Geometry Problems of the Day

The following problems were taken from the GEOMETRY (COMMON CORE) Regents Exam given on Thursday, January 26, 2017.
Previous problems can be found here.

Part 1

15. In circle O, secants ADB and AEC are drawn from external point A such that points D, B, E, and C are on circle O. If AD = 8, AE = 6, and EC is 12 more than BD, the length of BD is

(2) 22.

I sketched the image before solving so it isn't drawn to scale. This is a problem that you might want to work backward from the answers, plugging numbers into the equations.

The rule to remember is that (AD)(AB) = (AE)(AC)
or (8)(x + 8) = (6)(x + 12 + 6)
Simplify: 8x + 64 = 6x + 108
2x = 44
x = 22

Check: (8)(22 + 8) = (6)(22 + 12 + 6)
(8)(30) = (6)(40)
240 = 240

16. A parallelogram is always a rectangle if

(1) the diagonals are congruent
Congruent diagonals is true of rectangles and isosceles trapezoids (which aren't parallelograms, so they don't count here).
Opposite angles congruent and diagonals bisecting each other are true of all parallelograms, not just rectangles.
Diagonals intersecting at right angles proves that the quadrilateral is a rhombus, not a rectangle.




Continue to the next problems.