Given base b = 10 cm and Area = 60 cm². Area = (1/2) x base x height 60 = (1/2) x 10 x h 60 = 5h, which gives height h = 12 cm. The altitude to the base of an isosceles triangle bisects the base into two segments of length 10 / 2 = 5 cm. By Pythagoras theorem, each equal side acts as the hypotenuse:Read more
Given base b = 10 cm and Area = 60 cm².
Area = (1/2) x base x height
60 = (1/2) x 10 x h
60 = 5h, which gives height h = 12 cm.
The altitude to the base of an isosceles triangle bisects the base into two segments of length 10 / 2 = 5 cm.
By Pythagoras theorem, each equal side acts as the hypotenuse:
side² = 5² + 12² = 25 + 144 = 169
side = √169 = 13 cm.
Therefore, the lengths of the equal sides are 13 cm each.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Area of a right-angled triangle = (1/2) x leg1 x leg2. Given Area = 54 sq. cm and leg1 = 12 cm: 54 = (1/2) x 12 x leg2 = 6 x leg2 leg2 = 54 / 6 = 9 cm. By Baudhāyana-Pythagoras theorem, hypotenuse = √(leg1² + leg2²): hypotenuse = √(12² + 9²) = √(144 + 81) = √225 = 15 cm. Perimeter = leg1 + leg2 + hyRead more
Area of a right-angled triangle = (1/2) x leg1 x leg2.
Given Area = 54 sq. cm and leg1 = 12 cm:
54 = (1/2) x 12 x leg2 = 6 x leg2
leg2 = 54 / 6 = 9 cm.
By Baudhāyana-Pythagoras theorem, hypotenuse = √(leg1² + leg2²):
Let the side lengths be 2x, 3x and 4x. Perimeter = 2x + 3x + 4x = 9x = 45 cm, so x = 5 cm. The sides are a = 10 cm, b = 15 cm, c = 20 cm. Semi-perimeter s = 45 / 2 = 22.5 cm. Using Heron's formula: Area = √(s(s - a)(s - b)(s - c)) Area = √(22.5 x (22.5 - 10) x (22.5 - 15) x (22.5 - 20)) Area = √(22.Read more
Let the side lengths be 2x, 3x and 4x.
Perimeter = 2x + 3x + 4x = 9x = 45 cm, so x = 5 cm.
The sides are a = 10 cm, b = 15 cm, c = 20 cm.
Semi-perimeter s = 45 / 2 = 22.5 cm.
Using Heron’s formula:
Area = √(s(s – a)(s – b)(s – c))
Area = √(22.5 x (22.5 – 10) x (22.5 – 15) x (22.5 – 20))
Method 1 (Right Triangle Formula): Check sides with Pythagoras theorem: 7² + 24² = 49 + 576 = 625 = 25². This is a right-angled triangle with base 7 cm and height 24 cm. Area = (1/2) x base x height = (1/2) x 7 x 24 = 84 cm². Method 2 (Heron's Formula): Semi-perimeter s = (7 + 24 + 25) / 2 = 56 / 2Read more
The theatre club was 1. winding up their practice. Anil saw Sunny was 2. wearing a look of distress and not speaking to anyone. They had a disagreement in the morning and since then, Sunny was 3. lost in his thoughts. Anil did not want to 4. bring it up and disturb Sunny further but he finally decidRead more
The theatre club was 1. winding up their practice. Anil saw Sunny was 2. wearing a look of distress and not speaking to anyone. They had a disagreement in the morning and since then, Sunny was 3. lost in his thoughts. Anil did not want to 4. bring it up and disturb Sunny further but he finally decided to 5. bite the bullet and speak to Sunny. He was sure if he apologised first, his friend would 6. come around. With a lot of anxiety, he 7. found words to apologise. And finally, Sunny smiled! Everyone clapped and asked them to 8. throw a party to celebrate.
For more NCERT Solutions of Class 9 English Kaveri Chapter 6 Twin Melodies Question Answer (2026-27)
An isosceles triangle has base 10 cm and its area is 60 cm². What are the lengths of the equal sides?
Given base b = 10 cm and Area = 60 cm². Area = (1/2) x base x height 60 = (1/2) x 10 x h 60 = 5h, which gives height h = 12 cm. The altitude to the base of an isosceles triangle bisects the base into two segments of length 10 / 2 = 5 cm. By Pythagoras theorem, each equal side acts as the hypotenuse:Read more
Given base b = 10 cm and Area = 60 cm².
Area = (1/2) x base x height
60 = (1/2) x 10 x h
60 = 5h, which gives height h = 12 cm.
The altitude to the base of an isosceles triangle bisects the base into two segments of length 10 / 2 = 5 cm.
By Pythagoras theorem, each equal side acts as the hypotenuse:
side² = 5² + 12² = 25 + 144 = 169
side = √169 = 13 cm.
Therefore, the lengths of the equal sides are 13 cm each.
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/
See lessThe area of a right-angled triangle is 54 sq. cm. One of its legs has length 12 cm. Find its perimeter.
