ½(120) = 60

Area of trapezoid XYDC = ½(XY + DC)height

Height of both trapezoid = ½ PO = 4

base of triangles in trapezoid XYCD = 3

DC = 3+18+3 = 24

= ½(18+24)×4

= ½(168) = 84

Ratio of both areas

p:q = 60: 84

p:q = 5:7

p+q = 5+7

p+q = 12