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

  1. 1. Triangle ABC has side lengths AB = 8 cm, BC = 7 cm, and AC = 5 cm. Use Heron’s formula to find its area, correct to 3 significant figures.

  2. 2. Triangle PQR has sides PQ = 20 cm, QR = 7 cm, and PR = 15 cm. Show that its area is exactly 42 cm² using Heron’s formula.

  3. 3. In triangle ABC with AB = 9 cm, BC = 15 cm, and AC = 8 cm, the area is found to be approximately 29.9 cm². Explain how to use this area to find the height from vertex A to side BC.

  4. 4. Triangle XYZ has sides XY = 7 cm, YZ = 13 cm, and XZ = 10 cm. Calculate its area using Heron’s formula and give your answer correct to 3 significant figures.

  5. 5. Triangle DEF is isosceles with DE = EF = 8 cm and DF = 12 cm. Explain why the semi-perimeter is 14 cm and use it to find the area (correct to 3 sig. fig.).

Transcript
## Page 1 S5 Math CP Chapter 9 Applications of Trigonometry in 2-dimensional Problems Name: __________ Class ( ) 2 ③ 11. In each of the following, find the area of the triangle using Heron’s formula. (a) (b) (c) 11. (a) s=12(8+7+5) cm=10 cms = \frac{1}{2}(8 + 7 + 5)\,\text{cm} = 10\,\text{cm} By Heron’s formula, area of △ABC\triangle ABC =10(10−8)(10−7)(10−5) cm2= \sqrt{10(10 - 8)(10 - 7)(10 - 5)}\,\text{cm}^2 =17.3 cm2= 17.3\,\text{cm}^2, cor. to 3 sig. fig. (b) s=12(7+13+10) cm=15 cms = \frac{1}{2}(7 + 13 + 10)\,\text{cm} = 15\,\text{cm} By Heron’s formula, area of △XYZ\triangle XYZ =15(15−7)(15−13)(15−10) cm2= \sqrt{15(15 - 7)(15 - 13)(15 - 10)}\,\text{cm}^2 =34.6 cm2= 34.6\,\text{cm}^2, cor. to 3 sig. fig. (c) EF=DE=8 cmEF = DE = 8\,\text{cm} s=12(8+12+8) cm=14 cms = \frac{1}{2}(8 + 12 + 8)\,\text{cm} = 14\,\text{cm} By Heron’s formula, area of △DEF\triangle DEF =14(14−8)(14−12)(14−8) cm2= \sqrt{14(14 - 8)(14 - 12)(14 - 8)}\,\text{cm}^2 =31.7 cm2= 31.7\,\text{cm}^2, cor. to 3 sig. fig. ③ 12. In the figure, PQ=20 cmPQ = 20\,\text{cm}, QR=7 cmQR = 7\,\text{cm} and PR=15 cmPR = 15\,\text{cm}. (a) Find the area of △PQR\triangle PQR. 42 cm242\,\text{cm}^2 (b) Find the height hh of △PQR\triangle PQR. ⇨ Example 18 5.6 cm5.6\,\text{cm} 12. (a) s=12(7+15+20) cm=21 cms = \frac{1}{2}(7 + 15 + 20)\,\text{cm} = 21\,\text{cm} By Heron’s formula, area of △PQR\triangle PQR =21(21−7)(21−15)(21−20) cm2= \sqrt{21(21 - 7)(21 - 15)(21 - 20)}\,\text{cm}^2 =42 cm2= 42\,\text{cm}^2 (b) Area of △PQR=12×PR×h cm2\triangle PQR = \frac{1}{2} \times PR \times h\,\text{cm}^2 h=2×area of △PQRPRh = \frac{2 \times \text{area of } \triangle PQR}{PR} =2×4215 cm= \frac{2 \times 42}{15}\,\text{cm} =5.6 cm= 5.6\,\text{cm} ③ 13. In the figure, AB=9 cmAB = 9\,\text{cm}, BC=15 cmBC = 15\,\text{cm} and AC=8 cmAC = 8\,\text{cm}. (a) Find the area of △ABC\triangle ABC. 29.9 cm229.9\,\text{cm}^2 (b) Find the height from AA to BCBC. 3.99 cm3.99\,\text{cm} 13. (a) s=12(15+8+9) cm=16 cms = \frac{1}{2}(15 + 8 + 9)\,\text{cm} = 16\,\text{cm} By Heron’s formula, area of △ABC\triangle ABC =16(16−15)(16−8)(16−9) cm2= \sqrt{16(16 - 15)(16 - 8)(16 - 9)}\,\text{cm}^2 =29.9 cm2= 29.9\,\text{cm}^2, cor. to 3 sig. fig. 29.93329.933 (b) Let h cmh\,\text{cm} be the height from AA to BCBC. Area of △ABC=12×15×h cm2\triangle ABC = \frac{1}{2} \times 15 \times h\,\text{cm}^2 h=2×area of △ABCBCh = \frac{2 \times \text{area of } \triangle ABC}{BC} =2×29.93315= \frac{2 \times 29.933}{15} =3.99= 3.99, cor. to 3 sig. fig. ∴ The height from AA to BCBC is 3.99 cm3.99\,\text{cm}. P.2