An isosceles triangle has base 10 cm, and its area is 60 cm2. What are - End-of-Chapter Exercises

part 2 - Question 3 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar - Class 9
part 3 - Question 3 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar - Class 9 part 4 - Question 3 - End-of-Chapter Exercises - Chapter 6 Class 9 - Measuring Space: Perimeter and Area (Ganita Manjar - Class 9

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Question 3 An isosceles triangle has base 10 cm , and its area is 60ใ€–" " cmใ€—^2. What are the lengths of the equal sides? Since two sides are equal Let the equal sides be a So, a = a, b = 10, c = a We find Area using Herons formula Area of Triangle = โˆš(๐‘  (๐‘ โˆ’๐‘Ž)(๐‘ โˆ’๐‘)(๐‘ โˆ’๐‘)) Now, s = (๐’‚ + ๐’ƒ + ๐’„)/๐Ÿ = (๐‘Ž + 10 + ๐‘Ž)/2 = (2๐‘Ž + 10)/2 = 2๐‘Ž/2+10/2 = (a + 5) cm Now , Area of Triangle = โˆš(๐‘  (๐‘ โˆ’๐‘Ž)(๐‘ โˆ’๐‘)(๐‘ โˆ’๐‘)) 60 = โˆš((๐’‚+๐Ÿ“)(๐’‚+๐Ÿ“โˆ’๐’‚) ร— (๐’‚+๐Ÿ“โˆ’๐Ÿ๐ŸŽ) ร— (๐’‚+๐Ÿ“โˆ’๐’‚)) 60 = โˆš((๐‘Ž+5)(๐‘Ž+5โˆ’๐‘Ž) ร— (๐‘Ž+5โˆ’10) ร— (๐‘Ž+5โˆ’๐‘Ž)) = (๐‘Ž + 10 + ๐‘Ž)/2 = (2๐‘Ž + 10)/2 = 2๐‘Ž/2+10/2 = (a + 5) cm Now , Area of Triangle = โˆš(๐‘  (๐‘ โˆ’๐‘Ž)(๐‘ โˆ’๐‘)(๐‘ โˆ’๐‘)) 60 = โˆš((๐’‚+๐Ÿ“)(๐’‚+๐Ÿ“โˆ’๐’‚) ร— (๐’‚+๐Ÿ“โˆ’๐Ÿ๐ŸŽ) ร— (๐’‚+๐Ÿ“โˆ’๐’‚)) 60 = โˆš((๐‘Ž+5) ร— 5 ร— (๐‘Žโˆ’5) ร— 5) 60 = โˆš((๐’‚+๐Ÿ“) ร— (๐’‚โˆ’๐Ÿ“) ร— 5^2 ) 60 = โˆš((๐’‚^๐Ÿโˆ’๐Ÿ“^๐Ÿ) ร— 5^2 ) 60 = โˆš((๐‘Ž^2โˆ’25) ร— 5^2 ) 60 = โˆš((๐‘Ž^2โˆ’25) ) ร—โˆš(5^2 ) 60 = โˆš((๐’‚^๐Ÿโˆ’๐Ÿ๐Ÿ“) ) ร— ๐Ÿ“ 60/5=โˆš((๐‘Ž^2โˆ’25) ) 12=โˆš((๐‘Ž^2โˆ’25) ) Squaring both sides ๐Ÿ๐Ÿ^๐Ÿ=(๐’‚^๐Ÿโˆ’๐Ÿ๐Ÿ“) 144=๐‘Ž^2โˆ’25 144+25=๐‘Ž^2 169=๐‘Ž^2 ๐’‚^๐Ÿ=๐Ÿ๐Ÿ”๐Ÿ— ๐‘Ž^2=13^2 Cancelling squares ๐’‚=๐Ÿ๐Ÿ‘ Thus, length of equal sides is 13 cm

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