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Circle

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Huỳnh Anh Phương
04/10/2018 at 12:20
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A circle has the diameter and  the radius. Use D for diameter and R for radius, which one is corect?

A) D = R + 5   B) D =  R : 2   C) D = R x 2

GeometryCircle

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    Dao Trong Luan Coordinator 04/10/2018 at 14:24

    The correct answer is C

    Huỳnh Anh Phương selected this answer.

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Carter
18/04/2017 at 14:51
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 A circle passes through two adjacent vertices of a square and is tangent to one side of the square. If the side length of the square is 2, what is the radius of the circle?

Circle

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    Phan Thanh Tinh Coordinator 18/04/2017 at 15:16

    r 2 - r r O H A B K C D

    Name the points as shown

    Draw \(OH\perp BC\),then H is the midpoint of BC.So BH = 1 cm

    Quadrilateral ABHK has 3 right angles at A,B and H

    => ABHK is a rectangle => HK = AB = 2 cm 

    Let r be the radius of the circle,so OH = KH - OK = 2 - r (cm)

    \(\Delta OHB\) right at H has : OH2 + HB2 = OB2 (Pythagoras theorem)

    \(\Rightarrow\left(2-r\right)^2+1^2=r^2\Rightarrow4-4r+r^2+1-r^2=0\)
    \(\Rightarrow5-4r=0\Rightarrow r=\dfrac{5}{4}\)

    Carter selected this answer.
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    FA KAKALOTS 03/02/2018 at 12:37

    Name the points as shown

    Draw OH⊥BC

    ,then H is the midpoint of BC.So BH = 1 cm

    Quadrilateral ABHK has 3 right angles at A,B and H

    => ABHK is a rectangle => HK = AB = 2 cm 

    Let r be the radius of the circle,so OH = KH - OK = 2 - r (cm)

    ΔOHB

     right at H has : OH2 + HB2 = OB2 (Pythagoras theorem)

    ⇒(2−r)2+12=r2⇒4−4r+r2+1−r2=0


    ⇒5−4r=0⇒r=54


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Min Hoang moderators
16/03/2017 at 10:24
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A circle was inscribed within the triangle ABC. The circle is tangent to AB at point M with AM = 10 and MB = 6. The area of triangle ABC is \(120\sqrt{3}\) .

a) Find the perimeter of triangle ABC.

b) Find the measure of angle A.

Circle

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    mathlove 16/03/2017 at 19:06

    We have  \(AB=AM+BM=10+6=16;BC=BN+CN=6+x;CA=CP+AP=x+10\).

    The triangle ABC has his edges   \(a=10+6=16;b=6+x;c=x+10\) . Call  \(p\)  is the halp perimeter of \(ABC\) then  \(p-a=10,p-b=6,p-c=x\) and the area of ABC is

                                          \(\sqrt{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}=\sqrt{60x\left(16+x\right)}\) .

    By the assumptions we have    \(60x\left(x+16\right)=\left(120\sqrt{3}\right)^2\Leftrightarrow x^2+16x-720=0\Leftrightarrow x\in\left\{20;-36\right\}\).

    So, \(x=20;p=36\), the perimeter of  ABC is  72.

    Suppose CH is the height of CAB, then \(CH=\dfrac{2.120\sqrt{3}}{16}=15.\)  The triangle CHA has

                                      \(CHA=90^0;CH=15;CA=30\Rightarrow A=60^0\). 

    The measure of angle A is    \(60^0\) .

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    FA KAKALOTS 03/02/2018 at 12:37

    We have  AB=AM+BM=10+6=16;BC=BN+CN=6+x;CA=CP+AP=x+10

    .

    The triangle ABC has his edges   a=10+6=16;b=6+x;c=x+10

     . Call  p  is the halp perimeter of ABC then  p−a=10,p−b=6,p−c=x

     and the area of ABC is

                                          √p(p−a)(p−b)(p−c)=√60x(16+x)

     .

    By the assumptions we have    60x(x+16)=(120√3)2⇔x2+16x−720=0⇔x∈{20;−36}

    .

    So, x=20;p=36

    , the perimeter of  ABC is  72.

    Suppose CH is the height of CAB, then CH=2.120√316=15.

      The triangle CHA has

                                      CHA=900;CH=15;CA=30⇒A=600

    . 

    The measure of angle A is    600

     .


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