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Bunuel
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In the figure below, M is the midpoint of RS. What is the area of tria 28 Dec 2020, 07:00
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E
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87% (01:04) correct 13% (01:26) wrong based on 54 sessions

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In the figure below, M is the midpoint of RS. What is the area of triangle MOP?


(A) \(\sqrt{7}\)
(B) 3
(C) 3.5
(D) 4
(E) 4.5


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2020-12-21_21-11-52.png
2020-12-21_21-11-52.png [ 16.32 KiB | Viewed 1830 times ]
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E
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imeanup
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In the figure below, M is the midpoint of RS. What is the area of tria 28 Dec 2020, 07:46
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Formula:
Mid-Point \((x_m,y_m)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\)

Co-ordinate of \(M = (\frac{2+4}{2},\frac{4+2}{2}) =(3,3) \)

Area of \(\triangle OMP = \frac{1}{2}*MP*OP = \frac{1}{2}*3*3 = 4.5\)

IMO Ans E
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In the figure below, M is the midpoint of RS. What is the area of tria 28 Dec 2020, 08:12
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Bunuel
In the figure below, M is the midpoint of RS. What is the area of triangle MOP?


(A) \(\sqrt{7}\)
(B) 3
(C) 3.5
(D) 4
(E) 4.5


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Attachment:
2020-12-21_21-11-52.png
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Coordinates of M are (3 , 3) , so coordinates of P are ( 0 , 3) and O are ( 0 , 0 )

\(MP = \sqrt{(3 - 3)^2 + ( 0 - 3)^2} = 3\)
\(OP = \sqrt{(0 - 3)^2 + ( 0 - 0)^2} = 3\)

Hence area of MOP is \(\frac{1}{2}*3*3 = 4.5\), Answer must be (E)
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In the figure below, M is the midpoint of RS. What is the area of tria Updated on: 30 Dec 2020, 08:05
Originally posted by MHIKER on 29 Dec 2020, 11:48.
Last edited by MHIKER on 30 Dec 2020, 08:05, edited 1 time in total.
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Bunuel
In the figure below, M is the midpoint of RS. What is the area of triangle MOP?


(A) \(\sqrt{7}\)
(B) 3
(C) 3.5
(D) 4
(E) 4.5


Show Spoiler
Attachment:
2020-12-21_21-11-52.png
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Co-ordinate of the midpoint \(M = (\frac{2+4}{2},\frac{4+2}{2}) =(3,3) \), which means OP=3 and PM=3.

Area of the triangle \(= \frac{1}{2}*base*height\) [ Here base and height are equal]

Area of \(\triangle OMP = \frac{1}{2}*OP*PM= \frac{1}{2}*3*3 = 4.5\)

The answer is E.
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In the figure below, M is the midpoint of RS. What is the area of tria 29 Dec 2020, 20:32
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The Hypotenuse of the Triangle lies on the Line given by the Equation: Y = X

Concept: When Point (A, B) is reflected over the Line Y = X, the Image Point becomes (B , A)

Concept: the Mirror Line over which the Original Point is reflected to create the Image Point is always EQUIDISTANT from the Original Point and Image Point


Point M Lies on the Line: Y = X

the Mid-Point M of (2,4) and (4,2) is -----> (3 , 3) which is a Point on the Line Y = X

We know that (2,4) is the Original Point Reflected over the Hypotenuse (which lies on Line Y = X) to create its Image Point (4 , 2)


The Line Y = X creates a 45 Degree Angle with the X-Axis.

Therefore, the Angle <MOP = 45 Degrees and the Triangle is a 45-45-90 Degree Isosceles Right Triangle.


Lastly, since we know that Point P is Directly, Vertically Below Point M at (3 , 3):
Point P must share the Same X-Coordinate and since it lies on the X-Axis its Y Coordinate must be 0

Coordinates of Point P are: (3 , 0)

The Horizontal Distance from Point P to the Origin = 3 Units

the Vertical Distance from Point P to Point M = 3 Units


Area of Right Triangle = (1/2) * (3) * (3) = 9/2 = 4.5

(E)
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