← Math

Image_20261104204427_160_128.jpg

S3 · 中三 questions. Answer in your own words, then submit to get a score.

  1. 1. In quadrilateral ABCDABCD, AB=5AB = 5 cm, BC=15BC = 15 cm, CD=12CD = 12 cm, AC=18AC = 18 cm, and ∠ADC=118∘\angle ADC = 118^\circ. Explain how to find ∠DCA\angle DCA using the sine formula and angle sum of a triangle.

  2. 2. Quadrilateral ABCDABCD is split into △ABC\triangle ABC and △ACD\triangle ACD by diagonal ACAC. Given AB=5AB = 5 cm, BC=15BC = 15 cm, AC=18AC = 18 cm, CD=12CD = 12 cm, and ∠DCA=25.940∘\angle DCA = 25.940^\circ, explain why the total area of ABCDABCD is less than 80 cm280\ \text{cm}^2.

  3. 3. In sector OABCOABC with centre OO, OC=9OC = 9 cm and ∠AOC=90∘\angle AOC = 90^\circ. Point DD lies on OAOA such that OD=2ADOD = 2AD. Find ODOD and explain how ∠DCO\angle DCO is calculated.

  4. 4. Given that CDCD bisects ∠OCB\angle OCB in sector OABCOABC, and ∠DCO=33.690∘\angle DCO = 33.690^\circ, explain how ∠BOC\angle BOC is found using triangle angle sum in △OBC\triangle OBC.

  5. 5. The shaded region ABEDABED lies inside sector OABCOABC but outside △DOE\triangle DOE. Given OA=9OA = 9 cm, ∠AOB=∠DOE=44.760∘\angle AOB = \angle DOE = 44.760^\circ, OD=6OD = 6 cm, and OE≈5.0869OE \approx 5.0869 cm, write the expression for its area and explain each term.

