In a triangle A B C , from B and C , B Q and C P are drawn to meet A C and A B . area ( B C R ) = 1 0 sq.units , area ( B P R ) = 8 and area ( C Q R ) = 5 . Find area ( A P R Q ) .
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The areas of triangles are as follows:
[
C
B
R
]
=
1
0
,
[
C
B
Q
]
=
1
0
+
5
=
1
5
,
[
C
B
P
]
=
1
0
+
8
=
1
8
Since the breakdown into base and height is not given, we can use whatever is convenient. I have chosen to make the common base 2 , so that the heights of the triangles are simply equal to their areas.
Likewise it does not matter where the heights are located, so I have chosen ∠ A C B = 9 0 ∘ and placed the origin of my coordinate system at C
Equation of the line through Q B is y = 1 5 − 2 1 5 x
Point R is located on this line at y = 1 0 , so the coordinates of the point are R = ( 3 2 , 1 0 )
Line C P goes through point R and has the form y = k x . It is therefore given by equation y = 1 5 x
The point P is located on this line where y = 1 8 and has coordinates P = ( 5 6 , 1 8 )
Line A B has the format y = b − 2 b x and goes through point P , so it is y = 4 5 − 2 4 5 x
The y -intercept of this line is b = 4 5 so the area of the triangle A B C is [ A B C ] = 4 5
The area of the quadrilateral [ A Q R P ] = 4 5 − 1 0 − 5 − 8 = 2 2
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b + 8 a = 1 0 5 ( 1 )
a + 5 b = 1 0 8 ( 2 )
From ( 1 ) ,
a = 1 0 5 b + 4 0 ( 3 )
Substitute ( 3 ) in ( 2 ) , we have
1 0 5 b + 4 0 + 5 b = 1 0 8
1 0 5 b + 4 0 + 5 0 b = 1 0 8
1 b ( 5 b + 9 0 1 0 ) = 1 0 8
5 b + 9 0 1 0 b = 1 0 8
1 0 0 b = 4 0 b + 7 2 0
6 0 b = 7 2 0
b = 1 2
It follows that,
a = 1 0 5 ( 1 2 ) + 4 0 = 1 0 1 0 0 = 1 0
Finally,
a + b = 1 0 + 1 2 = 2 2