Page 1
Q u e s t i o n : 3 1
In the given figure, ?ABC is an equilateral triangle. Find m ?BEC.
S o l u t i o n :
It is given that, is an equilateral triangle
Page 2
Q u e s t i o n : 3 1
In the given figure, ?ABC is an equilateral triangle. Find m ?BEC.
S o l u t i o n :
It is given that, is an equilateral triangle
We have to find
Since is an equilateral triangle.
So
And
……
1
Since, quadrilateral BACE is a cyclic qualdrilateral
So , (Sum of opposite angles of cyclic quadrilateral is .)
Hence
Q u e s t i o n : 3 2
In the given figure, ?PQR is an isosceles triangle with PQ = PR and m ?PQR = 35°. Find m ?QSR and m ?QTR.
S o l u t i o n :
Disclaimer: Figure given in the book was showing m ?PQR as m ?SQR.
It is given that ?PQR is an isosceles triangle with PQ = PR and m ?PQR = 35°
We have to find the m ?QSR and m ?QTR
Since ?PQR is an isosceles triangle
So ?PQR = ?PRQ = 35°
Then
Since PQTR is a cyclic quadrilateral
So
In cyclic quadrilateral QSRT we have
Hence,
and
Q u e s t i o n : 3 3
In the given figure, O is the centre of the circle. If ?BOD = 160°, find the values of x and y.
Page 3
Q u e s t i o n : 3 1
In the given figure, ?ABC is an equilateral triangle. Find m ?BEC.
S o l u t i o n :
It is given that, is an equilateral triangle
We have to find
Since is an equilateral triangle.
So
And
……
1
Since, quadrilateral BACE is a cyclic qualdrilateral
So , (Sum of opposite angles of cyclic quadrilateral is .)
Hence
Q u e s t i o n : 3 2
In the given figure, ?PQR is an isosceles triangle with PQ = PR and m ?PQR = 35°. Find m ?QSR and m ?QTR.
S o l u t i o n :
Disclaimer: Figure given in the book was showing m ?PQR as m ?SQR.
It is given that ?PQR is an isosceles triangle with PQ = PR and m ?PQR = 35°
We have to find the m ?QSR and m ?QTR
Since ?PQR is an isosceles triangle
So ?PQR = ?PRQ = 35°
Then
Since PQTR is a cyclic quadrilateral
So
In cyclic quadrilateral QSRT we have
Hence,
and
Q u e s t i o n : 3 3
In the given figure, O is the centre of the circle. If ?BOD = 160°, find the values of x and y.
S o l u t i o n :
It is given that O is centre of the circle and ?BOD = 160°
We have to find the values of x and y.
As we know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,
Since, quadrilateral ABCD is a cyclic quadrilateral.
So,
x + y = 180° (Sum of opposite angles of a cyclic quadrilateral is 180°.)
Hence and
Q u e s t i o n : 3 4
In the given figure, ABCD is a cyclic quadrilateral. If ?BCD = 100° and ?ABD = 70°, find ?ADB.
S o l u t i o n :
It is given that ?BCD = 100° and ?ABD = 70°
We have to find the ?ADB
We have
?A + ?C = 180°
Oppositepairofangleofcyclicquadrilateral
So,
Now in is and
Therefore,
Hence,
Page 4
Q u e s t i o n : 3 1
In the given figure, ?ABC is an equilateral triangle. Find m ?BEC.
S o l u t i o n :
It is given that, is an equilateral triangle
We have to find
Since is an equilateral triangle.
So
And
……
1
Since, quadrilateral BACE is a cyclic qualdrilateral
So , (Sum of opposite angles of cyclic quadrilateral is .)
Hence
Q u e s t i o n : 3 2
In the given figure, ?PQR is an isosceles triangle with PQ = PR and m ?PQR = 35°. Find m ?QSR and m ?QTR.
S o l u t i o n :
Disclaimer: Figure given in the book was showing m ?PQR as m ?SQR.
It is given that ?PQR is an isosceles triangle with PQ = PR and m ?PQR = 35°
We have to find the m ?QSR and m ?QTR
Since ?PQR is an isosceles triangle
So ?PQR = ?PRQ = 35°
Then
Since PQTR is a cyclic quadrilateral
So
In cyclic quadrilateral QSRT we have
Hence,
and
Q u e s t i o n : 3 3
In the given figure, O is the centre of the circle. If ?BOD = 160°, find the values of x and y.
S o l u t i o n :
It is given that O is centre of the circle and ?BOD = 160°
We have to find the values of x and y.
As we know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,
Since, quadrilateral ABCD is a cyclic quadrilateral.
So,
x + y = 180° (Sum of opposite angles of a cyclic quadrilateral is 180°.)