Transcript
## Page 1 25. In the figure, ABCDABCD is a quadrilateral. AB=5AB = 5 cm, BC=15BC = 15 cm, CD=12CD = 12 cm, AC=18AC = 18 cm and ∠ADC=118∘\angle ADC = 118^\circ. (a) Find ∠DCA\angle DCA. 25.9∘25.9^\circ (Explain) (b) Is the area of ABCDABCD greater than 8080 cm2^2? Explain your answer. no 25. (a) In △ACD\triangle ACD, by the sine formula, 12 cmsin⁡∠DAC=18 cmsin⁡118∘\frac{12 \text{ cm}}{\sin \angle DAC} = \frac{18 \text{ cm}}{\sin 118^\circ} ∠DAC=36.060∘\angle DAC = 36.060^\circ or 180∘−36.060∘180^\circ - 36.060^\circ (rejected) =36.060∘= 36.060^\circ, cor. to 5 sig. fig. ∠DAC+∠DCA+∠ADC=180∘\angle DAC + \angle DCA + \angle ADC = 180^\circ 36.060∘+∠DCA+118∘=180∘36.060^\circ + \angle DCA + 118^\circ = 180^\circ ∠DCA=25.9∘\angle DCA = 25.9^\circ, cor. to 3 sig. fig. 25.940\boxed{25.940} (b) In △ABC\triangle ABC, s=12(5+15+18)s = \frac{1}{2}(5 + 15 + 18) cm =19= 19 cm By Heron’s formula, area of △ABC\triangle ABC =19(19−5)(19−15)(19−18)= \sqrt{19(19-5)(19-15)(19-18)} cm2^2 =32.619= 32.619 cm2^2, cor. to 5 sig. fig. In △ACD\triangle ACD, area of △ACD\triangle ACD =12×AC×CD×sin⁡∠ACD= \frac{1}{2} \times AC \times CD \times \sin \angle ACD =12×18×12×sin⁡25.940∘= \frac{1}{2} \times 18 \times 12 \times \sin 25.940^\circ cm2^2 =47.242= 47.242 cm2^2, cor. to 5 sig. fig. Area of ABCDABCD == area of △ABC+\triangle ABC + area of △ACD\triangle ACD =(32.619+47.242)= (32.619 + 47.242) cm2^2 =79.861= 79.861 cm2^2, cor. to 5 sig. fig. <80< 80 cm2^2 ∴\therefore The area of ABCDABCD is not greater than 8080 cm2^2. Ref: DSE 2019 Paper 2 Q33 27. In the figure, OO is the centre of the sector OABCOABC. DD is a point on OAOA such that OD=2ADOD = 2AD. CDCD and OBOB intersect at EE. OC=9OC = 9 cm and ∠AOC=90∘\angle AOC = 90^\circ. It is given that CDCD is the angle bisector of ∠OCB\angle OCB. (a) Find ∠OED\angle OED. 78.9∘78.9^\circ (b) Find the area of the shaded region ABEDABED. ⇒\Rightarrow Example 16 20.920.9 cm2^2 27. (a) OA=OC=9OA = OC = 9 cm OD+AD=OAOD + AD = OA 2AD+AD=92AD + AD = 9 cm AD=3AD = 3 cm ∴OD=OA−AD=(9−3)\therefore OD = OA - AD = (9 - 3) cm =6= 6 cm In △OCD\triangle OCD, tan⁡∠DCO=ODOC\tan \angle DCO = \frac{OD}{OC} =69= \frac{6}{9} ∠DCO=33.690∘\angle DCO = 33.690^\circ, cor. to 5 sig. fig. ∵CD\because CD is the angle bisector of ∠OCB\angle OCB. ∴∠BCD=∠DCO=33.690∘\therefore \angle BCD = \angle DCO = 33.690^\circ, cor. to 5 sig. fig. ∵OB=OC\because OB = OC ∴∠OBC=∠OCB\therefore \angle OBC = \angle OCB =2(33.690∘)= 2(33.690^\circ) =67.380∘= 67.380^\circ, cor. to 5 sig. fig. In △OBC\triangle OBC, ∠BOC+∠OCB+∠OBC=180∘\angle BOC + \angle OCB + \angle OBC = 180^\circ ∠BOC+2(67.380∘)=180∘\angle BOC + 2(67.380^\circ) = 180^\circ ∠BOC=45.240∘\angle BOC = 45.240^\circ, cor. to 5 sig. fig. In △OCE\triangle OCE, ∠OED=∠BOC+∠DCO\angle OED = \angle BOC + \angle DCO =45.240∘+33.690∘= 45.240^\circ + 33.690^\circ =78.9∘‾= \underline{78.9^\circ}, cor. to 3 sig. fig. 78.930\boxed{78.930} (b) ∠DOE=90∘−∠BOC\angle DOE = 90^\circ - \angle BOC =90∘−45.240∘= 90^\circ - 45.240^\circ =44.760∘= 44.760^\circ, cor. to 5 sig. fig. In △DOE\triangle DOE, tan⁡∠ODE=OCOD\tan \angle ODE = \frac{OC}{OD} =96= \frac{9}{6} ∠ODE=56.310∘\angle ODE = 56.310^\circ, cor. to 5 sig. fig. By the sine formula, OEsin⁡56.310∘=6 cmsin⁡78.930∘\frac{OE}{\sin 56.310^\circ} = \frac{6 \text{ cm}}{\sin 78.930^\circ} OE=6sin⁡56.310∘sin⁡78.930∘ cmOE = \frac{6 \sin 56.310^\circ}{\sin 78.930^\circ} \text{ cm} =5.0869= 5.0869 cm Area of the shaded region ABEDABED =π×OA2×∠AOB360∘−12×OD×OE×sin⁡∠AOB= \pi \times OA^2 \times \frac{\angle AOB}{360^\circ} - \frac{1}{2} \times OD \times OE \times \sin \angle AOB =(π×92×44.760∘360∘−12×6×5.0869×sin⁡44.760∘)cm2= \left( \pi \times 9^2 \times \frac{44.760^\circ}{360^\circ} - \frac{1}{2} \times 6 \times 5.0869 \times \sin 44.760^\circ \right) \text{cm}^2 =20.9‾ cm2= \underline{20.9} \text{ cm}^2, cor. to 3 sig. fig.