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An isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.
Given equal sides a = 15 cm, b = 15 cm and perimeter = 40 cm. Third side (base) c = 40 - (15 + 15) = 40 - 30 = 10 cm. Semi-perimeter s = 40 / 2 = 20 cm. Using Heron's formula: Area = √(s x (s - a) x (s - b) x (s - c)) Area = √(20 x (20 - 15) x (20 - 15) x (20 - 10)) Area = √(20 x 5 x 5 x 10) Area =Read more
Given equal sides a = 15 cm, b = 15 cm and perimeter = 40 cm.
Third side (base) c = 40 – (15 + 15) = 40 – 30 = 10 cm.
Semi-perimeter s = 40 / 2 = 20 cm.
Using Heron’s formula:
Area = √(s x (s – a) x (s – b) x (s – c))
Area = √(20 x (20 – 15) x (20 – 15) x (20 – 10))
Area = √(20 x 5 x 5 x 10)
Area = √(5000) = √(2500 x 2) = 50√2 cm².
In decimal form, 50 x 1.414 is approximately 70.71 cm².
Hence, the area of the triangle is 50√2 cm² (or 70.71 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 lessAn 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 less