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好急好急!!吾該好心幫幫忙!!中5 maths. in actionP243NO30

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http://www.hehaclub.com/account/heha00112504/heha00112504_m12.jpg呢個網系figurethe figure shows a circle C which touches the axes at(10,0)and(0,10)(a)find the equation of C(b)show that P(16,18) is a point lying on circle C(c)if A is the centre of circle C and L is a straight line passing though P and perpendicular... 顯示更多 http://www.hehaclub.com/account/heha00112504/heha00112504_m12.jpg 呢個網系figure the figure shows a circle C which touches the axes at(10,0)and(0,10) (a)find the equation of C (b)show that P(16,18) is a point lying on circle C (c)if A is the centre of circle C and L is a straight line passing though P and perpendicular to AP,find the equation of L (d)hence,find the area of the shaded region in terms of "pie"(計circle戈個[pie]吾識打)

最佳解答:

(a) centre of C=(10,10) radius of C=10 therefore equation of C is (x-10)^2+(y-10)^2=10^2 i.e. x^2+y^2-20x-20y+100=0 (b) Sub. x=16,y=18 into C, L.H.S.=16^2+18^2-20*16-20*18+100 =256+324-320-360+100 =0=R.H.S. therefore P is a point lying on C (c) A=(10,10) slope of AP=(18-10)/(16-10)=4/3 slope of L=-3/4 therefore equation of L is (y-18)/(x-16)=-3/4 i.e. 3x+4y-120=0 (d) When x=0,L becomes 4y-120=0 y=30 therefore y-intercept of L=30 When y=0,L becomes 3x-120=0 x=40 therefore x-intercept of L=40 area of shaded region=area of triangle formed between L and two axes - area of C =30*40/2-10^2*[pie] =600-100[pie]

其他解答:7638E7481407D16B
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