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09 Dec 2019, 03:03
Kudos
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A
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Difficulty:
25%
(medium)
Question Stats:
70% (01:28) correct
30%
(01:43)
wrong
based on 138
sessions
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Not Attempted Yet
In the xy-plane, the points P, Q and S have coordinates (14,10), (1,0), and (6,0), respectively. What is the area of triangular region PQS?
A. 25
B. 30
C. 35
D. 42
E. 50
A. 25
B. 30
C. 35
D. 42
E. 50
09 Dec 2019, 03:10
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Bunuel
Explanation:
Below Is the Image Attached For Reference.
Area= |14(0-0)+1(0-10)+6(10-0)/2|
=|50/2|
=25
IMO-A
Attachments
Capture12345.PNG [ 28.41 KiB | Viewed 6297 times ]
15 Jan 2020, 01:30
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rajatchopra1994
I'd just like to add my 2 cents to the above solution by Rajat. I'm not a big believer in needing to know more formulas than required, the good ol' 1/2*base*height also works just fine here.
In the figure provided by Rajat, simply drop a perpendicular to the X axis from the point (14,10), the height of this perpendicular is the height of the triangle, 10 in this case. The base is the distance between the points (1,0) and (6,0) which is 5.
So Area = 1/2*5*10 = 25.
In cases where one of the sides of the triangle does not lie solely along the X or Y axes, I would simply extend the vertices of the relevant side to form a right angled triangle and employ the Pythagoras theorem to find the length of that side.
