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Math Help - Help me out these easy geometry problems.?

  1. #1
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    Question Help me out these easy geometry problems.?

    I want the explanation too, cause i don't understand the question properly and then put it clearly, cause i am just 8th class student. best of answers will be given 5 *'s, plus a thumbs up.

    1.In ▲ PQR, PD ┴ QR and PO is the bisector of ∠ QPR. If ∠ PQR = 65 and ∠ PQR = 23.5,then
    ∠ DPO in degrees=
    2.In a parallelogram PQRS,the bisectors of ∠ P and ∠ Q meet on RS.If the perimeter PQRS is 13.5 cm, then the measure of QR =
    3.A circle is passing through three vertices of a rhombus of side 8cm and its centre is the forth vertex of the rhombus. Find the length of the longest diagonal of the rhombus.
    4.In a rhombus ABCD, the diagonals intersect each other at O.If ∠ A = 60 and OA = 2 cm, then the side of the rhombus is =
    5.In a rhombus PQRS,the diagonals intersect at O.Given that, ∠ P = 120 and OP = 3 cm. What is the side of the rhombus is =
    6.P is an interior point of quadrilateral ABCD and B = 3.5 cm,BC = 4 cm,CD = 4.8 cm and AD = 3.7 cm.Then the possible value of ( AP+BP+CP+DP) =
    7. MN and PS are 2 equal chords of a circle drawn on the either side of centre O of the circle. Both the chords are produced to meet at point A.If the radius of the circle is 10 cm, MN = 12 cm and OA= 17 cm, then NA =
    8. In ▲ ABC , if ∠ A < ∠ B < 45, then ▲ ABC is a/an .................................... triangle.
    9. If G is the centroid of ▲ ABC, then the area of ▲ BGC is ............................... thimes the area of the quadrilateral ABCG.

    If you can pr
    ovide a diagram also, it is good.... but if you can't it's not even a silly problem.. Please answer this question.I am trying for many Days ( nearly about 4 days.) Help me out.. ♣♣
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  2. #2
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    Lexington, MA (USA)
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    Re: Help me out these easy geometry problems.?

    Hello, kohila!

    Here is some help . . .


    3. A circle is passing through three vertices of a rhombus of side 8cm
    and its centre is the forth vertex of the rhombus.
    Find the length of the longest diagonal of the rhombus.

    We have rhombus ABCO.
    The center of the circle is O.
    Since OA = OB = OC, the rhombus
    . . is made of two equilateral triangles.
    Hence: \angle AOC = 120^o.
    Code:
                      B
                      *
               8   *  :  *   8
                *     :     *
             *        :        *
        A *  *  *  *  *  *  *  *  * C
             *        :        *
                *     :     *
               8   *  :  *   8
                      *
                      O
    Law of Cosines: . AC^2 \:=\:8^2 + 8^2 - 2(8)(8)\cos120^o

    . . . . . . . . . . . . AC^2\:=\: 64 + 64 - 128(\text{-}\tfrac{1}{2}) \:=\:192

    Therefore: . AC \:=\:\sqrt{192} \:=\:8\sqrt{3}




    4. In a rhombus ABCD, the diagonals intersect each other at O.
    If ∠ A = 60 and OA = 2 cm, find the length of a side of the rhombus.

    Once again, the rhombus is comprised of two equilateral triangles.
    Code:
                      B
                      *
                   *  :  *
                *     :     *
             *    2   :        *
        A *  -  -  -  +  -  -  -  * C
             *        :O       *
                *     :     *
               8   *  :  *   8
                      *
                      D
    The altitude of the equilateral triangle is 2.
    Code:
                *
               *|*
              * | * x
             *  |2 *
            *   |   *
           *    |    *
          * * * * * * *
                  x/2
    Pythagorus: . x^2 \;=\;\left(\tfrac{x}{2}\right)^2 + 2^2 \quad\Rightarrow\quad x^2 \:=\:\tfrac{x^2}{4} + 4

    . . . . . . . . . \tfrac{3}{4}x^2 \:=\:4 \quad\Rightarrow\quad x^2 \:=\:\tfrac{16}{3}

    . . . . . . . . . . x \:=\:\sqrt{\frac{16}{3}} \:=\:\frac{4}{\sqrt{3}} \:=\:\frac{4\sqrt{3}}{3}
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  3. #3
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    chennai
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    Re: Help me out these easy geometry problems.?

    thank you.. you can answer all but y u delayed.
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