Question

13(a). In the diagram, the radius of the sector of circle centre O, is 7 cm and ∠MON is 60°. Find, correct to one decimal place, the area of the shaded portion. [Take π=22/7]
13(b). The x and y intercepts of a straight line are -3/4 and 2/7 respectively. Find the equation of the line.

Accepted Answer

(a) Area of sector ONM=(θ/360)×πr²=(60/360)×(22/7)×7²=25.67 cm². From triangle OTN: |OT|=7cos60°=3.5cm. |TN|=7sin60°=6.062cm (using the height from the triangle). Area of triangle ONT=(1/2)×3.5×6.062=10.61 cm². Area of shaded portion=Area of sector - Area of triangle=25.67-10.61=15.06≈15.1 cm² (1 d.p.).

(b) Using x/a+y/b=1 where a=-3/4 (x-intercept) and b=2/7 (y-intercept): x/(-3/4)+y/(2/7)=1 → -4x/3+7y/2=1. Multiplying through by 6: -8x+21y=6, or 21y=8x+6, giving y=(8x+6)/21 or 8x-21y+6=0.

WAEC Mathematics 2023 Theory — Question 13

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