Area of a right-angled triangle = (1/2) x leg1 x leg2. Given Area = 54 sq. cm and leg1 = 12 cm: 54 = (1/2) x 12 x leg2 = 6 x leg2 leg2 = 54 / 6 = 9 cm. By Baudhāyana-Pythagoras theorem, hypotenuse = √(leg1² + leg2²): hypotenuse = √(12² + 9²) = √(144 + 81) = √225 = 15 cm. Perimeter = leg1 + leg2 + hyRead more
Area of a right-angled triangle = (1/2) x leg1 x leg2.
Given Area = 54 sq. cm and leg1 = 12 cm:
54 = (1/2) x 12 x leg2 = 6 x leg2
leg2 = 54 / 6 = 9 cm.
By Baudhāyana-Pythagoras theorem, hypotenuse = √(leg1² + leg2²):
hypotenuse = √(12² + 9²) = √(144 + 81) = √225 = 15 cm.
Perimeter = leg1 + leg2 + hypotenuse = 12 + 9 + 15 = 36 cm.
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/
See lessThe sides of a triangle are in the ratio 2: 3: 4 and its perimeter is 45 cm. Find its area.
Let the side lengths be 2x, 3x and 4x. Perimeter = 2x + 3x + 4x = 9x = 45 cm, so x = 5 cm. The sides are a = 10 cm, b = 15 cm, c = 20 cm. Semi-perimeter s = 45 / 2 = 22.5 cm. Using Heron's formula: Area = √(s(s - a)(s - b)(s - c)) Area = √(22.5 x (22.5 - 10) x (22.5 - 15) x (22.5 - 20)) Area = √(22.Read more
Let the side lengths be 2x, 3x and 4x.
Perimeter = 2x + 3x + 4x = 9x = 45 cm, so x = 5 cm.
The sides are a = 10 cm, b = 15 cm, c = 20 cm.
Semi-perimeter s = 45 / 2 = 22.5 cm.
Using Heron’s formula:
Area = √(s(s – a)(s – b)(s – c))
Area = √(22.5 x (22.5 – 10) x (22.5 – 15) x (22.5 – 20))
Area = √(22.5 x 12.5 x 7.5 x 2.5)
Area = √[(45/2) x (25/2) x (15/2) x (5/2)]
Area = √(84375 / 16) = (75 / 4)√15 cm² (approximately 72.62 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/
See lessThe sides of a triangle have lengths 7 cm, 24 cm, 25 cm. Find the area of the triangle in two different ways.
Method 1 (Right Triangle Formula): Check sides with Pythagoras theorem: 7² + 24² = 49 + 576 = 625 = 25². This is a right-angled triangle with base 7 cm and height 24 cm. Area = (1/2) x base x height = (1/2) x 7 x 24 = 84 cm². Method 2 (Heron's Formula): Semi-perimeter s = (7 + 24 + 25) / 2 = 56 / 2Read more
Method 1 (Right Triangle Formula):
Check sides with Pythagoras theorem: 7² + 24² = 49 + 576 = 625 = 25².
This is a right-angled triangle with base 7 cm and height 24 cm.
Area = (1/2) x base x height = (1/2) x 7 x 24 = 84 cm².
Method 2 (Heron’s Formula):
Semi-perimeter s = (7 + 24 + 25) / 2 = 56 / 2 = 28 cm.
Area = √(28 x (28 – 7) x (28 – 24) x (28 – 25))
= √(28 x 21 x 4 x 3)
= √(7056) = 84 cm².
Both methods give 84 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/
See lessWork in pairs to complete the paragraph given on the next page by choosing the correct phrases given in the box. Discuss with your teacher whether the following phrases have a literal meaning or a figurative one. Paragraph: The theatre club was 1. ________ their practice. Anil saw Sunny was 2. ________ and not speaking to anyone. They had a disagreement in the morning and since then, Sunny was 3. ________. Anil did not want to 4. ________ and disturb Sunny further but he finally decided to 5. ________ and speak to Sunny. He was sure if he apologised first, his friend would 6. ________. With a lot of anxiety, he 7. ________ to apologise. And finally, Sunny smiled! Everyone clapped and asked them to 8. ________ to celebrate.
The theatre club was 1. winding up their practice. Anil saw Sunny was 2. wearing a look of distress and not speaking to anyone. They had a disagreement in the morning and since then, Sunny was 3. lost in his thoughts. Anil did not want to 4. bring it up and disturb Sunny further but he finally decidRead more
The theatre club was 1. winding up their practice. Anil saw Sunny was 2. wearing a look of distress and not speaking to anyone. They had a disagreement in the morning and since then, Sunny was 3. lost in his thoughts. Anil did not want to 4. bring it up and disturb Sunny further but he finally decided to 5. bite the bullet and speak to Sunny. He was sure if he apologised first, his friend would 6. come around. With a lot of anxiety, he 7. found words to apologise. And finally, Sunny smiled! Everyone clapped and asked them to 8. throw a party to celebrate.
For more NCERT Solutions of Class 9 English Kaveri Chapter 6 Twin Melodies Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/english/kaveri-chapter-6/
See less