標題:
F.4 maths
發問:
APQ is a semicircle.QP is produced to B and BA is a tangent to the circle at A.PC:CQ=1:4 and AB = 8cm http://i240.photobucket.com/albums/ff164/walkinginthesang/e9afdcf8.jpg?t=1197301824 AC^2=PC x QC (proved) 1.show that AB^2=BPxBQ 2.If BP=y and PC=k,show that k=3/5y 3. find BP 4.Find radius
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最佳解答:
APQ is a semicircle.QP is produced to B and BA is a tangent to the circle at A.PC:CQ=1:4 and AB = 8cm AC^2=PC x QC (proved) angle PAQ is a right angle (interior angle of a semi-circle) ∴triangle APC is similar to triangle ACQ AC/QC = PC/AC AC^2 = PC*QC (1) show that AB^2=BP*BQ Let O be the center of the semi-circle and ∠AQC be θ° consider triangle ABP and triangle ABQ then ∠ BAQ = ∠ BAO + ∠OAQ = 90° + θ° (triangle OAQ is an isosceles triangle) ∠ ABQ =180° - 90° - θ° - θ° = 90° - 2θ° ∠BAP = ∠ BAQ -∠PAQ =90° + θ°-90° =θ° ∠ABP =180° - 90° + 2θ° - θ° = 90° +θ° ∴triangle ABP and triangle ABQ is similar to each other. AB/BP = BQ/AB AB^2=BP*BQ (2)if BP=y and PC=k,show that k=(3/5)y ∵AC^2=PC*QC and PC:CQ=1:4 if PC=k, then CQ=4k ,PQ = 5k and AC=2k ∵AB^2=BP*BQ if BP=y, then BQ=64/y consider the right angle triangles ABC and ACO, O is the center of the semi-circle and ∠AQC is θ° ∠COA = 2θ°(exterior angle of an isosceles triangle) ∠CAO = 90°-2θ° recall, ∠ ABQ = 90° - 2θ° ∴ the right angle triangles ABC and ACO is similar to each other BC/8 = AC/OA (OA is the radius of semi-circle) 2OA =PQ (∵PQ is the diameter of semi-circle) OA = 5k/2 OC=OP-PC =5k/2-k 3k/2 BC/AC = AC/OA (y+k)/2K = 2k/(3k/2) y/(2k)+1/2=4/3 y/(2k)=5/6 y/k = 5/3 k=(3/5)y (3) find BP. consider the right angle triangle ABC By Pythagorean theorem, AB^2 = AC^2 + BC^2 8*8 = (y+0.6y)^2+(1.2y)^2 64=(2.56+1.44)y^2 y=√(64/4) =4 or -4(rejected) The length of BP is 4 cm (4) find radius AO=2.5k =2.5*0.6y =1.5*4 =6 The radius of the semi-circle is 6 cm
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