2012 AMC 10B Problem 12

Question

Point

B

is due east of point

A.

Point

C

is due north of point

B.

The distance between points

A

and

C

is

10\sqrt 2

meters, and

\angle BAC = 45^\circ.

Point

D

is

20

meters due north of point

C.

The distance

AD

is between which two integers?

30

and

31

31

and

32

32

and

33

33

and

34

34

and

35

Answer: B

Concepts:Pythagorean Theorem special right triangle estimation

Difficulty rating: 1140

Small hint:

Triangle

ABC

is

45

-

45

-

90

Big hint:

Find

AB

and

BD

, then use the Pythagorean Theorem

Solution:

We know

AB

and

BC

are perpendicular, so

AB^2 + BC^2 = (10\sqrt 2)^2 = 200.

Also, as

\angle BAC = 45^\circ,

we know that

\triangle ABC

is an isosceles right triangle, so

AB = BC ,

making

2AB^2 = 200.

Thus,

AB = BC = 10.

As such, we know that

BD= BC+CD = 30.

Thus, by the Pythagorean Theorem, we have that

\begin{aligned}AD^2 &= AB^2+BD^2 \\&= 10^2+30^2\\&=1000\end{aligned}

Thus, since

31^2 \lt AD^2 \lt 32^2,

we have

31 \lt AD\lt 32

Thus, the correct answer is B .

2012 AMC 10B Problem 12

Answer: B

Difficulty rating: 1140

Small hint:

Big hint:

Solution:

Problem 12 in other years