Q1. ABC is an isosceles triangle in
which AB = AC, circumscribed about a circle. Prove that the base is bisected by the point of
contact. i.e. if AB = AC, prove that BE = EC. CBSE 2008
Solution:-
Since the tangents drawn from
any exterior point to a circle are equal in length, therefore:-
BD = BE - ( Tangents
from B)
CE = CF - ( Tangents
from C)
Now,
AB = AC
AB- AD = AC- AD -
(Subtracting AD from both sides)
AB- AD = AC – AF -
(Since, AD=AF)
BD = CF
BE = CF
BE = CE , Hence proved.
Q2. A circle touches all the four sides
of a quadrilateral ABCD. Prove that AB + CD = BC + DA.
CBSE 2008,2009
Solution:-
Since the tangents drawn from any
exterior point to a circle are equal in length, therefore:-
AP = AS -
(Tangents from A) - (i)
BP = BQ - (
Tangents from B) - (ii)
CR = CQ - ( Tangents
from C) - (iii)
DR = CQ - ( Tangents
from D) - (iv)
Adding (i) , (ii) , (iii) , (iv) ,
we get :-
AP + BP + CR + DR = AS + BQ + CQ +
DS
(AP + BP) + (CR + DR) = (AS + DS) +
(BQ + CQ)
AB + CD = AD + BC
AB + CD = BC + DA , Hence proved.
Q3. Prove that a parallelogram
circumscribing a circle ia a rhombus. CBSE 2002, 2008
Solution:-
Let ABCD be a parallelogram such
that its sides touch a circle with centre O.
Since the tangents to a circle
from an exterior point are equal in length , therefore:-
AP = AS - (Tangents from A)
….......(i)
BP = BQ - (Tangents from B)
…........(ii).
CR = CQ - (Tangents from C)
….........(iii)
DR = DS - (Tangents from D)
…...........(iv).
Adding (i) , (ii) , (iii) , (iv) ,
we get:-
AP + BP + CR + DR = AS + BQ + CQ +
DS
(AP + BP) + (CR + DR) = (AS + DS) +
(BQ + CQ)
AB + CD = AD + BC
2AB = 2BC
….....................(Since, ABCD is a parallelogram, therefore AB
= CD and BC = AD)
AB = BC
AB = BC = CD = AD
Therefore, ABCD is a rhombus, Hence
proved.
Q4. A circle is inscribed in a triangle
ABC having sides 8cm , 10cm and 12cm. Find AD , BE and CF. CBSE 2001
Solution:-
Since, the tangents drawn from an
external point to a circle are equal , therefore:-
AD = AF = say , x
Now,
AB = 12cm , BC = 8cm and CA
= 10cm
Therefore,
x + y = 12 , y + z = 8
and z + x = 10
(x+y)+(y+z)+(z+x) = 12
+ 8 + 10
2(x+y+z) = 30
x+y+z = 15
Now,
x+y = 12 and x+y+z = 15
12 + z = 15
z = 15 – 12 = 3
Similarly,
y+z = 8 and x+y+z =
15
x+8 = 15
x = 15 – 8 = 7
Similarly,
z+x = 10 and x+y+z
= 15
y+10 = 15
y = 15 – 10 = 5
Therefore, x = 7cm , y = 5cm ad z
= 3cm i.e. AD = 7cm , BE = 5cm and CF = 3cm.
Q5. Prove that the tangents at the
extremities of any chord make equal angles with the chord.
CBSE 2000,2001,2002
Solution:-
Let AB be a chord of a circle
with centre O , and let AP and BP be the tangents at A and B.
Suppose the tangents meet at P.
Join OP. Suppose OP meets AB at C .
We need to prove that PAC
= PBC.
In triangles PCA and PCB , we
have:-
PA = PB ….......(Since,
tangents from an external points are equal)
APC =
BPC ….........(Since, PA and PB are equally inclined to OP)
PC = PC ….......(Common)
therefore, by SAS criterion of
congruence , we have:-
triangle PAC = triangle PBC
PAC
= PBC
, Hence proved.
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problem in last question...........
ReplyDelete@gurvinder
ReplyDeletemake the diagram of 5th question yourself and give this question a try...still if u found any problem in solving the ques then do tell us..!!!
done! actually i was confused with diagram that was given
ReplyDelete