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How would you use the following figure to justify the statement that the angle in a semicircle is 90°?
Let the endpoints of the diameter be B and C, and let A be a point on the semicircle. Join OA. Since OA, OB and OC are radii, OA = OB = OC. Therefore, the two triangles formed with OA are isosceles. If the angles at B and C are a and b, the corresponding central angles are 180° - 2a and 180° - 2b. SRead more
Let the endpoints of the diameter be B and C, and let A be a point on the semicircle. Join OA. Since OA, OB and OC are radii, OA = OB = OC. Therefore, the two triangles formed with OA are isosceles. If the angles at B and C are a and b, the corresponding central angles are 180° – 2a and 180° – 2b. Since BOC is a straight angle, a + b = 90°. Hence, ∠BAC = 90°.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/
See lessLet A be any point within a given circle with centre O. Show that the shortest chord of the circle that passes through point A is the one that is perpendicular to OA.
Let a chord through A meet the circle at B and C. The perpendicular distance of this chord from O determines its length. Since A lies inside the circle, the greatest possible distance of a chord through A from O is OA itself. This occurs when the chord is perpendicular to OA. A chord farther from thRead more
Let a chord through A meet the circle at B and C. The perpendicular distance of this chord from O determines its length. Since A lies inside the circle, the greatest possible distance of a chord through A from O is OA itself. This occurs when the chord is perpendicular to OA. A chord farther from the centre is shorter. Therefore, among all chords passing through A, the shortest chord is the one perpendicular to OA. Hence proved.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/
See lessThere is no chord of a circle that is longer than its diameter. How do you justify this statement?
Let the radius of the circle be r. The diameter has length 2r and passes through the centre. For any other chord, the perpendicular from the centre bisects the chord. If its distance from the centre is d, then half the chord is less than r whenever d is greater than zero. Thus, its complete length iRead more
Let the radius of the circle be r. The diameter has length 2r and passes through the centre. For any other chord, the perpendicular from the centre bisects the chord. If its distance from the centre is d, then half the chord is less than r whenever d is greater than zero. Thus, its complete length is less than 2r. Therefore, the diameter is the longest chord, and no chord can be longer than it.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/
See lessLet ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., ∠CDE = ∠ABC, where E is a point on the extension of side CD).
Let ABCD be a cyclic quadrilateral and let CD be extended to E. Then ∠CDE and ∠CDA form a linear pair, so their sum is 180°. Also, opposite angles of a cyclic quadrilateral are supplementary. Therefore, ∠ABC + ∠CDA = 180° and ∠CDE + ∠CDA = 180°. Hence, ∠CDE = ∠ABC. Thus, the exterior angle of a cyclRead more
Let ABCD be a cyclic quadrilateral and let CD be extended to E. Then ∠CDE and ∠CDA form a linear pair, so their sum is 180°. Also, opposite angles of a cyclic quadrilateral are supplementary.
Therefore,
∠ABC + ∠CDA = 180°
and
∠CDE + ∠CDA = 180°.
Hence,
∠CDE = ∠ABC.
Thus, the exterior angle of a cyclic quadrilateral is equal to its interior opposite angle.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/
See lessA quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about ∠MOP and ∠MNP? Explain your reasoning.
Since MNOP is inscribed in a circle, both ∠MOP and ∠MNP subtend the same arc MP. ∠MOP is formed at the centre of the circle, whereas ∠MNP is formed at a point on the circumference. According to the theorem, the angle subtended by an arc at the centre is twice the angle subtended by the same arc at aRead more
Since MNOP is inscribed in a circle, both ∠MOP and ∠MNP subtend the same arc MP. ∠MOP is formed at the centre of the circle, whereas ∠MNP is formed at a point on the circumference. According to the theorem, the angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the remaining circle.
Therefore,
∠MOP = 2∠MNP.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-5/
See less