consider the following figure:
consider the following figure:two circles with centres q and r touch each other externally at x. another circle, with centre p, touches the other two circles internally at s and t. if pq = 8 cm, qr = 11 cm and pr = 9 cm, find the radius of circle r.
consider the following figure:two circles with centres q and r touch each other externally at x. another circle, with centre p, touches the other two circles internally at s and t. if pq = 8 cm, qr = 11 cm and pr = 9 cm, find the radius of circle r.
let c1 and c2 are two concentric circles with center o and radius of outer circle is 5 cm.given : ac is a chord with length 8 cm that is tangent to inner circle .to find : radius of inner circle i.e. odac is a tangent for c1 at d so,od ⏊ ac [tangent at a point on the circle is perpendicular to the radius through point of contact ]so, oad is a right-angled triangle at d .also it implies that od is perpendicular to the chord ac in c2so we have,ad = dc [perpendicular from the center to the chord bisects the chord]ac = ad + dc8 = ad + adad = 4 cmin △oad by pythagoras theorem(oa)2 = (od)2 + (ad)2(5)2 = (od)2+ (4)225 = (od)2 + 16(od)2= 9od = 3 cm you can find the radius in the following way. 
write the equation of the line that is perpendicular to the given tangent line. this perpendicular goes through the stated center, and intersects the tangent; the perpendicular's slope is the negative reciprocal of the stated tangent line.
find the intersection (point t) of the tangent and radius line--then use the distance formula to find the radius, which is a segment from the stated center to the point t. given: volume of cylinder = 1617 cm3let r be the radius of base and h be the height of coner:h = 2:3∴ r/h = 2/33r = 2hh = 3r/2 …(i)formula: volume of cylinder = πr2h∴ 1617 = (22/7) r2hr2h = 514.5using (i) we have∴ r2 × (3r/2) = 514.53r3 = 1029r3 = 343r = 7 cmh = 21/2 cmtotal surface area of cylinder = 2πr2 + 2πrh= 2 × (22/7) × 72 + 2 × (22/7) × 7 × (21/2)= 308 + 462= 770 cm2therefore total surface area of cylinder = 770 cm2 hint: assume the co-ordinates of $r$ as $(x,y)$ and as it is given that $p,q,r$ are collinear points. apply distance condition of collinear points which is if $p,q,r$ are three collinear points then they must satisfy $pq+qr=pr$ and get one equation. apply distance formula which states that if $a({{x}\_{1}},{{y}\_{1}})$ and $b({{x}\_{2}},{{y}\_{2}})$ are two points then the distance $ab=\sqrt{{{({{x}\_{2}}-{{x}\_{1}})}^{2}}+{{({{y}\_{2}}-{{y}\_{1}})}^{2}}}$ and get the second equation by using $pr=10$ , solve these two equations and get the co-ordinates of $r$. complete step-by-step solution:it is given in the question that $p, q, r$ are collinear points.collinear means lying on the same line which means that $p, q, r$ lies on the same straight line.let us suppose that the co-ordinates of $r$ is $(x,y)$. now, we have $p(7,7)$, $q(3,4)$, $r(x,y)$ and also $pr=10$.we have to use distance formula in order to find $pq,qr,pr$. so, according to distance formula if $a({{x}\_{1}},{{y}\_{1}})$ and $b({{x}\_{2}},{{y}\_{2}})$ then $ab=\sqrt{{{({{x}\_{2}}-{{x}\_{1}})}^{2}}+{{({{y}\_{2}}-{{y}\_{1}})}^{2}}}$.now, $p,q,r$ are three collinear points then they must satisfy,$\therefore pq+qr=pr$using distance formula, we get$\rightarrow \sqrt{{{(7-3)}^{2}}+{{(7-4)}^{2}}}+\sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=10$$\rightarrow \sqrt{16+9}+\sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=10$$\rightarrow 5+\sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=10$\[\rightarrow \sqrt{{{(3-x)}^{2}}+{{(4-y)}^{2}}}=5\]squaring both sides, we get$\rightarrow {{(3-x)}^{2}}+{{(4-y)}^{2}}=25$$\rightarrow {{x}^{2}}+{{y}^{2}}-6x-8y+25=25$cancelling 25 from both sides, we get$\rightarrow {{x}^{2}}+{{y}^{2}}=6x+8y...................(1)$also, it is given that $pr=10$, using this we get$\therefore pr=10$\[\rightarrow \sqrt{{{(7-x)}^{2}}+{{(7-y)}^{2}}}=10\]squaring both sides, we get\[\rightarrow {{(7-x)}^{2}}+{{(7-y)}^{2}}=100\]$\rightarrow {{x}^{2}}+{{y}^{2}}-14x-14y+98=100$putting the value of ${{x}^{2}}+{{y}^{2}}$ from equation (1), we get$\rightarrow 6x+8y-14x-14y=100-98$$\rightarrow -8x-6y=2$now, expressing $y$ in terms of $x$, we get$\rightarrow y=\dfrac{-(2+8x)}{6}=\dfrac{-(1+4x)}{3}..................(2)$putting the value of $y$ from equation (2) in equation (1), we get$\rightarrow {{x}^{2}}+{{\left( \dfrac{-(1+4x)}{3} \right)}^{2}}=6x+\dfrac{-(1+4x)}{3}\times 8$$\rightarrow \dfrac{9{{x}^{2}}+1+8x+16{{x}^{2}}}{9}=\dfrac{18x-8-32x}{3}$$\rightarrow 25{{x}^{2}}+8x+1=3(-14x-8)$$\rightarrow 25{{x}^{2}}+8x+1+42x+24=0$$\rightarrow 25{{x}^{2}}+50x+25=0$$\rightarrow 25({{x}^{2}}+2x+1)=0$$\rightarrow {{(x+1)}^{2}}=0$$\rightarrow x+1=0\rightarrow x=-1$now, putting the value of $x$ in equation (2), we get$\rightarrow y=\dfrac{-(1+4\times -1)}{3}$$\rightarrow y=\dfrac{3}{3}=1$so, $x=-1$ and $y=1$.hence the co-ordinates of $r$ is $(-1,1)$. note: this question is simple and there are many other alternative ways to solve it but the tricky part is to choose which method to solve the question. there are three different conditions of collinearity as there is slope condition, distance condition, and area condition. as in the other two conditions distances are not involved, only the distance condition involves the use of distances. in the question, distance is mentioned so we choose that method to solve it. hence, students should carefully read the question and choose which method is suitable to solve the question according to the values given. 圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872668990.jpg
 
