## Discussion

In the figure shown, ∆

*This question is included in Nova Math - Problem Set I: Geometry, question #2

*PST*is an isosceles right triangle, and*PS*= 2. What is the area of the shaded region URST?(A) | 4 |

(B) | ... |

(C) | ... |

(D) | ... |

(E) | ... |

(F) | ... |

The solution is

There is something wrong with this problem. The area of the bigger triangle is 2 (.5 x 2 x 2) and the smaller one is .5 as u mentioned therefore the shaded area must be equal to 1.5 or 3/2 as thats the difference of the bigger triangle minus the smaller one. Can someone please explain why u chose the answer given?

Mariam, here's how I approach this problem.

The area of URST = area PST - area PRU, as you noted. To find area PST, we need to know what TS is. Since PS is 2, by Phytagorean theorem, PT^2 + TS^2 = PS^2. Since PT = TS (isosceles triangle), we can substitute PT with TS, hence 2 x TS^2 = 2^2 =4. TS = √2.

Now we can calculate the area of PST, i.e., 1/2 x √2 x √2 = 1.

Next, let's calculate the area of the smaller triangle PUR. It is also an isosceles, since it is congruent with PTS. Hence PU = QR = UR = 1. Area PUR = 1/2 x 1 x 1 = 1/2.

Hence, the area of URTS = area PST - area PUR = 1 - 1/2 = 1/2, which is E.

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The area of URST = area PST - area PRU, as you noted. To find area PST, we need to know what TS is. Since PS is 2, by Phytagorean theorem, PT^2 + TS^2 = PS^2. Since PT = TS (isosceles triangle), we can substitute PT with TS, hence 2 x TS^2 = 2^2 =4. TS = √2.

Now we can calculate the area of PST, i.e., 1/2 x √2 x √2 = 1.

Next, let's calculate the area of the smaller triangle PUR. It is also an isosceles, since it is congruent with PTS. Hence PU = QR = UR = 1. Area PUR = 1/2 x 1 x 1 = 1/2.

Hence, the area of URTS = area PST - area PUR = 1 - 1/2 = 1/2, which is E.

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It will help us keep this service going. Thanks.

If your interested in a logical approach used to verify and will at least show you there's nothing wrong with the problem :)

The square root of 2 is often used as an example of an irrational number therefore this root is commonly known, at least it's approximate of 1.4

Breaking down the shaded area will give us (1/2) for the rectangle and (1/2)bh = (1/2)*(1/2)*(2-1.4) = (1/4) * (3/5) = (3/20) now we add the rectangle (1/2) and get (3/20) + (10/20) = (13/20)

Now we use 'eye balling' going along the answers since (1/2) is the only one which comes close to our approx. of (13/20).

Niels

[EDIT]

Since my goal isn't to give you confusion but take it away:

The logical conclusion is also that UT must be less than (1/2)

You can verify PT with 2*sin(45 degrees) = +/- 1.4 = sqrt(2) which comes from the mnemonic SOH sin = opposite / hypotenuse

The square root of 2 is often used as an example of an irrational number therefore this root is commonly known, at least it's approximate of 1.4

Breaking down the shaded area will give us (1/2) for the rectangle and (1/2)bh = (1/2)*(1/2)*(2-1.4) = (1/4) * (3/5) = (3/20) now we add the rectangle (1/2) and get (3/20) + (10/20) = (13/20)

Now we use 'eye balling' going along the answers since (1/2) is the only one which comes close to our approx. of (13/20).

Niels

[EDIT]

Since my goal isn't to give you confusion but take it away:

The logical conclusion is also that UT must be less than (1/2)

You can verify PT with 2*sin(45 degrees) = +/- 1.4 = sqrt(2) which comes from the mnemonic SOH sin = opposite / hypotenuse

Posted: 01/19/2012 17:30

What's tha answer bra

Posted: 01/26/2012 11:01

Tay, please take a look at previous posts from me and Niels.

I feel the answer to this question be 0.7 .....I have attached my workspace here..... I applied the formula of the area of trapezium......and UT comes out to be 0.4......ST =2.4 and UR= 1...... Then applying the formula......we get 0.2*3.4=0.68

Jayneel,

The area we want is equal to the difference of triangles

1/2(PT*TS) - 1/2(PU*UR)

1/2(PT*TS) - 1/2(1*1)

1/2(PT*TS) - 1/2(1)

1/2(PT*TS) - 1/2

Phytagoras says a^2 + b^2 = c^2 since we're dealing here with an isosceles triangle we take a^2 + a^2 = c^2 thus

2a^2 = c^2

2a^2 = 2^2

2a^2 = 4

a^2 = (1/2)4

a^2 = 2

a = sqrt(2) = PT = TS

Now we continue

1/2(sqrt(2)*sqrt(2)) - 1/2

1/2(sqrt(4)) - 1/2

1/2(2) - 1/2

1 - 1/2

1/2

Niels

The area we want is equal to the difference of triangles

1/2(PT*TS) - 1/2(PU*UR)

1/2(PT*TS) - 1/2(1*1)

1/2(PT*TS) - 1/2(1)

1/2(PT*TS) - 1/2

Phytagoras says a^2 + b^2 = c^2 since we're dealing here with an isosceles triangle we take a^2 + a^2 = c^2 thus

2a^2 = c^2

2a^2 = 2^2

2a^2 = 4

a^2 = (1/2)4

a^2 = 2

a = sqrt(2) = PT = TS

Now we continue

1/2(sqrt(2)*sqrt(2)) - 1/2

1/2(sqrt(4)) - 1/2

1/2(2) - 1/2

1 - 1/2

1/2

Niels

I feel the answer to this question be 0.7 .....I have attached my workspace here..... I applied the formula of the area of trapezium......and UT comes out to be 0.4......ST =2.4 and UR= 1...... Then applying the formula......we get 0.2*3.4=0.68

I feel the answer to this question be 0.7 .....I have attached my workspace here..... I applied the formula of the area of trapezium......and UT comes out to be 0.4......ST =2.4 and UR= 1...... Then applying the formula......we get 0.2*3.4=0.68