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  1. Asked: In:

    The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.

    Piyush365

    Let the parallel sides be a = 40 cm and b = 20 cm and each equal non-parallel leg be c = 26 cm. Dropping perpendicular heights h from the upper base to the lower base forms two symmetrical right-angled triangles at the sides. Base of each right triangle = (40 - 20) / 2 = 20 / 2 = 10 cm. By PythagoraRead more

    Let the parallel sides be a = 40 cm and b = 20 cm and each equal non-parallel leg be c = 26 cm.

    Dropping perpendicular heights h from the upper base to the lower base forms two symmetrical right-angled triangles at the sides.

    Base of each right triangle = (40 – 20) / 2 = 20 / 2 = 10 cm.

    By Pythagoras theorem:

    h² + 10² = 26²

    h² = 676 – 100 = 576

    h = 24 cm.

    Area of trapezium = (1/2) x (sum of parallel sides) x height = (1/2) x (40 + 20) x 24 = 30 x 24 = 720 cm².

     

    For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/

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  2. Asked: In:

    Find the area of a triangle, given that its sides are 8 cm and 11 cm long, and its perimeter is 32 cm.

    Piyush365

    Given perimeter = 32 cm and two sides a = 8 cm, b = 11 cm. Third side c = 32 - (8 + 11) = 32 - 19 = 13 cm. Semi-perimeter s = perimeter / 2 = 32 / 2 = 16 cm. Now apply Heron's formula: Area = √(s x (s - a) x (s - b) x (s - c)) Area = √(16 x (16 - 8) x (16 - 11) x (16 - 13)) Area = √(16 x 8 x 5 x 3)Read more

    Given perimeter = 32 cm and two sides a = 8 cm, b = 11 cm.

    Third side c = 32 – (8 + 11) = 32 – 19 = 13 cm.

    Semi-perimeter s = perimeter / 2 = 32 / 2 = 16 cm.

    Now apply Heron’s formula:

    Area = √(s x (s – a) x (s – b) x (s – c))

    Area = √(16 x (16 – 8) x (16 – 11) x (16 – 13))

    Area = √(16 x 8 x 5 x 3)

    Area = √(16 x 4 x 2 x 15)

    Area = 4 x 2 x √30 = 8√30 cm² (approximately 43.82 cm²).

     

    For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/

     

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  3. Asked: In:

    The sides of a triangular plot are in the ratio 3: 5: 7; its perimeter is 300 m. Find its area.

    Piyush365

    Let the sides be 3x, 5x and 7x meters. Perimeter = 3x + 5x + 7x = 300 m. 15x = 300, which gives x = 20 m. The side lengths are: a = 3 x 20 = 60 m b = 5 x 20 = 100 m c = 7 x 20 = 140 m. Semi-perimeter s = 300 / 2 = 150 m. Using Heron's formula: Area = √(150 x (150 - 60) x (150 - 100) x (150 - 140)) ARead more

    Let the sides be 3x, 5x and 7x meters.

    Perimeter = 3x + 5x + 7x = 300 m.

    15x = 300, which gives x = 20 m.

    The side lengths are:

    a = 3 x 20 = 60 m

    b = 5 x 20 = 100 m

    c = 7 x 20 = 140 m.

    Semi-perimeter s = 300 / 2 = 150 m.

    Using Heron’s formula:

    Area = √(150 x (150 – 60) x (150 – 100) x (150 – 140))

    Area = √(150 x 90 x 50 x 10)

    Area = √(67,500,000) = 1500√3 m² (approx 2598.08 m²).

     

    For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/

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    • 16
  5. Asked: In:

    I throw a pair of 6-sided dice. Write down an event that has a probability of 0 and an outcome that has a probability of 1.

    Sonal Chauhan

    When throwing two 6-sided dice, the sum of numbers ranges from 1 + 1 = 2 to 6 + 6 = 12. Event with probability 0: 'Getting a sum of 13' (or getting a negative sum). This is an impossible event because the highest possible total is 12. Event with probability 1: 'Getting a sum between 2 and 12 inclusiRead more

    When throwing two 6-sided dice, the sum of numbers ranges from 1 + 1 = 2 to 6 + 6 = 12.

    Event with probability 0: ‘Getting a sum of 13’ (or getting a negative sum).

    This is an impossible event because the highest possible total is 12.

    Event with probability 1: ‘Getting a sum between 2 and 12 inclusive’ (or getting a sum less than 13). This is a certain event because every possible outcome satisfies it.

     

    For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/

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    • 22
  6. Asked: In:

    A child has 2 shirts (one red and one blue) and 3 types of pants (jeans, khakis and shorts). List all the possible combinations of outfits consisting of one shirt and one pair of pants. Display your answer in a table format.

    Sonal Chauhan

    The child has shirts {Red, Blue} and pants {Jeans, Khakis, Shorts}. Total number of outfit combinations is 2 x 3 = 6. Outfit 1 is (Red Shirt, Jeans), Outfit 2 is (Red Shirt, Khakis), Outfit 3 is (Red Shirt, Shorts), Outfit 4 is (Blue Shirt, Jeans), Outfit 5 is (Blue Shirt, Khakis) and Outfit 6 is (BRead more

    The child has shirts {Red, Blue} and pants {Jeans, Khakis, Shorts}. Total number of outfit combinations is 2 x 3 = 6.

    Outfit 1 is (Red Shirt, Jeans),

    Outfit 2 is (Red Shirt, Khakis),

    Outfit 3 is (Red Shirt, Shorts),

    Outfit 4 is (Blue Shirt, Jeans),

    Outfit 5 is (Blue Shirt, Khakis) and

    Outfit 6 is (Blue Shirt, Shorts).

    Thus, the sample space comprises exactly these 6 outfit combinations.

     

    For more NCERT Solutions of Class 9 Ganita Manjari Chapter 7 The Mathematics of Maybe: Introduction to Probability Question Answer (2026-27)

    https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-7/

     

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    • 22
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