I'm Up and Down, and Round and Round — Important Questions
CBSE · Class 9 · Mathematics
90 important questions from I'm Up and Down, and Round and Round for CBSE Class 9 Mathematics, with answers. Part of the CBSE Class 9 Mathematics syllabus.
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Important Questions from I'm Up and Down, and Round and Round
A right-angled triangle has legs of 6 cm and 8 cm. Find the distance between the circumcentre and each vertex.
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5 cm and 5 cm for the two acute vertices, 5 cm for the right angle vertex too
First find the hypotenuse: √(6² + 8²) = √(36 + 64) = √100 = 10 cm. The circumcentre of a right triangle is the midpoint of the hypotenuse, so the circumradius is 10 ÷ 2 = 5 cm. The circumcentre is 5 cm from all three vertices.
A circle has radius 17 cm. A chord is 16 cm long. What is the distance of the chord from the centre?
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15 cm
Half the chord is 8 cm. Distance = √(17² - 8²) = √(289 - 64) = √225 = 15 cm.
A chord is 30 cm long in a circle of radius 25 cm. Find the perpendicular distance of the chord from the centre.
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20 cm
Half the chord is 15 cm. Distance = √(25² - 15²) = √(625 - 225) = √400 = 20 cm.
Two chords of a circle are 18 cm and 24 cm long. Which chord is nearer to the centre, and by how many centimetres if the radius is 15 cm?
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The 24 cm chord, by 3 cm
For 18 cm, half is 9 cm, so distance = √(15² - 9²) = √(225 - 81) = √144 = 12 cm. For 24 cm, half is 12 cm, so distance = √(225 - 144) = √81 = 9 cm. The 24 cm chord is nearer by 12 - 9 = 3 cm.
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Sources & Official References
- NCERT Official — ncert.nic.in
- CBSE Academic — cbseacademic.nic.in
- CBSE Official — cbse.gov.in
- National Education Policy 2020 — education.gov.in
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