41 - 48 of 48 Questions
# | Question | Ans |
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41. |
The figure FGHK is a rhombus. What is the value of angle X? A. 30o B. 90o C. 150o D. 120o E. 60o Detailed Solution< HKF = 60o, < KFG = 120o< KFG = < KHG = x(opposite angles) x = 120o |
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42. |
PQRS is a desk of dimensions 2m x 0.8 which is inclined at 30o to the horizontal. Find the inclination of the diagonal PR to the horizontal A. 30o 35' B. 30o 32' C. 15o 36' D. 10o E. 10o 42' Detailed Solutiontan\(\theta\) = \(\frac{0.8}{2}\) = (0.4)\(\theta\) = tan-1(0.4) From the diagram, the inclination of the diagonal PR to the horizontal is 10o 42' |
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43. |
If O is the centre of the circle in the figure, find the value of x A. 50 B. 260 C. 100 D. 65 E. 130 Detailed SolutionFrom the diagram; The value of x = 360o - 2(130o)= 360 - 260 = 100o |
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44. |
PQRS isa cyclic quadrilateral which PQ = PS. PT is a tangent to the circle and PQ makes an angle of 50o with the tangent as shown in the figure. What is the size of QRS? A. 50o B. 40o C. 110v D. 80o E. 100o Detailed Solution< SPQ = 80o< SPQ + < SRQ = 180(Supplementary) 80 + < QRS = 180o - 80o = 100o |
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45. |
The farm yield of four crops on a piece of land in Ondo are represented on the pie chart. what is the angle of the sector occupies by Okro in the chart? A. 11o B. 19\(\frac{1}{3}\)o C. 33\(\frac{1}{3}\)o D. 91\(\frac{1}{3}\)o E. 91o Detailed SolutionAdding the values of all the items together, it gives 70Okro sector = \(\frac{145}{270}\) x \(\frac{360^o}{1}\) = 19.33o = 19\(\frac{1}{3}\)o |
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46. |
In the figure, PQR is a straight line. Find the values of x and y A. x = 22.5o, y = 33.75o B. x = 15o, y = 52.5o C. x = 22.5o, y = 45.0o D. x = 56.25o, y = 11.5o Detailed Solution\(\frac{3}{2}\)x + 3y + 45° = 180°3x + 6y + 90° = 360° 3x + 6y = 270.......(i) x 2, 5x + y + y = 180° 5x + 2y = 180° ... (ii) x 6 6x + 12y = 540 ... (iii) 30x + 12y = 1080 ... (iv) eqn(iv) - eqn(iii) 24x = 540 x = 22.5° and y = 33.75° |
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47. |
In the diagram, if PS = SR and PQ||SQ, what is the size of PQR? A. 25o B. 50o C. 55o D. 65o E. 75o Detailed Solution< PQR + , PSR 180(Opposite angles in a Quadrilateral are supplementary)< PQR + 130o = 180 < PQR = 180 - 130 = 50o |
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48. |
In the figure PT is a tangent to the circle with centre at O. If PQT = 30o, find the value of PTO A. 30o B. 50o C. 24o D. 12o E. 60o Detailed SolutionFROM the diagram, PQT = 50oPTQ = 50o(opposite angles are supplementary) |
41. |
The figure FGHK is a rhombus. What is the value of angle X? A. 30o B. 90o C. 150o D. 120o E. 60o Detailed Solution< HKF = 60o, < KFG = 120o< KFG = < KHG = x(opposite angles) x = 120o |
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42. |
PQRS is a desk of dimensions 2m x 0.8 which is inclined at 30o to the horizontal. Find the inclination of the diagonal PR to the horizontal A. 30o 35' B. 30o 32' C. 15o 36' D. 10o E. 10o 42' Detailed Solutiontan\(\theta\) = \(\frac{0.8}{2}\) = (0.4)\(\theta\) = tan-1(0.4) From the diagram, the inclination of the diagonal PR to the horizontal is 10o 42' |
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43. |
If O is the centre of the circle in the figure, find the value of x A. 50 B. 260 C. 100 D. 65 E. 130 Detailed SolutionFrom the diagram; The value of x = 360o - 2(130o)= 360 - 260 = 100o |
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44. |
PQRS isa cyclic quadrilateral which PQ = PS. PT is a tangent to the circle and PQ makes an angle of 50o with the tangent as shown in the figure. What is the size of QRS? A. 50o B. 40o C. 110v D. 80o E. 100o Detailed Solution< SPQ = 80o< SPQ + < SRQ = 180(Supplementary) 80 + < QRS = 180o - 80o = 100o |
45. |
The farm yield of four crops on a piece of land in Ondo are represented on the pie chart. what is the angle of the sector occupies by Okro in the chart? A. 11o B. 19\(\frac{1}{3}\)o C. 33\(\frac{1}{3}\)o D. 91\(\frac{1}{3}\)o E. 91o Detailed SolutionAdding the values of all the items together, it gives 70Okro sector = \(\frac{145}{270}\) x \(\frac{360^o}{1}\) = 19.33o = 19\(\frac{1}{3}\)o |
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46. |
In the figure, PQR is a straight line. Find the values of x and y A. x = 22.5o, y = 33.75o B. x = 15o, y = 52.5o C. x = 22.5o, y = 45.0o D. x = 56.25o, y = 11.5o Detailed Solution\(\frac{3}{2}\)x + 3y + 45° = 180°3x + 6y + 90° = 360° 3x + 6y = 270.......(i) x 2, 5x + y + y = 180° 5x + 2y = 180° ... (ii) x 6 6x + 12y = 540 ... (iii) 30x + 12y = 1080 ... (iv) eqn(iv) - eqn(iii) 24x = 540 x = 22.5° and y = 33.75° |
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47. |
In the diagram, if PS = SR and PQ||SQ, what is the size of PQR? A. 25o B. 50o C. 55o D. 65o E. 75o Detailed Solution< PQR + , PSR 180(Opposite angles in a Quadrilateral are supplementary)< PQR + 130o = 180 < PQR = 180 - 130 = 50o |
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48. |
In the figure PT is a tangent to the circle with centre at O. If PQT = 30o, find the value of PTO A. 30o B. 50o C. 24o D. 12o E. 60o Detailed SolutionFROM the diagram, PQT = 50oPTQ = 50o(opposite angles are supplementary) |