Bunuel wrote:

In the figure above, what is the perimeter of triangle OPQ?

(A) 4 + 2√2

(B) 8 + 4√2

(C) 6 + 2√5

(D) 6 + 6√2

(E) 6√2 + 2√10

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To describe this method takes much longer than to use it.

(This problem took one minute.)

Draw a box / rectangle around the triangle.

There are three new right triangles around the original triangle.

If you can calculate the lengths of the legs of the new triangles in your head, no need for coordinates.

If not, see footnote.*

The hypotenuse of each new right triangle is one side of the original triangle.

Find the length of those original sides with the Pythagorean theorem or special triangle properties.

1) Length of side OP

Side lengths of right ∆ KPO

KO = 4

KP = 4

OP = ?

A right isosceles triangle has side lengths in ratio

\(x : x: x\sqrt{2}\)\(4\) corresponds with \(x\)

OP corresponds with

\(x\sqrt{2}\)Length of OP is

\(4\sqrt{2}\)OR:

\(4^2 + 4^2 = OP^2\)

\(32 = OP^2\)

\(\sqrt{16 * 2} = \sqrt{OP^2}\)

\(OP = 4\sqrt{2}\)2) Length of side OQ

Right ∆ MOQ is isosceles, too

Side lengths in ratio

\(x : x: x\sqrt{2}\)Its sides MO and MQ = \(2\)

Side

\(OQ = 2\sqrt{2}\)OR

\(2^2 + 2^2 = OQ^2\)

\(8 = OQ^2\)

\(\sqrt{4 * 2} = \sqrt{OQ^2}\)

\(OQ = 2\sqrt{2}\)3) Length of side PQ

One leg, LQ, of right ∆ LPQ = \(2\)

Other leg, LP = \(6\)

PQ = ?

\(2^2 + 6^2 = PQ^2\)

\(4 + 36 = PQ^2\)

\(40 = PQ^2\)

\(\sqrt{4 * 10} = \sqrt{PQ^2}\)

\(PQ = 2\sqrt{10}\)Perimeter of original ∆ OPQ = sum of lengths of three sides calculated above

\(4\sqrt{2} + 2\sqrt{2} + 2\sqrt{10}=\)

\(6\sqrt{2} +2\sqrt{10}\)

ANSWER E

*

Coordinates of vertices of drawn rectangle

K has the same x-coordinate as P, and the same y-coordinate as O

--K is on the same vertical line as P

--K is on the same horizontal line as O

L has the same x-coordinate as Q, and the same y-coordinate as P

--L is on the same vertical line as Q

--L is on the same horizontal line as P

M has the same x-coordinate as Q, and the same y-coordinate as O

-- M is on the same vertical line as Q

-- M is on the same horizontal line as O

To calculate side lengths, generally:

Each end point is a vertex, and each vertex has one coordinate that is equal to the corresponding coordinate for the other vertex.

Ignore the identical number (it means they're on the same horizontal or vertical line).

Subtract the coordinate numbers that are different - the absolute value of the difference is the length of the line

Length of KO: Its endpoints lie on the x-axis, y = 0.

Both have y-coordinates of 0. Ignore them.

Subtract x-coordinate of K from x-coordinate of O

0 - (-4) = 4 = KO

Length of KP

Both have x-coordinates of -4. Ignore them.

Subtract: (4 - 0) = 4

Length of KP = 4

And so on . . .