WAEC 2020 · Paper 2 · Q4

  1. (a)

    In the diagram, RR, SS, TT, UU are points on a circle and QRSQRS is a straight line. ∠STU=124∘\angle STU = 124^\circ and ∠QUR=31∘\angle QUR = 31^\circ. Find ∠RQU\angle RQU.

    31°124°UTSRQ
  2. (b)

    The ratio of the length of an arc of a circle to the circumference of the circle is 3:73 : 7. If the diameter of the circle is 14 cm14\text{ cm}, calculate, correct to three significant figures, the: (i) perimeter of the minor sector; (ii) area of the minor sector. [Take π=227]\left[\text{Take }\pi = \frac{22}{7}\right]

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

  1. RSTURSTU is a cyclic quadrilateral, so opposite angles add up to 180∘180^\circ: ∠URS=180∘−124∘\angle URS = 180^\circ - 124^\circ
    =56∘= 56^\circ.
  2. QRSQRS is a straight line, so ∠URQ=180∘−56∘\angle URQ = 180^\circ - 56^\circ
    =124∘= 124^\circ.
  3. In triangle QURQUR: ∠RQU=180∘−124∘−31∘\angle RQU = 180^\circ - 124^\circ - 31^\circ
    =25∘= 25^\circ.

(b)(i)

  1. The radius is 7 cm.
  2. The arc is 37\frac37 of the circumference: 37×2×227×7≈18.86\frac37 \times 2 \times \frac{22}{7} \times 7 \approx 18.86 cm.
  3. Perimeter of the sector =18.86+7+7≈32.9= 18.86 + 7 + 7 \approx 32.9 cm.

(ii)

  1. The sector is the same fraction 37\frac37 of the circle's area: 37×227×72=66.0 cm2\frac37 \times \frac{22}{7} \times 7^2 = 66.0\text{ cm}^2.

Report a problem with this question