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

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

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




    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: .$\displaystyle x^2 \;=\;\left(\tfrac{x}{2}\right)^2 + 2^2 \quad\Rightarrow\quad x^2 \:=\:\tfrac{x^2}{4} + 4$

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

    . . . . . . . . . . $\displaystyle x \:=\:\sqrt{\frac{16}{3}} \:=\:\frac{4}{\sqrt{3}} \:=\:\frac{4\sqrt{3}}{3}$
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  3. #3
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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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