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18 Apr 2019, 00:36
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
32% (03:10) correct
68%
(03:06)
wrong
based on 57
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Three unit squares and two line segments connecting two pairs of vertices are shown. What is the area of triangle ABC?
(A) 1/6
(B) 1/5
(C) 2/9
(D) 1/3
(E) \(\frac{\sqrt{2}}{4}\)
Attachment:
344ad68ce1c966777c56c860c4fc9b0da8bb06cb.png [ 5.12 KiB | Viewed 5914 times ]
21 Apr 2019, 21:42
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There are a few ways to solve - for example, if you look at the long line containing BC, it rises 2 units when we go across 1 unit. The line containing AC falls 1 unit when we go across 2 units, the perfect reverse of what BC does. That's what it means for two lines to be perpendicular (in coordinate geometry terms, the slope of BC is 2, and the slope of AC is -1/2, which are negative reciprocals). So ABC is a right triangle, with a right angle at C.
But if we look at the largest triangle that contains ABC, the triangle using the entire horizontal line at the top of the picture, and the vertical line at the far right of the picture, that triangle also has a 90 degree angle and has the same angle at A that ABC has, so it shares two angles with ABC, and therefore is similar to triangle ABC. Since the legs of this large triangle are clearly in a 2 to 1 ratio, so are the legs of ABC.
So if we just call the two legs of ABC 'b' and '2b', the area of ABC is (b)(2b)/2 = b^2, which is what we want to find. And since the hypotenuse of ABC is 1, we can solve for b^2 using Pythagoras:
b^2 + (2b)^2 = 1
5b^2 = 1
b^2 = 1/5
Another approach which is longer, but which is convenient and flexible: you can just put everything in an xy-plane, with the point (0, 0) at the bottom left of the diagram. Then by finding the equations of each line, you can solve for their intersection point, and from there it's easy to find the area of ABC.
But if we look at the largest triangle that contains ABC, the triangle using the entire horizontal line at the top of the picture, and the vertical line at the far right of the picture, that triangle also has a 90 degree angle and has the same angle at A that ABC has, so it shares two angles with ABC, and therefore is similar to triangle ABC. Since the legs of this large triangle are clearly in a 2 to 1 ratio, so are the legs of ABC.
So if we just call the two legs of ABC 'b' and '2b', the area of ABC is (b)(2b)/2 = b^2, which is what we want to find. And since the hypotenuse of ABC is 1, we can solve for b^2 using Pythagoras:
b^2 + (2b)^2 = 1
5b^2 = 1
b^2 = 1/5
Another approach which is longer, but which is convenient and flexible: you can just put everything in an xy-plane, with the point (0, 0) at the bottom left of the diagram. Then by finding the equations of each line, you can solve for their intersection point, and from there it's easy to find the area of ABC.
21 Apr 2019, 22:27
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We just need to figure out the length of perpendicular drawn from C to segment AB.
Let the bottom left corner as origin(0,0). Now we can figure out the co-ordinates of C wrt. origin which is intersection point of 2 diagonal lines.
Eq of the diagonal that passes through our origin is y=2x
Ex. of other diagonal line is y-2=(-1/2)x
We can find the intersection point of above mentioned 2 line by equating them
2x=2-(-1/2)x
or x=4/5
and hence y=8/5
Now the length of perpendicular drawn from C to segment AB=2-(8/5)=2/5
Area of ABC triangle= (1/2)*(2/5)*1=1/5
Let the bottom left corner as origin(0,0). Now we can figure out the co-ordinates of C wrt. origin which is intersection point of 2 diagonal lines.
Eq of the diagonal that passes through our origin is y=2x
Ex. of other diagonal line is y-2=(-1/2)x
We can find the intersection point of above mentioned 2 line by equating them
2x=2-(-1/2)x
or x=4/5
and hence y=8/5
Now the length of perpendicular drawn from C to segment AB=2-(8/5)=2/5
Area of ABC triangle= (1/2)*(2/5)*1=1/5
