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15 Jul 2021, 05:41
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The key to solving the problem is finding the coordinates where the equation meets the axes with the extremems we can get the value of the figure
|x/2| + |Y/2| = 5
so when x = 0, y= |10|
when y=0 , x=|10|
we get the extrremums as (0,10),(10,0),(0,-10) and (-10,0).
it's a square whose one side has a value can be traced to as 10*2^1/2
area of the figure being 100*2 = 200
Hence IMO D
|x/2| + |Y/2| = 5
so when x = 0, y= |10|
when y=0 , x=|10|
we get the extrremums as (0,10),(10,0),(0,-10) and (-10,0).
it's a square whose one side has a value can be traced to as 10*2^1/2
area of the figure being 100*2 = 200
Hence IMO D
rajshreeasati
Joined: 11 Jul 2018
Last visit: 16 Mar 2025
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Schools: ISB '27 (A)
29 Apr 2022, 04:45
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The best approach to such Coordinate Geometry Questions is always diagrams-
we have |x/2| + |y/2| = 5,
we will get 4 coordinates. x = 0, y = 10, y = 0, x = 10, x = 0, y = -10 and y = 0 and x = -10.
by drawing a diagram we get a figure where we have 4 triangles with 10 as the height and 10 as the base.
so area of region will be = 4 * 1/2 * 10*10 => 200.
we have |x/2| + |y/2| = 5,
we will get 4 coordinates. x = 0, y = 10, y = 0, x = 10, x = 0, y = -10 and y = 0 and x = -10.
by drawing a diagram we get a figure where we have 4 triangles with 10 as the height and 10 as the base.
so area of region will be = 4 * 1/2 * 10*10 => 200.
17 Nov 2022, 01:28
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Equation of a line whose intercept on X-axis is (a,0) and Y-axis is (0,b) is
\(\frac{x}{a}+\frac{y}{b}=1\)
We get intercept points as (10,0) and (0,10).
Area of a triangle formed by these points and origin = \(\frac{(10*10)}{2}=50\)
Area of the region = 50*4 = 200
\(\frac{x}{a}+\frac{y}{b}=1\)
We get intercept points as (10,0) and (0,10).
Area of a triangle formed by these points and origin = \(\frac{(10*10)}{2}=50\)
Area of the region = 50*4 = 200
