F3-F4 mathematics(20點)－dvhtlbz的部落格

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F3-F4 mathematics(20點)

發問:

1))In the figure, PM is the median of trianglePQR, S is the foot of the perpendicular from P to QR.Find (area of trianglePMR)/(area of trianglePQR).http://ihs.meric.hk/rforum.php/219267.jpg2))In the figure, triangleABC is equilateral, AD is an altitude, O is the circumcentre of the triangle. If AB=18cm, find... 顯示更多 1))In the figure, PM is the median of trianglePQR, S is the foot of the perpendicular from P to QR.Find (area of trianglePMR)/(area of trianglePQR). http://ihs.meric.hk/rforum.php/219267.jpg 2))In the figure, triangleABC is equilateral, AD is an altitude, O is the circumcentre of the triangle. If AB=18cm, find OA+OB+OC, leaving your answer in surd form. http://ihs.meric.hk/rforum.php/219271.jpg 3))In the figure, AB=AC, D and E are the points on AB and AC respectively such that AD=AE, CD and BE meet at F. Prove that BF=CF. http://ihs.meric.hk/rforum.php/219274.jpg 4))In the figure, X is the mid-point of AB and PQ, Y is the mid-point of AC and RQ. Prove that PR//BC. http://ihs.meric.hk/rforum.php/219275.jpg 5))In the figure, the area of triangleABC is 56cm^2, and is twice the area of the parallelogram CDEF. Find the shaded area. http://ihs.meric.hk/rforum.php/219277.jpg

最佳解答:

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Q.1 Area of triangle PMR = (PS x MR)/2. Area of triangle PQR = (PS x QR)/2. But QR=2MR because PM is the median. Therefore, Area of triangle PMR/Area of triangle PQR = [(PS x MR)/2]/[(PS x 2MR)/2] = 1/2. Q.2 AB=18, BD = 9. By Pythagoras thm, AD^2 + BD^2 = AB^2 AD^2 = 18^2 - 9^2 = sqrt243= 9sqrt3. Let OA = OB = OC = x. Therefore, OD = 9sqrt3 - x. By Pythagoras thm, OB^2 = OD^2 + BD^2 x^2 = (9sqrt3-x)^2 + 9^2 x^2 = 243 + x^2 - (18sqrt3)x + 81 (18sqrt3)x = 324 x = 324/18sqrt3 = 18/sqrt3 = 6sqrt3= OB Therefore, OA + OB + OC = 3x = 18sqrt3. Q.3 For triangle ABE and triangle ACD AB = AC (given) AE = AD (given) angle BAE = angle CAD (common) therefore, triangle ABE congruent triangle ACD (SAS) therefore, angle ABE = angle ACD.......(1) Angle ABC = angle ACB (base angle of isos. triangle ABC) That is angle ABE + angle FBC = angle ACD + angle FCB Using the result of (1), angle FBC = angle FCB , therefore, triangle FBC is an isos. triangle. Therefore, BF = FC ( side of isos. triangle FBC). Q.4 AX = XB and AY = YC , therefore, XY//BC (mid-point theorem)...................(1) PX = XQ and RY=YQ, therefore, XY//PR (mid-point thm)..................(2) From (1) and (2), BC//PR. 2008-07-25 10:38:49 補充： Q.5 Answer should be 14cm^2. I am trying to find a simple way to explain it.

其他解答:

The answer by eelyw is correct.7638E7481407D16B