Answer this question please!
Question no.8
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Answer this question please!
Question no.8
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Verified answer
:Given :
X and Y are the mid points on BC and CD respectively of a parallelogram ABCD.
:
ar(Δ AXY) = ar( ABCD )
:
:
Join BD,
draw mid point () on AD and join XP
=> ar(Δ CXY) = (Δ BCD)
[ X and Y are mid points ]
So,
=> ar(Δ CXY) = [ ar( ABCD ) ]
=> ar(Δ CXY) = ar( ABCD ) ___(i)
Now,
=> ar(Δ ABX) = ar( ABXP )
=> ar(Δ ABX) = ar [ ar( ABCD ) ]
=> ar(Δ ABX) = ar( ABCD )___(ii)
Similarly,
=> ar(Δ ADY) = ar ( ABCD )___(iii)
From figure,
=> ar(Δ AXY) = ar( ABCD ) - ar(Δ ABX) - ar(Δ CXY) - ar(ΔADY)
=> ar(Δ AXY) = ar( ABCD ) - ar( ABCD) - ar( ABCD ) - ar ( ABCD)
=> ar(Δ AXY) = ar( ABCD ) [ 1 - - - ]
=> ar(Δ AXY) = ar( ABCD) [ ]
=> ar(Δ AXY) = ar( ABCD)
=> ar(Δ AXY) = ar( ABCD)
Answer:
Tobeproved :
ar(Δ AXY) = \frac{3}{8}
8
3
ar( \parallel^{\normalsize{gm}}∥
gm
ABCD )
\bf{\underline {proof}}
proof
:
\bf{ Construction }Construction :
Join BD,
draw mid point (\bf{P}P ) on AD and join XP
=> ar(Δ CXY) = \frac{1}{4}
4
1
(Δ BCD)
[ \therefore∴ X and Y are mid points ]
So,
=> ar(Δ CXY) = \frac{1}{4}
4
1
[ ar(\frac{1}{2}
2
1
\parallel^{\normalsize{gm}}∥
gm
ABCD ) ]
=> ar(Δ CXY) = \frac{1}{8}
8
1
ar( \parallel^{\normalsize{gm}}∥
gm
ABCD ) ___(i)
Now,
=> ar(Δ ABX) = \frac{1}{2}
2
1
ar( \parallel^{\normalsize{gm}}∥
gm
ABXP )
=> ar(Δ ABX) = \frac{1}{2}
2
1
ar [ \frac{1}{2}
2
1
ar( \parallel^{\normalsize{gm}}∥
gm
ABCD ) ]
=> ar(Δ ABX) = \frac{1}{4}
4
1
ar( \parallel^{\normalsize{gm}}∥
gm
ABCD )___(ii)
Similarly,
=> ar(Δ ADY) = \frac{1}{4}
4
1
ar ( \parallel^{\normalsize{gm}}∥
gm
ABCD )___(iii)
From figure,
=> ar(Δ AXY) = ar( \parallel^{\normalsize{gm}}∥
gm
ABCD ) - ar(Δ ABX) - ar(Δ CXY) - ar(ΔADY)
=> ar(Δ AXY) = ar( \parallel^{\normalsize{gm}}∥
gm
ABCD ) - \frac{1}{4}
4
1
ar( \parallel^{\normalsize{gm}}∥
gm
ABCD) - \frac{1}{8}
8
1
ar( \parallel^{\normalsize{gm}}∥
gm
ABCD ) - \frac{1}{4}
4
1
ar ( \parallel^{\normalsize{gm}}∥
gm
ABCD)
=> ar(Δ AXY) = ar( \parallel^{\normalsize{gm}}∥
gm
ABCD ) [ 1 - \frac{1}{4}
4
1
- \frac{1}{8}
8
1
- \frac{1}{4}
4
1
]
=> ar(Δ AXY) = ar( \parallel^{\normalsize{gm}}∥
gm
ABCD) [ \frac{8-2-1-2}{8}
8
8−2−1−2
]
=> ar(Δ AXY) = ar( \parallel^{\normalsize{gm}}∥
gm
ABCD) \frac{3}{8}
8
3
=> ar(Δ AXY) = \frac{3}{8}
8
3
ar( \parallel^{\normalsize{gm}}∥
gm
ABCD)
\bf{Hence,\:proved }Hence,proved