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The length of a side of square PQRS above is 2 cm. If ST = 2 cm and PT [#permalink]
PST=PTO - SOP

now, PTO = \(\frac{1}{2}*\sqrt{2}*(2+\sqrt{2}) = \sqrt{2} + 1\)
And, SOP = \(\frac{1}{2}*\sqrt{2}*\sqrt{2} = 1\)

Therefore, PST = \(\sqrt{2} + 1 -1 = \sqrt{2}\)

Answer: E
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Re: The length of a side of square PQRS above is 2 cm. If ST = 2 cm and PT [#permalink]
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Bunuel wrote:

The length of a side of square PQRS above is 2 cm. If ST = 2 cm and PT = RT, what is the area of ∆PST?

A. 3 cm^2
B. 2√2 cm^2
C. 2 cm^2
D. √3 cm^2
E. √2 cm^2




We see that the area of triangle PST is equal to the area of triangle POT minus the area of triangle POS.

Notice that both triangles POT and POS are right triangles. The base of both triangles is PO, which has a length of ½ x 2√2 = √2. The height of triangle POS is OS, which also has a length of √2. The height of triangle POT is OT, which has a length of 2 + √2. So, the area of triangle POT is ½ x √2(2 + √2) and that of triangle POS is ½ x √2(√2). Therefore, the area of triangle PST is:

½ x √2(2 + √2) - ½ x √2(√2) = ½ x √2(2 + √2 - √2) = ½ x √2(2) = √2

Answer: E
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The length of a side of square PQRS above is 2 cm. If ST = 2 cm and PT [#permalink]
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Hello!

Hopefully you see that the area of triangle PST can be found by subtracting the area of triangle POS from the area of triangle OTP.

To solve the problem, you have to find the area of triangle OTP and of triangle POS.

In a problem like this, you want to start with the information given to you.

Knowing that the sides of the square are 2, we can derive the length of the diagonal PR and of OP and OS.

This will also give us the length of OT because we know that ST is 2. This will allow us to calculate the area of triangle OTP.

We can also use that information to calculate the area of triangle POS and subtract it from OTP to get the area of triangle PST.

The key here is knowing some rules about squares.

If a square has side length x, the diagonal length will always be x\(\sqrt{2}\)

You can also derive the length of the diagonal PR using the Pythagorean theorem

\(a^2 + b^2 = c^2\)

\(2^2 + 2^2 = c^2\)

\(4 + 4 = c^2\)

\(8 = c^2\\
\)

\(c = \sqrt{8} = 2\sqrt{2}\) because \(\sqrt{8} = \sqrt{4}\) x \(\sqrt{2} = 2\sqrt{2}\)

The diagonals of a square bisect each other, meaning each diagonal divides the other into two equal halves.

That means that PO = OR = OS = \(\frac{2\sqrt{2}}{2}\) = \( \sqrt{2}\)

If OS = \(\sqrt{2}\) and we are given that ST = 2 cm, OT = 2 + \(\sqrt{2}\)

If OP = \(\sqrt{2}\), use the formula for the area of a triangle \(\frac{1}{2}\)(base)(height) to find the area:

\(\frac{1}{2}\)(\(\sqrt{2}\))(2 + \(\sqrt{2}\)) = \(\frac{1}{2}(2\sqrt{2} + 2) = \sqrt{2} + 1\)

The area of triangle OTP is \(\sqrt{2} + 1\)

If we can find the area of triangle POS, we can subtract it from the area of OTP to find the area of triangle PST

Not only do the diagonals of a square bisect each other, they are perpendicular to each other and create four right angles at the center. This means that angle POS is a right angle and triangle POS is a right triangle.

We know that OS = OP = \(\sqrt{2}\)

The area of POS is therefore \(\frac{1}{2}(\sqrt{2})(\sqrt{2})\)\( = \frac{1}{2}(2) = 1\)

Area of PST = Area of OTP - Area of POS

Area of PST = \(\sqrt{2} + 1 - 1 = \sqrt{2}\)

The answer is (E) \(\sqrt{2}\) cm\(^2\)
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The length of a side of square PQRS above is 2 cm. If ST = 2 cm and PT [#permalink]
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