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
F3-F4 mathematics(20點)
2008-07-24 4:01 am
回答 (2)
2008-07-25 6:08 pm
✔ 最佳答案
Q.1Area 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.
2008-07-31 9:38 pm
The answer by eelyw is correct.
收錄日期: 2021-04-25 22:37:27
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