pls ans ques 7..its from ch mid point theorem
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nancyyy:
its from ch Rectilinear figures
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Heya Friend !!
Here's ur answer wid solution :-
1 )
Given :-
ABCD is a square
EBC is an equilateral triangle
To find :- x
Solution :-
As we know that the side of square are equal.
and side of equilateral triangle are also equal.
Here Equilateral triangle is surmounted on square
So,
Side of square = side of equilateral ∆
DC = EC
hence ,
∆ECD is an isosceles triangle.
and ,
|_ CED = |_CDE
And all angle of square and equilateral triangle are 90° and 60° respectively.
So,
|_ ECD = |_BCE + |_BCD
|_ECD = 60° + 90°
|_ ECD = 150°
In ∆ECD using angle sum property of Triangle.
|_ECD + |_CED + |_CDE = 180
150 + 2 |_CED = 180
2|_CED = 180 - 150
2|_CED = 30
|_CED = 15
and
Now in ∆ BEC
|_BEC = 60°
|_BED + |_CED = 60°
x + 15 = 60
x = 60 - 15
x = 45
2)
Given :- |_ ECD = 146
As we know that diagonal of rectangle are equal and bisect each other.
AC = BD
AC/2 = BD/2
OC = OD
So, ∆ DOC is an isosceles triangle
|_ ODC = |_ OCD .. ( i )
|_ECD + |_ OCD = 180 ( by linear pair )
146 + |_COD = 180
|_OCD = 180 - 146
|_OCD = 34
|_ODC = 34 ... ( from i )
In ∆ DOC using exterior angle property
|_DOC + |_ODC = |_ECD
|_DOC = 146 - 34
|_DOC = 112
In ∆ AOB
|_AOB = |_DOC ( vertically opposite angle)
|_AOB = 112
Since opposite sides of rectangle are parallel
DC || AB
|_DCO = |_OAB ( alternate interior angle)
34 = |_OAB
|_CDO = |_OBA ( alternate interior angle )
34 = |_OBA
3 )
As we know that the diagonal of rhombus intersect at 90°
|_AOB = 90
|_DOA = 90
Let the angle be x
then
|_OAB = 3x
|_OBA = 2x
In ∆AOB using Angle sum property
|_OAB + |_OBA + |_AOB = 180
3x + 2x + 90 = 180
5x = 90
x = 18
|_OAB = 3x = 54
|_OBA = 2x = 36
As we know that diagonal of rhombus are angle bisector
|_ADO = |_ODC .... ( i )
|_DAO = |_OAB .....( ii )
From ii
|_DAO = 54
And opposite side of rhombus are parallel
So,
|_OBA = |_ CDO ( alternate interior angle)
36 = |_CDO
|_ADO = |_CDO ....( from i )
|_ADO = 36
In ∆AOD
|_ADO = 36
|_DOA = 90
|_DAO = 54
@Altaf
Here's ur answer wid solution :-
1 )
Given :-
ABCD is a square
EBC is an equilateral triangle
To find :- x
Solution :-
As we know that the side of square are equal.
and side of equilateral triangle are also equal.
Here Equilateral triangle is surmounted on square
So,
Side of square = side of equilateral ∆
DC = EC
hence ,
∆ECD is an isosceles triangle.
and ,
|_ CED = |_CDE
And all angle of square and equilateral triangle are 90° and 60° respectively.
So,
|_ ECD = |_BCE + |_BCD
|_ECD = 60° + 90°
|_ ECD = 150°
In ∆ECD using angle sum property of Triangle.
|_ECD + |_CED + |_CDE = 180
150 + 2 |_CED = 180
2|_CED = 180 - 150
2|_CED = 30
|_CED = 15
and
Now in ∆ BEC
|_BEC = 60°
|_BED + |_CED = 60°
x + 15 = 60
x = 60 - 15
x = 45
2)
Given :- |_ ECD = 146
As we know that diagonal of rectangle are equal and bisect each other.
AC = BD
AC/2 = BD/2
OC = OD
So, ∆ DOC is an isosceles triangle
|_ ODC = |_ OCD .. ( i )
|_ECD + |_ OCD = 180 ( by linear pair )
146 + |_COD = 180
|_OCD = 180 - 146
|_OCD = 34
|_ODC = 34 ... ( from i )
In ∆ DOC using exterior angle property
|_DOC + |_ODC = |_ECD
|_DOC = 146 - 34
|_DOC = 112
In ∆ AOB
|_AOB = |_DOC ( vertically opposite angle)
|_AOB = 112
Since opposite sides of rectangle are parallel
DC || AB
|_DCO = |_OAB ( alternate interior angle)
34 = |_OAB
|_CDO = |_OBA ( alternate interior angle )
34 = |_OBA
3 )
As we know that the diagonal of rhombus intersect at 90°
|_AOB = 90
|_DOA = 90
Let the angle be x
then
|_OAB = 3x
|_OBA = 2x
In ∆AOB using Angle sum property
|_OAB + |_OBA + |_AOB = 180
3x + 2x + 90 = 180
5x = 90
x = 18
|_OAB = 3x = 54
|_OBA = 2x = 36
As we know that diagonal of rhombus are angle bisector
|_ADO = |_ODC .... ( i )
|_DAO = |_OAB .....( ii )
From ii
|_DAO = 54
And opposite side of rhombus are parallel
So,
|_OBA = |_ CDO ( alternate interior angle)
36 = |_CDO
|_ADO = |_CDO ....( from i )
|_ADO = 36
In ∆AOD
|_ADO = 36
|_DOA = 90
|_DAO = 54
@Altaf
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