41 - 45 of 45 Questions
# | Question | Ans |
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41. |
In the diagram above, EFGH is a circle center O. Fh is a diameter and GE is a chord which meets FH at right angle at the point N. If NH = 8cm and EG = 24cm, calculate FH. A. 16 cm B. 20 cm C. 26 cm D. 32 cm Detailed SolutionThe locus of X is the perpendicular bisector of PQ |
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42. |
In the diagram, If < RPS = 50o, < RPQ = 30o and pq = QR, Find the value of < prs A. 50o B. 60o C. 70o D. 80o Detailed Solution< QPR = < PRQ = 30o (PQ = QR)< SPQ + < QRS = 180o (supplementary) 80 + < QRS = 180o = < QRS = 180o - 80o = 100o < QRP + < PRS = 100o = 30o + < PRS = 100o = < PRS = 100o - 30o = 70o |
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43. |
In the diagram, EFGH is a circle centre O. FH is a diameter and GE is a chord which meets FH at right angle at the point N. If NH = 8cm and EG = 24cm, calculate FH A. 32cm B. 26cm C. 20cm D. 16cm Detailed SolutionOE = OG = r(radii) (r - 8)2 + 122 = r2 r2 - 16r + 64 + 144 = r2 16r = 64 + 144 = r2 16r = 64 + 144 = 208 \(\frac{208}{16}\) = 13cm FH = 2FO = 2OH = 2R = 2 x 13cm = 26cm |
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44. |
If the diagram is the graph of y = x2, the shaded area is A. 64 square units B. \(\frac{126}{4}\) square units C. 32 square units D. \(\frac{64}{3}\) square units Detailed SolutionA = \(\int^{x = 4}_{x = 0}\) ydx = \(\int^{4}_{0}\) x2dx = [\(\frac{x^3}{3} + c]^{4}_{0}\)= \(\frac{64}{3}\) square units |
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45. |
The cumulative frequency curve represents the ages of ages of students in a school. What age group do 70% of the students belongs? A. 17.5 - 20.5 B. 16.5 - 19.5 C. 15.5 - 19.5 D. 15.5 - 18.5 |
A |
41. |
In the diagram above, EFGH is a circle center O. Fh is a diameter and GE is a chord which meets FH at right angle at the point N. If NH = 8cm and EG = 24cm, calculate FH. A. 16 cm B. 20 cm C. 26 cm D. 32 cm Detailed SolutionThe locus of X is the perpendicular bisector of PQ |
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42. |
In the diagram, If < RPS = 50o, < RPQ = 30o and pq = QR, Find the value of < prs A. 50o B. 60o C. 70o D. 80o Detailed Solution< QPR = < PRQ = 30o (PQ = QR)< SPQ + < QRS = 180o (supplementary) 80 + < QRS = 180o = < QRS = 180o - 80o = 100o < QRP + < PRS = 100o = 30o + < PRS = 100o = < PRS = 100o - 30o = 70o |
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43. |
In the diagram, EFGH is a circle centre O. FH is a diameter and GE is a chord which meets FH at right angle at the point N. If NH = 8cm and EG = 24cm, calculate FH A. 32cm B. 26cm C. 20cm D. 16cm Detailed SolutionOE = OG = r(radii) (r - 8)2 + 122 = r2 r2 - 16r + 64 + 144 = r2 16r = 64 + 144 = r2 16r = 64 + 144 = 208 \(\frac{208}{16}\) = 13cm FH = 2FO = 2OH = 2R = 2 x 13cm = 26cm |
44. |
If the diagram is the graph of y = x2, the shaded area is A. 64 square units B. \(\frac{126}{4}\) square units C. 32 square units D. \(\frac{64}{3}\) square units Detailed SolutionA = \(\int^{x = 4}_{x = 0}\) ydx = \(\int^{4}_{0}\) x2dx = [\(\frac{x^3}{3} + c]^{4}_{0}\)= \(\frac{64}{3}\) square units |
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45. |
The cumulative frequency curve represents the ages of ages of students in a school. What age group do 70% of the students belongs? A. 17.5 - 20.5 B. 16.5 - 19.5 C. 15.5 - 19.5 D. 15.5 - 18.5 |
A |