首先請先 複習下列公式
 
設扇形的中心角θ弳，半徑為r，弧長為s，面積為a，則
s=rθ；a= r2θ= rs
sol: 設: ∠aob=θ 弳 (弧度) od=x
(x+2.5) θ =6 ……………………………..(1)
求黃色部分面積= (x+5)2 θ - x2θ
= θ ( x2+10x+25-x2 )
= θ (10x+25)
=5θ( x + 2.5)=30



圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872669011.jpg
 

圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872669012.jpg
 

圖片參考：http://imgcld.yimg.com/8/n/ab00135862/o/161006040753513872669033.jpg
 



 2010-06-04 23:25:06 補充：
y= |cosx|

恕不再多畫 將以文字說明

1. 當cosx >0 時 y=cosx
2. 當cosx <0 時 y = -cosx

 2010-06-04 23:26:21 補充：
注意

y=3sin2(x-π)+2

之 a , b 步驟為同一圖形

 2010-06-05 13:06:49 補充：
奇怪 1/2 通通不見了

更正

s=rθ；a=0.5 r2θ=0.5 rs

求黃色部分面積=0.5(x+5)2 θ - 0.5x2 θ
=0.5θ ( x2+10x+25-x2 )
=θ (10x+25)
=5θ( x + 2.5)=30

p.s. x2 即為 x平方

 2010-08-28 11:35:07 補充：
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rid of the given equations arex2 + y2 = 16 …(i)and y2 = 6x …(ii)on solving the equation (i) and (ii),or x2 + 6x = 16or x2 + 6x – 16 = 0or (x + 8)(x – 2) = 0or x = 2 or – 8 is not possible solution∴ when x = 2, y = ± = ± = ±equation (i) is a circle with centre (0, 0) and meets axes at (±4,0), (0,±4)equation (ii) represents a parabola with axis as x - axis.points of intersection are a () and c (2,)these are shown in the graph below:area bounded by the circle and parabola= 2[area (oado) + area (adba)]area of circle = π(r) 2= π(4)2= 16π sq. unitsthus, required areathe area of the circle x2 + y2 = 16 which is exterior the parabola y2 = 6x is you don't say that the area must be a rectangle. the true largest area would be a circle, with the circumference equal to 400m and the area would be 12,732.4 m^2. however, evidently, the area is restricted to a rectangle, in which case the maximum area is a square, and the side of the square will be 
400/4 = 100 m. the area is 100x100 = 10,000 m^2.
algebraically,
let x = one pair of opposite sides
(400-2x)/2 = 200-x = the other pair of opposite sides

area = a(x) = x(200 - x) = -x^2 + 200x
x-coordinate of vertex = -b/2a = 200/2 = 100
a(100) = 10,000

that's it! ;) the sides of the triangle are 6 cm, 8 cm and 10 cm. the longest side is 10 cm. (10)2 = 100 ….(i) now, the sum of the squares of the other two sides will be, (6)2 + (8)2 = 36 + 64 = 100 ….(ii) (10)2 = (6)2 + (8)2 ….. from (i) and (ii) by the converse of pythagoras theorem: the given sides form a right angled triangle. given (pq – prq) = (pq×1 – pq×r)taking pq common,(pq – prq) = pq(1 – r) sin(α β) = sinαcosβ cosαsinβ 
sin(α-β) = sinαcosβ - cosαsinβ


cos(α β) = cosαcosβ - sinαsinβ
cos(α-β) = cosαcosβ sinαsinβ


tan(α β) = tanα tanβ / 1-tanα tanβ
tan(α-β) = tanα-tanβ / 1 tanα tanβ


sin2θ = 2sinθcosθ = 2tanθ /1 tanθtanθ


cos2θ = cos平方θ−sin平方θ = 2cos平方θ−1 = 1−2sin平方θ


△abc=海龍公式 √s(s-a)(s-b)(s-c) 
[s=a b c/2 ，就是周長/2 ]



△abc=1/2 ａ.b*sinc = axbxc / 4r = rxs 

 [r=外接圓半徑，r=內接圓半徑]




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