Question 4 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar
Last updated at June 3, 2026 by Teachoo
End-of-Chapter Exercises
Question 1
Important
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Class 9
Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar
- Definition (and Quesitons)
- Perimeter of Different Shapes
- Perimeter of a Circle - C/D Ratio
- Pi is Irrational
- Length of an Arc of a Circle
- Problems, Puzzles, and Paradoxes on Perimeter
- Exercise Set 6.1
- Area of Square & Rectangle
- Area of a Parallelogram
- Area of a Triangle
- Heron's Formula
- Circumcircle and Incircle of a Triangle
- Brahmagupta’s Formula for the Area of a Cyclic 4-gon
- Squaring a Rectangle
- Exercise Set 6.2
- Area of a circle
- Exercise Set 6.3
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End-of-Chapter Exercises
Transcript
Question 4 The area of a right-angled triangle is 54 sq.cm. One of its legs has length 12 cm. Find its perimeter. Let the other two sides of right triangle be a & c We need to find perimeter Thus, we need to find all 3 sides Given Base = 12 cm Area = 54 cm2 Now, Area of triangle = 1/2 × Base × Height 54 = 𝟏/𝟐 × 12 × a 54 = 6 × a 54/6 = a 9 = a a = 9 cm Since it is a right angled triangle By Pythagoras Theorem 𝑯𝒚𝒑𝒐𝒕𝒆𝒏𝒖𝒔𝒆^𝟐=𝑩𝒂𝒔𝒆^𝟐+𝑯𝒆𝒊𝒈𝒉𝒕^𝟐 𝑐^2=𝑎^2+12^2 𝒄^𝟐=𝟗^𝟐+〖𝟏𝟐〗^𝟐 𝑐^2=81+144 𝒄^𝟐=𝟐𝟐𝟓 𝑐^2=15^2 Cancelling squares 𝑐^2=15^2 9^2+12^2 We need to find perimeter Thus, we need to find all 3 sides 𝒄^𝟐=𝟐𝟐𝟓 𝑐^2=15^2 Cancelling squares 𝐜=𝟏𝟓 cm Now, Perimeter = a + 12 + c = 9 + 12 + 15 = 36 cm
