Q-The cast of 3 pens and copild is ₹110 represent this sitnation withe a linear equation in two variables also find the cost of one copy if one per cost ₹20
Q- In triangle DEF,A, B and C are mid points of sides EF,DF and DE respectively.If ar (triangleBAF)=24 cm 2means square ,find ar (AFBC)
Q- prove that if two chords of a circle bisect each other then the two chords are diameter of the given circle
Q-Draw a line segmen SR of length 10 cm divide it into 4 equal parts using compass and ruler
Q-construct triangl ABC surch that BC=3.2cm angle B=45° and AC-AB=2.1cm.
Q-AB and CD are two parallel chords of a circle such that lengths of AB and CD are 10cm and 24cm resputively.If the chord are on opposite sides of the center and the distance between then is17cm find the radius of the circle.




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Price/Cost of a pen = Rs X
Price/Cost of a copy = Rs y

   Total cost =  3 X + Y = Rs 110
  a pen costs Rs 20.  =>  X = Rs 20
   So Y = 110 - 3 * 20 = Rs 50.
 cost of a copy = Rs 50.
see diagram.

       When we join the mid points of the sides of the triangle DEF, we have the triangle DEF divided into 4 equal triangles.  Thus the triangles ABC, BAF, AEC and BCD are all identical congruent triangles.  So area of the triangle BAF = 24 cm² = 1/4 th of area of triangle DEF.
     Area of parallelogram AFBC = 2 * area of triangle BAF = 48 cm²

see diagram.
AC and  BD are chords bisecting each other at O.
  Hence,  AO = OC  and  BO = OD.  Also, the angles AOB and COD are same (vertical angles at O).  Hence, the two triangles AOB and COD are congruent.
   Hence,  AB = CD.

   Similarly, in triangles AOD and BOC, the angles AOD = angle BOC (vertical angles).  As, AO = OC and BO = OD, the two triangles are congruent.
     Hence,  AD = BC.

   The two triangles ACB and ADB are congruent as all the three corresponding sides are equal to one another.
   So angle D = angle B.
   Similarly, angle C = angle A.

   In a cyclic quadrilateral ABCD, the opposite angles are supplementary.  Hence,  angle B = angle D = 90°,
           same way, angle C = angle A = 90°.

In a circle, if a chord AC (or BD) subtends an angle 90° at a point B (or A) on the circumference, then it is the diameter of the circle.

  hence, AC and BD are diameters.
    Draw a line segment SR of length 10 cm horizontally using the ruler.  Then using the compass, measure about 6 to 7 cm.  With S as the center draw an arc above SR and an arc below SR.  With R as the center and with the same radius, draw an arc above SR and an arc below SR , intersecting the earlier arcs drawn.
    The points are intersection of arcs can be joined to get the perpendicular bisector of SR.  Let it intersect SR at P.  So P is the mid point of SR.

    Follow the above procedure for the line segments SP and PR.  Measure 3 to 4 cm with the compass. Draw arcs with centers S and P above and below SP.  Join their points of intersection.  Similarly, draw arcs with centers as P and R above and below PR.  Join their intersection points.

   Join/Draw the perpendicular bisectors for SP and PR.  Now the line segment SR is divided into 4 equal parts.
construct ABC: 
   draw a line segment horizontally BC of length 3.2 cm.
   Now draw a line BD (long one) making the angle CBD = 45°.
   We need to find the point A such that AC = AB + 2.1 cm.

see diagram.

AB || CD.  AB and CD are chords of the circle.    Let R be the radius of the circle.
AB = 10 cm      CD = 24 cm.  Center of circle is O.  Draw perpendicular bisectors for AB and CD.  They will join at center O.  Given, EF = EO + OF = 17 cm.
    we have,  AE = EB = 5 cm.    CF = FD = 12 cm.
   Using Pythagoras theorem in triangle AEO,  EO = √(R² - 5²)
     Using Pythagoras theorem in triangle BOF,  OF = √(R² - 12²)

   Hence,  EO + OF = √(R² - 25) + √(R² - 144) = 17
       √(R² - 144) = 17 - √(R² - 25)
        R² - 144 =  17² + (R² - 25) - 2 * 17 * √(R² - 25)
        - 408 = - 2 * 17 * √(R² - 25)
         12 = √(R² - 25)
         R² = 144+25 = 169
         R = 13 cm.  = radius of the circle.
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