Exercise-9.3
Maths - Exercise
Q1. In Fig. 9.23, A,B and C are three points on a circle with centre O such that \(\angle BOC = 30°\) and \(\angle AOB = 60°\). If D is a point on the circle other than the arc ABC, find \(\angle ADC.\)
Solution:
Given: Points \(A\), \(B\), and \(C\) lie on a circle with centre \(O\), \(\angle BOC = 30^\circ\), \(\angle AOB = 60^\circ\), and \(D\) is a point on the circle lying on the arc opposite to arc \(ABC\). Find \(\angle ADC\).
First, calculate the angle at the centre subtended by arc \(AC\):
\[ \angle AOC = \angle AOB + \angle BOC = 60^\circ + 30^\circ = 90^\circ \]
The angle at the centre (\(\angle AOC\)) is twice the angle (\(\angle ADC\)) subtended on the remaining circle (opposite arc) by the same arc (\(AC\)), using the property of the circle:
\[ \angle AOC = 2\angle ADC \]
Therefore, \[ \angle ADC = \frac{1}{2} \angle AOC = \frac{1}{2} \times 90^\circ = 45^\circ \]
Q2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Solution:
Let the circle have centre \(O\) and radius \(r\). Let chord \(AB\) be such that \(AB = r\). Join \(OA\) and \(OB\). In \(\triangle OAB\), the sides \(OA = OB = AB = r\), making it an equilateral triangle.
Therefore, each angle in \(\triangle OAB\) is \(60^\circ\), so the angle subtended by chord \(AB\) at the centre is:
\[ \angle AOB = 60^\circ \]
The angle subtended by chord \(AB\) at a point on the minor arc (say, at point \(C\)) is half the angle at the centre, by the circle theorems:
\[ \text{Angle on minor arc} = \frac{1}{2} \times 60^\circ = 30^\circ \]
Let \(D\) be a point on the major arc. By the cyclic quadrilateral property (angles subtended by the same chord in the same segment are supplementary):
\[ \text{Angle on major arc} = 180^\circ - 30^\circ = 150^\circ \]
Q3. In Fig. 9.24, \(\angle PQR = 100°\), where P, Q and R are points on a circle with centre O. Find \(\angle OPR.\)
Given: \( \triangle PQR \) is inscribed in a circle with centre \(O\) and \( \angle PQR = 100^\circ \). To find: \( \angle OPR \).
The angle subtended by the chord \(PR\) at the centre (\( \angle POR \)) is twice the angle subtended at the remaining part of the circle (\( \angle PQR \)), by the circle theorem.
\[\begin{aligned} \angle POR &= 2 \times \angle PQR \\&= 2 \times 100^\circ \\&= 200^\circ \end{aligned}\]
Since the central angle is greater than \(180^\circ\), the reflex angle \( \angle POR = 200^\circ \) and the acute angle at the centre is:
\[ \text{Acute } \angle POR = 360^\circ - 200^\circ = 160^\circ \]
Now, \( OP = OR \) (radii of the circle), so \( \triangle POR \) is isosceles with \( OP = OR \). Let \( \angle OPR = \angle ORP = x \).
Using the angle sum property of a triangle: \[\begin{aligned} x + x + 160^\circ &= 180^\circ \\\implies 2x &= 20^\circ \\\implies x &= 10^\circ \end{aligned}\]
Q4. In Fig. 9.25, \(\angle ABC = 69°, \;\angle ACB = 31°\), find \(\angle BDC.\)
Solution:
Given: In the circle, \( \angle ABC = 69^\circ \), \( \angle ACB = 31^\circ \). To find: \( \angle BDC \).
First, use the angle sum property in \( \triangle ABC \):
\[ \angle ABC + \angle ACB + \angle BAC = 180^\circ \]
Substitute values: \[\begin{aligned} 69^\circ + 31^\circ + \angle BAC &= 180^\circ \\\implies \angle BAC &= 180^\circ - (69^\circ + 31^\circ) \\&= 180^\circ - 100^\circ \\&= 80^\circ \end{aligned}\]
Now, \( \angle BDC \) subtends the same chord \( BC \) as \( \angle BAC \), but from the opposite segment (circle theorem). So:
\[ \angle BDC = \angle BAC = 80^\circ \]
Q5. In Fig. 9.26, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that \(\angle BEC = 130°\) and \(\angle ECD = 20°.\) Find \(\angle BAC.\)
Solution:
Given: In the circle, AC and BD intersect at point E, \( \angle BEC = 130^\circ \), \( \angle ECD =
20^\circ \).
To find:
\( \angle BAC \).
