Let ABCD be a rectangle inscribed in a circle. Its diagonals AC and BD are equal and bisect each other. Since AC and BD are equal chords passing through the centre, they are diameters of the circle. The diagonals of the rectangle intersect at their common midpoint. The centre of a circle is the midpRead more
Let ABCD be a rectangle inscribed in a circle. Its diagonals AC and BD are equal and bisect each other. Since AC and BD are equal chords passing through the centre, they are diameters of the circle. The diagonals of the rectangle intersect at their common midpoint. The centre of a circle is the midpoint of every diameter. Therefore, the point where the diagonals of the rectangle intersect must be the centre of the circle.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
Let ABCD be a parallelogram inscribed in a circle. Since it is cyclic, its opposite angles are supplementary. Also, opposite angles of a parallelogram are equal. Therefore, ∠A + ∠C = 180° and ∠A = ∠C. Thus, 2∠A = 180°, so ∠A = 90°. Similarly, all four angles are 90°. Hence, the parallelogram is a reRead more
Let ABCD be a parallelogram inscribed in a circle. Since it is cyclic, its opposite angles are supplementary.
Also, opposite angles of a parallelogram are equal.
Therefore,
∠A + ∠C = 180°
and ∠A = ∠C.
Thus, 2∠A = 180°, so ∠A = 90°. Similarly, all four angles are 90°.
Hence, the parallelogram is a rectangle. Therefore, a rectangle is the only parallelogram that can be inscribed in a circle.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I’m Up and Down and Round and Round (2026-27):
To factor this six-term polynomial, we look at the signs of the products to determine which base is negative. The perfect square bases are 3a, 2b and c. We notice that the terms 12ab and 4bc are negative, while 6ac is positive. Since the negative terms both contain the variable b, the base associateRead more
To factor this six-term polynomial, we look at the signs of the products to determine which base is negative. The perfect square bases are 3a, 2b and c. We notice that the terms 12ab and 4bc are negative, while 6ac is positive. Since the negative terms both contain the variable b, the base associated with b must carry the negative sign. This results in the factor (3a – 2b + c) square.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
We use the three-term square identity to expand the given expression. By setting a equal to p, b equal to 3q and c equal to 7r, we square each term to get p square, 9q square and 49r square. Then we calculate the three double cross-product terms: two times p times 3q gives 6pq; two times 3q times 7rRead more
We use the three-term square identity to expand the given expression. By setting a equal to p, b equal to 3q and c equal to 7r, we square each term to get p square, 9q square and 49r square. Then we calculate the three double cross-product terms: two times p times 3q gives 6pq; two times 3q times 7r gives 42qr; two times 7r times p gives 14rp. Combining them gives the full expansion.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
We expand this expression using the three-term standard identity. Here, our terms are 3x, minus 2y and 4z. Squaring these individual parts results in 9x square, 4y square and 16z square. When calculating the paired products, any term multiplied by minus 2y becomes negative, resulting in minus 12xy aRead more
We expand this expression using the three-term standard identity. Here, our terms are 3x, minus 2y and 4z. Squaring these individual parts results in 9x square, 4y square and 16z square. When calculating the paired products, any term multiplied by minus 2y becomes negative, resulting in minus 12xy and minus 16yz. The final product two times 4z times 3x stays positive at 24zx.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.
Let ABCD be a rectangle inscribed in a circle. Its diagonals AC and BD are equal and bisect each other. Since AC and BD are equal chords passing through the centre, they are diameters of the circle. The diagonals of the rectangle intersect at their common midpoint. The centre of a circle is the midpRead more
Let ABCD be a rectangle inscribed in a circle. Its diagonals AC and BD are equal and bisect each other. Since AC and BD are equal chords passing through the centre, they are diameters of the circle. The diagonals of the rectangle intersect at their common midpoint. The centre of a circle is the midpoint of every diameter. Therefore, the point where the diagonals of the rectangle intersect must be the centre of the circle.
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 lessShow that rectangle is the only parallelogram that can be inscribed in a circle.
Let ABCD be a parallelogram inscribed in a circle. Since it is cyclic, its opposite angles are supplementary. Also, opposite angles of a parallelogram are equal. Therefore, ∠A + ∠C = 180° and ∠A = ∠C. Thus, 2∠A = 180°, so ∠A = 90°. Similarly, all four angles are 90°. Hence, the parallelogram is a reRead more
Let ABCD be a parallelogram inscribed in a circle. Since it is cyclic, its opposite angles are supplementary.
Also, opposite angles of a parallelogram are equal.
Therefore,
∠A + ∠C = 180°
and ∠A = ∠C.
Thus, 2∠A = 180°, so ∠A = 90°. Similarly, all four angles are 90°.
Hence, the parallelogram is a rectangle. Therefore, a rectangle is the only parallelogram that can be inscribed in a circle.
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 lessFactor using suitable identities: 9a square + 4b square + c square – 12ab + 6ac – 4bc
To factor this six-term polynomial, we look at the signs of the products to determine which base is negative. The perfect square bases are 3a, 2b and c. We notice that the terms 12ab and 4bc are negative, while 6ac is positive. Since the negative terms both contain the variable b, the base associateRead more
To factor this six-term polynomial, we look at the signs of the products to determine which base is negative. The perfect square bases are 3a, 2b and c. We notice that the terms 12ab and 4bc are negative, while 6ac is positive. Since the negative terms both contain the variable b, the base associated with b must carry the negative sign. This results in the factor (3a – 2b + c) square.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessExpand using the identity (a + b + c) square: (p + 3q + 7r) square
We use the three-term square identity to expand the given expression. By setting a equal to p, b equal to 3q and c equal to 7r, we square each term to get p square, 9q square and 49r square. Then we calculate the three double cross-product terms: two times p times 3q gives 6pq; two times 3q times 7rRead more
We use the three-term square identity to expand the given expression. By setting a equal to p, b equal to 3q and c equal to 7r, we square each term to get p square, 9q square and 49r square. Then we calculate the three double cross-product terms: two times p times 3q gives 6pq; two times 3q times 7r gives 42qr; two times 7r times p gives 14rp. Combining them gives the full expansion.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See lessExpand using the identity (a + b + c) square: (3x – 2y + 4z) square
We expand this expression using the three-term standard identity. Here, our terms are 3x, minus 2y and 4z. Squaring these individual parts results in 9x square, 4y square and 16z square. When calculating the paired products, any term multiplied by minus 2y becomes negative, resulting in minus 12xy aRead more
We expand this expression using the three-term standard identity. Here, our terms are 3x, minus 2y and 4z. Squaring these individual parts results in 9x square, 4y square and 16z square. When calculating the paired products, any term multiplied by minus 2y becomes negative, resulting in minus 12xy and minus 16yz. The final product two times 4z times 3x stays positive at 24zx.
For more NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 4 Exploring Algebraic Identities (2026-27):
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-4/
See less