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# Weekly Dose 22 - Maths Olympiad Practice - Area

Weekly Dose 22 - Maths Olympiad Practice - Area

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### Weekly Dose 22 - Maths Olympiad Practice - Area

1. 1. Given the area for triangle ABC is 1, BE = 2AB, BC = CD, find the area for triangle BDE. Solution: Draw a line connecting CE, making AB and BE as the base for △ABC and △BEC S△BEC = 2 S△ABC = 2 And △BEC and △CED are of same base and same height S△CED = S△BEC = 2 S△BDE = S△BEC + S△CED = 4
2. 2. Find the area of the shaded region (diameter for big circle is 2cm). Solution: Flip the bottom half of the diagram to look like this: The area of the shaded region = area of big circle – area of small circle Radius of big circle = 2 ÷ 2 = 1 Radius for small circle = 1 ÷ 2 = 0.5 Area of shaded region = 3.14 × (12 − 0.52) = 2.355 cm2
3. 3. Given the area for triangle ABE, ADF and rectangle AECF are the same, find the area for triangle AEF. Solution: S □ ABCD = 9× 6 = 54 Since S □ AECF = S△ABE = S△AFD, ∴ S □ AECF + S△ABE + S△AFD = area for rectangle ABCD S □ AECF = S△ABE = S△AFD = 54 × 1 3 = 18 S△ABE = 1 2 × 6 × (9 – EC) = 18 EC = 9 – 18 × 2 ÷ 6 = 3 S△AFD = 1 2 × 9 × (6 – FC) = 18 FC = 6 – 18 × 2 ÷ 9 = 2 S△ECF = 1 2 × 3 × 2 = 3 S△AEF = S □ AECF – S△ECF =18 – 3 = 15
4. 4. The diagram shows an isosceles right triangle ABC with side 10cm, The area for the shaded region 甲and 乙 are the same. Find the area of circle which AEF is part of. Solution: AC = BC, ∠A = ∠B = 45º S△ABC = 1 2 × 10 × 10 = 50 Since S甲 = S乙 ，area for AEF = S△ABC Area of AEF Area of the circle which it is part of = 45º 360º Area of the circle = Area of AEF 45 × 360 = S△ABC 45 × 360 = S△ABC × 8 = 400 cm2
5. 5. Cut a rectangle with the width of ½ meter out of a square board. The area of the remaining board is 65/18 m2. Find the area of the rectangle. Solution: Redraw the diagram as below (we called “弦图”in Chinese): As you can see, the width of the big square is x + y, the width of the small square is x – y = 1 2 ---- ① Also, the area of the big square = (x + y)2 = 65 18 × 4 + 1 2 × 1 2 = 529 36 x + y = 23 6 ---- ② From ① + ② , x = ( 23 6 + 1 2 ) ÷ 2 = 13 6 The area of the remaining board = 13 6 × 1 2 = 1 1 12
6. 6. A rectangle board, with the length cut 4m, and width cut 1m, it becomes a square, as shown in diagram below. The new area now is 49m2 less than original area. Find the original area of the rectangle. Solution: Let the width of the square as x. (x + 4) × 1 + 4x = 49 x + 4 + 4x = 49 x = 9 The area of the rectangle = 9 × 9 + 49 = 130 m2
7. 7. Given any rectangle ABCD, where AE = 2 3 AB, BF = 2 3 BC, CG = 2 3 CD, DH = 2 3 DA, joining E, F, C, H. Find the ratio of area for EFGH to area of ABCD. Solution: Connect ED and BD. We know S△AEH = 1 3 S△AED , S△AED = 2 3 S△ADB ∴ S△AEH= 1 3 × 2 3 S△ADB = 2 9 S△ADB Similarly, S△CGF = 2 9 S△BCD S△AEH + S△CGF = 2 9 S△ADB + 2 9 S△BCD = 2 9 (S△ADB + S△BCD) = 2 9 S □ ABCD Similarly, S△BFE + S△DHG = 2 9 S □ ABCD S△AEH + S△CGF + S△BFE + S△DHG = 4 9 S □ ABCD Therefore, S □ EFGH = (1 - 4 9 ) S □ ABCD = 5 9 S □ ABCD S □ EFGH : S □ ABCD = 5 : 9
8. 8. As shown in diagram, the three heights of the triangle ABC meet at point P, prove that 𝑃𝐷 𝐴𝐷 + 𝑃𝐸 𝐴𝐸 + 𝑃𝐹 𝐴𝐹 = 1 Solution: Since △PBC and △ABC share the same base, , the ratio of their area is the same as the ratio of their heights △PBC △ABC = 𝑃𝐷 𝐴𝐷 Similarly, △PCA △ABC = 𝑃𝐸 𝐵𝐸 , △PAB △ABC = 𝑃𝐹 𝐶𝐹 △PBC △ABC + △PCA △ABC + △PAB △ABC = 𝑃𝐷 𝐴𝐷 + 𝑃𝐸 𝐵𝐸 + 𝑃𝐹 𝐶𝐹 △PBC + △PCA + △PAB △ABC = △ABC △ABC = 1 𝑃𝐷 𝐴𝐷 + 𝑃𝐸 𝐵𝐸 + 𝑃𝐹 𝐶𝐹 = 1