In the figure the sum of the distances AD and BD is

1987 AMC 12 Problem 8

Attempt Problem 8 of the 1987 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1987 AMC 12 solutions, or check the answer key.

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In the figure the sum of the distances

AD

and

BD

between

10

and

11

12

between

15

and

16

between

16

and

17

17

Use the

3

4

5

right triangle to find

BD

The horizontal and vertical displacements from

A

D

are

10

and

4

Triangle

BCD

is a

3

4

5

right triangle, so

BD=5.

From

A

D,

the horizontal displacement is

13-3=10

and the vertical displacement is

4,

hence

AD=\sqrt{10^2+4^2}=\sqrt{116}.

Since

10\lt\sqrt{116}\lt11,

the sum

AD+BD

is between

15

and

16.

Thus the correct answer is C.