Question

Answer this question please!

Question no.8


thanks!

Answer 1

$\mathbb{ SOLUTION }$ :

$<b>$Given :
X and Y are the mid points on BC and CD respectively of a parallelogram ABCD.

$\bf{To\:be \:proved }$ :
ar(Δ AXY) = $\frac{3}{8}$ ar( $\parallel^{\normalsize{gm}}$ ABCD )

$\bf{\underline {proof}}$ :

$\bf{ Construction }$ :
Join BD,
draw mid point ($\bf{P}$) on AD and join XP

=> ar(Δ CXY) = $\frac{1}{4}$ (Δ BCD)

[ $\therefore$ X and Y are mid points ]

So,

=> ar(Δ CXY) = $\frac{1}{4}$ [ ar($\frac{1}{2}$ $\parallel^{\normalsize{gm}}$ ABCD ) ]

=> ar(Δ CXY) = $\frac{1}{8}$ ar( $\parallel^{\normalsize{gm}}$ ABCD ) ___(i)

Now,

=> ar(Δ ABX) = $\frac{1}{2}$ ar( $\parallel^{\normalsize{gm}}$ ABXP )

=> ar(Δ ABX) = $\frac{1}{2}$ ar [ $\frac{1}{2}$ ar( $\parallel^{\normalsize{gm}}$ ABCD ) ]

=> ar(Δ ABX) = $\frac{1}{4}$ ar( $\parallel^{\normalsize{gm}}$ ABCD )___(ii)

Similarly,

=> ar(Δ ADY) = $\frac{1}{4}$ ar ( $\parallel^{\normalsize{gm}}$ ABCD )___(iii)

From figure,

=> ar(Δ AXY) = ar( $\parallel^{\normalsize{gm}}$ ABCD ) - ar(Δ ABX) - ar(Δ CXY) - ar(ΔADY)

=> ar(Δ AXY) = ar( $\parallel^{\normalsize{gm}}$ ABCD ) - $\frac{1}{4}$ ar( $\parallel^{\normalsize{gm}}$ ABCD) - $\frac{1}{8}$ ar( $\parallel^{\normalsize{gm}}$ ABCD ) - $\frac{1}{4}$ ar ( $\parallel^{\normalsize{gm}}$ ABCD)

=> ar(Δ AXY) = ar( $\parallel^{\normalsize{gm}}$ ABCD ) [ 1 - $\frac{1}{4}$ - $\frac{1}{8}$ - $\frac{1}{4}$ ]

=> ar(Δ AXY) = ar( $\parallel^{\normalsize{gm}}$ ABCD) [ $\frac{8-2-1-2}{8}$ ]

=> ar(Δ AXY) = ar( $\parallel^{\normalsize{gm}}$ ABCD) $\frac{3}{8}$

=> ar(Δ AXY) = $\frac{3}{8}$ ar( $\parallel^{\normalsize{gm}}$ ABCD)

$\bf{Hence,\:proved }$

Answer 2

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