modal ____________ i bookyour ticket along with mine?
Answers
Answer:
AnswEr :
⋆ Refrence of Image is in the Diagram :
• Let's Head to the Question Now :
Diagonals of Parallelogram Bisects each other, Here at O.
Mid Points of Both Diagonal will be Equal.
\begin{lgathered}\implies \tt Mid \:Point \:of \:Diagonal \:AC=Mid \:Point \:of \:Diagonal \:BD \\ \\\implies \tt\bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg) = \bigg(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}\bigg)\\ \\\implies \tt\bigg(\dfrac{-2+4}{2},\dfrac{1+b}{2}\bigg) = \bigg(\dfrac{a+1}{2},\dfrac{0+2}{2}\bigg) \\ \\\implies \tt\bigg( \cancel\dfrac{2}{2},\dfrac{1+b}{2}\bigg) = \bigg(\dfrac{a+1}{2},\cancel\dfrac{2}{2}\bigg) \\ \\\implies \tt\bigg( 1,\dfrac{1+b}{2}\bigg) = \bigg(\dfrac{a+1}{2},1\bigg) \\ \\\implies \tt1 = \dfrac{a+1}{2} \quad and \quad\dfrac{1+b}{2} = 1 \\ \\\implies \tt2 = a + 1\quad and \quad1 + b = 2 \\ \\\implies \red{\tt a = 1\quad} \tt and \quad \red{ b = 1}\end{lgathered}
⟹MidPointofDiagonalAC=MidPointofDiagonalBD
⟹(
2
x
1
+x
2
,
2
y
1
+y
2
)=(
2
x
1
+x
2
,
2
y
1
+y
2
)
⟹(
2
−2+4
,
2
1+b
)=(
2
a+1
,
2
0+2
)
⟹(
2
2
,
2
1+b
)=(
2
a+1
,
2
2
)
⟹(1,
2
1+b
)=(
2
a+1
,1)
⟹1=
2
a+1
and
2
1+b
=1
⟹2=a+1and1+b=2
⟹a=1andb=1
⠀
∴ Therefore, Value of a and, b will be 1.
• CALCULATION⠀OF⠀SIDES :
\begin{lgathered}\longrightarrow \tt AB = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}} \\ \\\longrightarrow \tt AB = \sqrt{(a-( - 2))^{2}+(0 - 1)^{2}} \\ \\\longrightarrow \tt AB = \sqrt{(1 + 2)^{2}+(1)^{2}} \\ \\\longrightarrow \tt AB = \sqrt{(3)^{2}+(1)^{2}} \\ \\\longrightarrow \tt AB = \sqrt{9 + 1} \\ \\\longrightarrow \green{\tt AB = \sqrt{10} \:unit}\end{lgathered}
⟶AB=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
⟶AB=
(a−(−2))
2
+(0−1)
2
⟶AB=
(1+2)
2
+(1)
2
⟶AB=
(3)
2
+(1)
2
⟶AB=
9+1
⟶AB=
10
unit
\begin{lgathered}\longrightarrow \tt BC = \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}} \\ \\\longrightarrow \tt BC = \sqrt{(4-a)^{2}+(b-0)^{2}} \\ \\\longrightarrow \tt BC = \sqrt{(4-1)^{2}+(1-0)^{2}} \\ \\\longrightarrow \tt BC = \sqrt{(3)^{2}+(1)^{2}} \\ \\\longrightarrow \tt BC = \sqrt{9 + 1} \\\\\longrightarrow \green{\tt BC = \sqrt{10} \:unit}\end{lgathered}
⟶BC=
(x
2
−x
1
)
2
+(y
2
−y
1
)
2
⟶BC=
(4−a)
2
+(b−0)
2
⟶BC=
(4−1)
2
+(1−0)
2
⟶BC=
(3)
2
+(1)
2
⟶BC=
9+1
⟶BC=
10
unit
◗ In Parallelogram opposite sides are equal
⇒ AB = CD = √10 Unit
⇒ BC = AD = √10 Unit
⠀
∴ Therefore, All Sides will be √10 unit.
#answerwithquality #BAL
May I book your ticket along with mine?