Since AC and BD intersect at E, consider the angles formed at E. At point E, the sum of the angles around the point is \(360^\circ\):
\[\scriptsize \angle BEC + \angle CED + \angle DEB + \angle BEA = 360^\circ \]
By vertical angles,
\[ \angle BEA = \angle CED \] and \[ \angle BEC = \angle DEB \]
Substitute given values:
\[
\angle BEC = 130^\circ, \\\angle CED = \angle BEA \\
\angle ECD = 20^\circ
\]
Let \( x = \angle BEA = \angle CED \).
\[\small 130^\circ + 20^\circ + 130^\circ + x = 360^\circ \]
\[\scriptsize\begin{aligned} 2 \times 130^\circ + 20^\circ + x &= 360^\circ \\\implies 260^\circ + 20^\circ + x &= 360^\circ \\\implies x &= 360^\circ - 280^\circ \\&= 80^\circ \end{aligned}\]
Now, consider triangle \( AEB \): \[\scriptsize \angle BAC + \angle ABE + \angle BEA = 180^\circ \]
Given:
\( \angle ABE = \angle ECD = 20^\circ \), \( \angle BEA = 80^\circ \):
\[\small\begin{aligned}
\angle BAC + 20^\circ + 80^\circ &= 180^\circ \\\implies \angle BAC &= 180^\circ - 100^\circ \\&=
80^\circ
\end{aligned}\]
Q6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If \(\angle DBC = 70°,\; \angle BAC \text{ is } 30°\), find \(\angle BCD.\) Further, if AB = BC, find \(\angle ECD.\)
Solution:
Given:
Cyclic quadrilateral \(ABCD\) with diagonals intersecting at \(E\), \(
\angle DBC = 70^\circ \),
\( \angle BAC = 30^\circ \).
To find:
\( \angle BCD \) and if \(AB = BC\),
\( \angle ECD \).
Since \(ABCD\) is cyclic, opposite angles sum to \(180^\circ\): \[ \angle BCD + \angle BAD = 180^\circ \]
From triangle \(ABD\),
\( \angle BAD = \angle BAC + \angle DAC \).
Also, by the Inscribed Angle
Theorem:
- \( \angle BAC \) and \( \angle BDC \)
subtend chord \(BC\), so
\( \angle BDC = \angle BAC = 30^\circ \). - \( \angle DBC \) and \( \angle DAC \)
subtend chord \(DC\), so
\( \angle DAC = \angle DBC = 70^\circ \).
Sum for \( \angle BAD \): \[\scriptsize\begin{aligned} \angle BAD &= \angle BAC + \angle DAC \\&= 30^\circ + 70^\circ \\&= 100^\circ \end{aligned}\]
Now, solve for \( \angle BCD \): \[\scriptsize\begin{aligned} \angle BCD &= 180^\circ - \angle BAD \\&= 180^\circ - 100^\circ \\&= 80^\circ \end{aligned}\]
If \(AB = BC\), triangle \(ABC\) is isosceles:
- \( \angle BAC = \angle BCA = 30^\circ \)
From the quadrilateral, sum around point \(C\): \[\scriptsize \angle DBC + \angle BAC + \angle BCA + \angle ECD = 180^\circ \]
Substitute known angles: \[\scriptsize\begin{aligned} 70^\circ + 30^\circ + 30^\circ + \angle ECD &= 180^\circ \\\implies \angle ECD &= 180^\circ - 130^\circ \\&= 50^\circ \end{aligned}\]
Q7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Solution:
Let \(ABCD\) be a cyclic quadrilateral such that its diagonals \(AC\) and \(BD\) are also diameters of the circle. To prove: \(ABCD\) is a rectangle.
Since \(AC\) and \(BD\) are diameters, each vertex of the quadrilateral \(A\), \(B\), \(C\), and \(D\) lies on the circle.
By the circle theorem, the angle subtended by a diameter at the circumference is a right angle (\(90^\circ\)): \[ \text{Angle in a semicircle} = 90^\circ \] Thus, \[\tiny \angle ABC = \angle BCD = \angle CDA = \angle DAB = 90^\circ \]
All four interior angles are \(90^\circ\), so \(ABCD\) is a quadrilateral with all right angles.
Therefore, a cyclic quadrilateral whose diagonals are diameters must be a rectangle.
Q8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Solution:
Given:
In trapezium \(ABCD\), \(AB \parallel CD\) and non-parallel sides \(AD =
BC\).
To prove:
\(ABCD\)
is a cyclic quadrilateral.
Extend \(AD\) and \(BC\) to meet at \(E\).
Draw \(BE \parallel AD\) such that \(AD = BE\),
so
by
construction, \(BE = AD = BC\).
Thus, triangle \(BEC\) is isosceles with \(BE = BC\).