Hence and
Q u e s t i o n : 3 4
In the given figure, ABCD is a cyclic quadrilateral. If ?BCD = 100° and ?ABD = 70°, find ?ADB.
S o l u t i o n :
It is given that ?BCD = 100° and ?ABD = 70°
We have to find the ?ADB
We have
?A + ?C = 180°
Oppositepairofangleofcyclicquadrilateral
So,
Now in is and
Therefore,
Hence,
Q u e s t i o n : 3 5
If ABCD is a cyclic quadrilateral in which AD || BC Inthegivenfigure
. Prove that ?B = ?C.
S o l u t i o n :
It is given that, ABCD is cyclic quadrilateral in which AD || BC
We have to prove
Since, ABCD is a cyclic quadrilateral
So,
and ..…
1
and
Sumofpairofconsecutiveinterioranglesis180° …… 2
From equation
1 and
2 we have
……
3
……
4
Hence Proved
Q u e s t i o n : 3 6
In the given figure, O is the centre of the circle. Find ?CBD.
S o l u t i o n :
It is given that,
We have to find
Since,
Given
So,
(The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.)
Now,
Oppositepairofangleofcyclicquadrilateral
Page 5
Q u e s t i o n : 3 1
In the given figure, ?ABC is an equilateral triangle. Find m ?BEC.
S o l u t i o n :
It is given that, is an equilateral triangle
We have to find
Since is an equilateral triangle.
So
And
……
1
Since, quadrilateral BACE is a cyclic qualdrilateral
So , (Sum of opposite angles of cyclic quadrilateral is .)
Hence
Q u e s t i o n : 3 2
In the given figure, ?PQR is an isosceles triangle with PQ = PR and m ?PQR = 35°. Find m ?QSR and m ?QTR.
S o l u t i o n :
Disclaimer: Figure given in the book was showing m ?PQR as m ?SQR.
It is given that ?PQR is an isosceles triangle with PQ = PR and m ?PQR = 35°
We have to find the m ?QSR and m ?QTR
Since ?PQR is an isosceles triangle
So ?PQR = ?PRQ = 35°
Then
Since PQTR is a cyclic quadrilateral
So
In cyclic quadrilateral QSRT we have
Hence,
and
Q u e s t i o n : 3 3
In the given figure, O is the centre of the circle. If ?BOD = 160°, find the values of x and y.
S o l u t i o n :
It is given that O is centre of the circle and ?BOD = 160°
We have to find the values of x and y.
As we know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,
Since, quadrilateral ABCD is a cyclic quadrilateral.
So,
x + y = 180° (Sum of opposite angles of a cyclic quadrilateral is 180°.)
Hence and
Q u e s t i o n : 3 4
In the given figure, ABCD is a cyclic quadrilateral. If ?BCD = 100° and ?ABD = 70°, find ?ADB.
S o l u t i o n :
It is given that ?BCD = 100° and ?ABD = 70°
We have to find the ?ADB
We have
?A + ?C = 180°
Oppositepairofangleofcyclicquadrilateral
So,
Now in is and
Therefore,
Hence,
Q u e s t i o n : 3 5
If ABCD is a cyclic quadrilateral in which AD || BC Inthegivenfigure
. Prove that ?B = ?C.
S o l u t i o n :
It is given that, ABCD is cyclic quadrilateral in which AD || BC
We have to prove
Since, ABCD is a cyclic quadrilateral
So,
and ..…
1
and
Sumofpairofconsecutiveinterioranglesis180° …… 2
From equation
1 and
2 we have
……
3
……
4
Hence Proved
Q u e s t i o n : 3 6
In the given figure, O is the centre of the circle. Find ?CBD.
S o l u t i o n :
It is given that,
We have to find
Since,
Given
So,
(The angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.)
Now,
Oppositepairofangleofcyclicquadrilateral
So,
……
1
(Linear pair)
( )
Hence
Q u e s t i o n : 3 7
In the given figure, AB and CD are diameters of a circle with centre O. If ?OBD = 50°, find ?AOC.
S o l u t i o n :
It is given that, AB and CD are diameter with center O and
We have to find
Construction: Join the point A and D to form line AD
Clearly arc AD subtends at B and at the centre.
Therefore,
?AOD = 2 ?ABD = 100° ……
1
Since CD is a straight line then
?DOA + ?AOC = 180° (Linear pair)
Hence
Q u e s t i o n : 3 8
On a semi-circle with AB as diameter, a point C is taken, so that m ( ?CAB) = 30°. Find m ( ?ACB) and m ( ?ABC).
S o l u t i o n :
It is given that, as diameter, is centre and
We have to find and
Since angle in a semi-circle is a right angle therefore
In we have
Given
Angleinsemi -circleisrightangle
Now in we have
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