Question

Trapezoid $ABCD$ has parallel sides $\overline{AB}$ of length $33$ and $\overline {CD}$ of length $21 .$ The other two sides are of lengths $10$ and $14 .$ The angles $A$ and $B$ are acute. What is the length of the shorter diagonal of $ABCD ?$

$10\sqrt{6}$

$25$

$8\sqrt{10}$

$18\sqrt{2}$

$26$

Answer: B

Concepts:trapezoid Pythagorean Theorem

Difficulty rating: 1790

Solution:

Let the feet of the perpendiculars from $D$ and $C$ to $AB$ be $E$ and $F$ , respectively. As drawn, take $AD=10$ and $BC=14$ ; interchanging the two legs only reflects the trapezoid.

Let $AE=x$ and let the altitude be $h$ . Since $EF=CD=21$ and $AB=33$ , we have $FB=12-x$ .

The two right triangles give $10^2=x^2+h^2$ and $14^2=(12-x)^2+h^2.$ Subtracting yields $96=144-24x$ , so $x=2$ and $h^2=96$ .

The shorter diagonal is $AC$ , whose horizontal displacement is $AE+EF=2+21=23$ . Therefore $AC=\sqrt{23^2+96}=\sqrt{625}=25.$

Thus, the correct answer is B .

2014 AMC 10B Problem 21

Answer: B

Difficulty rating: 1790

Solution:

Problem 21 in Other Years