Discussion
In the figure shown, ∆PST is an isosceles right triangle, and PS = 2. What is the area of the shaded region URST?
*This question is included in Nova Math - Problem Set I: Geometry, question #2
| (A) | 4 |
| (B) | ... |
| (C) | ... |
| (D) | ... |
| (E) | ... |
| (F) | ... |
The solution is
Posted: 01/12/2012 13:46
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?
Posted: 01/12/2012 14:21
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.
Posted: 01/12/2012 19:44
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.
Posted: 11/17/2012 03:00
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
Posted: 11/17/2012 05:45
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
Posted: 11/17/2012 03:00
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
Posted: 11/17/2012 03:01
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