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KVPY2013solution4 contains the explanation for the question 7, 8 and 9 of KVPY 2013 in detail.

7) Tangents to a circle at points P and Q on the circle intersect at a point R. If PQ=6 and PR=5 then the radius of the circle is
   A) \(\frac{13}{3}\)       B)4
   C) \(\frac{15}{4}\)       D)\(\frac{16}{5}\)

      Let 'O' be the center of the circle PQ is a chord, RP and RQ is the tangents to the circle.
                  RA \(\perp\) PQ
                  RO \(\perp\) PQ
       A is the mid point of PQ.
                  PA = 3
                  RM = \(\sqrt {PR²-PM²}\)
                      = \(\sqrt {25-9}\)
                      = 4
       \(\triangle RMP ~ \triangle PMO\) by AA similarity i.e., \(\angle RMP = \angle PMO = 90°\) and \(\angle MRP = \angle MPO\). 
           \(\frac{PR}{OP} = \frac{RM}{PM}\)
           \(\frac{5}{OP} = \frac{4}{3}\)
           OP = \(\frac{15}{4}\)
        Therefore the radius of the circle is \(\frac{15}{4}\). Hence the correct option is (C).

Solution 8

8) In an acute-angled triangle ABC, the altitudes from A, B, C when extended intersect the circumcircle again at points A1, B1, C1 respectively. If \(\angle ABC\)=45° , then \(\angle A1B1C1\) equals
         A) 45°    B)60°     C)90°     D)135°

    In \(\triangle DBE\), \(\angle ABC = 45°\), \(\angle BDC = 45°\), so by the angle sum property \(\angle DCB = 45°\). 
    In \(\triangle A_1EC\), \(\angle A_1EC = 90°\), \(\angle CA_1E = 45°) as segment angles are equal here AC is the chord and \(\angle ABC = \angle AA_1C = 45°\), then by angle sum property \(\angle A_1CE = 45°\).
    With chord \(A_1C_1 and \angleC_1B_1A_1 = \angleC_1CA_1 = 45°+45° = 90°\),.
    Therefore \(\angleA_1B_1C_1 = 90°\). So the correct option is (C). 

9) In a rectangle ABCD, points X and Y are the midpoints of AD and DC, respectively. Lines BX and CD when extended intersect at E, lines BY and AD when extended intersect at F. If the area of ABCD is 60 then the area of BEF is
     A) 60      B)80   C)90       D)120

Let  AB = p and BC = q .

Area of rectangle ABCD = pq = 60.

 We know that \(\triangle ABX \cong \triangle DEX\) and \(\triangle DFY \cong \triangle CBY\) by ASA Property.

Now from this we get DE= AB =p  and DF = BC = q .

\(Area\triangle DEF \) = \(\frac{1}{2}*p*q\) = \(\frac{1 }{2}*60\) = 30

\( Area of \triangle BEF \) = Area of Rectangle ABCD + Area of \(triangle DEF\)

                               = 60 + 30 = 90 .

Hence option (C) is the answer.


These are the explanations for the questions 7, 8, and 9 of KVPY 2013 question paper . To go to the previous solutions click on BACK and move to further solutions click on NEXT links below.

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