Understanding Quadrilaterals Test - 25

$${l} p=\frac { { 3\overline { d }  } }{ 4 }  \\ lets, \\ AC=\lambda \, (\overline { b } +\overline { d } ) \\ PB=\frac { { 3\overline { d }  } }{ 4 } +\gamma (\overline { b } -\frac { { 3\overline { d }  } }{ 4 } ) \\ then,co-ordinate\, are, \\ AC=PB \\ \lambda \, (\overline { b } +\overline { d } )=\frac { { 3\overline { d }  } }{ 4 } +\gamma (\overline { b } -\frac { { 3\overline { d }  } }{ 4 } ) \\ \lambda \, (\overline { b } +\overline { d } )=\frac { { 3\overline { d }  } }{ 4 } +\gamma \overline { b } -\frac { { 3\gamma \overline { d }  } }{ 4 }  \\ Now,compare\, both\, \, side, \\ \lambda =\gamma ,\, \, \, \, \, \, \, \lambda =\frac { 3 }{ 4 } -\frac { { 3\gamma  } }{ 4 }  \\ \, \, \, \, \, \, \, \, \lambda =\frac { 3 }{ 4 } -\frac { { 3\lambda  } }{ 4 }  \\ \, \, \, \, \frac { { 7\lambda  } }{ 4 } =\frac { 3 }{ 4 }  \\ \therefore \, \, \, \, \lambda =\frac { 3 }{ 7 }  \\ it\, means,\, Q\, po{ { int } }\, is\, \frac { 3 }{ 7 } (\overline { b } +\overline { c } ) \\ So,the\, AQ:QC\, is\, 3:4 \\ and,\, the\, \, correct\, \, option\, is\, A. $$

In $$\triangle BCD,$$

$$XY\parallel BD$$ and $$XY=\dfrac{1}{2}BD$$           {From the mid point theorem}

$$\Rightarrow$$  $$ar(\triangle CYX)=\dfrac{1}{4}ar(\triangle DBC)$$    {property of triangle having mid points}

$$\Rightarrow$$  $$ar(\triangle CYX)=\dfrac{1}{8}ar(\parallel^{gm}\,ABCD)$$      [ Area of parallelogram is twice the area of triangle made by diagona ]    --- ( 1 )

Since, parallelogram $$ABCD$$ and $$\triangle ABX$$ are between the same parallel lines $$AD$$ and $$BC$$ and $$BX=\dfrac{1}{2}BC.$$

$$\Rightarrow$$  $$ar(\triangle ABX)=\dfrac{1}{4}ar(\parallel^{gm}\,ABCD)$$     ----- ( 2 )

Similarly, $$ar(\triangle AYD)=\dfrac{1}{4}ar(\parallel^{gm} ABCD)$$       ----- ( 3 )

Now, $$ar(\triangle AXY)=ar(\parallel^{gm}\,ABCD)-[ar(\triangle ABX)+ar(\triangle AYD)+ar(\triangle CYX)]$$

                                $$=ar(\parallel^{gn} ABCD)-[\dfrac{1}{4}ar(\parallel^{gm}ABCD+\dfrac{1}{4}ar(\parallel^{gm}ABCD)+\dfrac{1}{8}ar(\parallel^{gm}ABCD)]$$              [ From ( 1 ) ( 2 ) and ( 3 ) ]

                                $$=ar(\parallel^{gm}ABCD)-\dfrac{5}{8}(\parallel^{gm}ABCD)$$

$$\Rightarrow$$  $$ar(\triangle AXY)=\dfrac{3}{8}ar(\parallel^{gm}ABCD)$$

$$\therefore$$  $$ar(\parallel^{gm}ABCD)=\dfrac{8}{3}\times ar(\triangle AXY)$$

perimeter of parallelogram $$=38 \text{ cm}$$

length of longer  side $$=11 \text{ cm}$$ 

To find: shorter side 

since, perimeter of parallelogram $$=2(l + h)$$

$$\implies 38 =  2(l + b)$$

$$\implies 38 = 2(11 + b)$$

$$\implies 38 = 22 + 2b$$

$$\implies b=\dfrac{{16}}{2} $$

$$\implies b = 8\text{ cm}$$

$$\therefore$$ The length of the shorter side is $$8\text{ cm}$$

Let $$ABCD$$ be the quadrilateral in which $$AC$$ and $$BD$$ are the diagonals that intersect at $$O$$.

Now, 

$$\dfrac{AO}{OC} =\dfrac{1}{2}$$ 

We know that diagonals of a parallelogram bisect eah other. 

So, for $$ABCD$$ to be a parallelogram, 

$$\dfrac{AO}{OC} = \dfrac{1}{1}$$ 

But here 

 $$\dfrac{AO}{OC} =\dfrac{1}{2}$$ 

So, the quadrilateral can't be a parallelogram in which the point of intersection of the diagonals divide one diagonal in the ratio of $$1 : 2$$.