In triangle \(BEC\),
\(\angle EBC = \angle ECB\)
(base angles of isosceles triangle).
Since \(AB
\parallel CD\) and \(BE \parallel AD\), corresponding angles hold:
\[ \angle DAB = \angle ECB \]
The sum of opposite angles of quadrilateral \(ABCD\): \[\scriptsize \angle DAB + \angle BCD = \angle ECB + \angle BCD \] But at point \(C\), \(\angle ECB + \angle BCD = 180^\circ\) (linear pair). \[\scriptsize \implies \angle DAB + \angle BCD = 180^\circ \]
Since the sum of a pair of opposite angles is \(180^\circ\), \(ABCD\) is cyclic.
Q9.Two circles intersect at two points B and C. Through B, two line segments ABD and PBQare drawn to intersect the circles at A, D and P, Q respectively (see Fig. 9.27). Prove that \(\angle ACP = \angle QCD.\)
Solution:
Given: Two circles intersect at points \(B\) and \(C\). Lines \(ABD\) and \(PBQ\) pass through \(B\) and intersect the circles at \(A, D\) and \(P, Q\) respectively. To prove: \( \angle ACP = \angle QCD \).
In the first circle, \( \angle ACP \) and \( \angle ABP \) both subtend the same chord \(AP\) from points \(C\) and \(B\), so: \[ \angle ACP = \angle ABP \]
At intersection point \(B\), \( \angle ABP \) and \( \angle DBQ \) are vertically opposite angles, so: \[ \angle ABP = \angle DBQ \]
In the second circle, \( \angle DBQ \) and \( \angle QCD \) subtend the same chord \(QD\) from points \(B\) and \(C\), therefore: \[ \angle DBQ = \angle QCD \]
Thus, by transitivity, \[ \angle ACP = \angle QCD \]
Hence proved.
Q10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Solution:
Let \( \triangle ABC \) be given. Draw two circles: one with diameter \( AB \) and the other with diameter \( AC \). Let these circles intersect at a point \( M \) (other than \( A \)).
Since \( AB \) is a diameter of the first circle, by the angle in a semicircle theorem: \[ \angle AMB = 90^{\circ} \] Similarly, since \( AC \) is a diameter of the second circle: \[ \angle AMC = 90^{\circ} \]
Therefore, at point \( M \): \[ \angle AMB + \angle AMC = 90^\circ + 90^\circ = 180^\circ \]
This means that points \( B \), \( M \), and \( C \) are collinear, so \( M \) lies on \( BC \).
Hence, the point of intersection of these circles lies on the third side of the triangle.
Q11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that \(\angle CAD = \angle CBD.\)
Solution:
Consider right triangles \(ABC\) and \(ADC\) sharing the hypotenuse \(AC\). In each triangle, the right angle is at \(B\) and \(D\) respectively, so: \[ \angle ABC = 90^\circ, \\ \angle ADC = 90^\circ \]
In triangle \(ADC\), by the angle sum property: \[\scriptsize\begin{aligned} \angle CAD + \angle DCA + 90^\circ &= 180^\circ \\\implies \angle CAD + \angle DCA &= 90^\circ \end{aligned}\] Similarly, in triangle \(ABC\): \[\scriptsize\begin{aligned} \angle BAC + \angle BCA + 90^\circ &= 180^\circ \\\implies \angle BAC + \angle BCA &= 90^\circ \end{aligned}\]
Notice that the points \(A\), \(B\), \(C\), and \(D\) all lie on a circle (since each right angle is subtended by diameter \(AC\)). So, the quadrilateral \(ABCD\) is cyclic.
The property of cyclic quadrilaterals states that angles subtended by the same chord at the circumference are equal. Thus, angles \( \angle CAD \) and \( \angle CBD \) are subtended by chord \(CD\) from points \(A\) and \(B\): \[ \angle CAD = \angle CBD \]
Hence proved.
Q12. Prove that a cyclic parallelogram is a rectangle.
Solution:
Let \(ABCD\) be a cyclic parallelogram. To prove: \(ABCD\) is a rectangle.
In any cyclic quadrilateral, the sum of the measures of each pair of opposite angles is \(180^\circ\): \[ \angle A + \angle C = 180^\circ, \\ \angle B + \angle D = 180^\circ \]
In a parallelogram, opposite angles are equal: \[ \angle A = \angle C, \\ \angle B = \angle D \]
Combining these, since \( \angle A = \angle C \), we have: \[\begin{aligned} 2\angle A &= 180^\circ \\\implies \angle A &= 90^\circ \end{aligned}\] Similarly, \( \angle B = 90^\circ \).
Therefore, all angles of parallelogram \(ABCD\) are \(90^\circ\), so \(ABCD\) is a rectangle.
