Elitmus
Exam
Numerical Ability
Geometry
O and O’ are centers of the two circles with radii 7cm and 9cm respectively. The distance between the centers is 20cm. If PQ is the transverse common tangent to the circles which cuts OO’ at X, what is the length of O’X in cm?
a. 35/4
b. 45/4
c. 10
d. 11
Read Solution (Total 7)
-
- b. 45/4
traingle's PO'X & QOX are similar
=> PO'/QO = O'X/OX
=> 9/7 = O'X/(20-O'X)
=> O'X = 45/4
- 10 years agoHelpfull: Yes(40) No(0)
- b.45/4
when you draw the figure the radius of the two circles,the distance between their centres,and the tangent will form two similar triangles,
pox and qo'x
let the length of ox be y,then o'x will be 20-y
from the property of similar triangles:
(po/ox)=(qo'/o'x)
(7/y)=(9/20-y)
=> y=35/4
since,o'x is 20-y
o'x will be 20-(35/4)=45/4
(7/y)=(9/20-y)
- 10 years agoHelpfull: Yes(5) No(1)
- option (b) is the correct ans....because if the circles would have been of same radius...then length of O'X woulb have been 11(9+4)cm....bt as second circle is greater so tranverse common tangent will cut the OO' nearer to smaller circle ....so length of O'X will b greater than 11.....which is supported by only opt (b)
- 10 years agoHelpfull: Yes(1) No(6)
- OQ/O'P = OX/O'X
=> OX = (7/9)*O'X
And since, OX + O'X = 20
=> (7/9)*O'X + O'X = 20
=> O'X = 45/4 - 10 years agoHelpfull: Yes(1) No(0)
- b,let the common tangent is meet at the point p on 7 radii circle and q on 9 radii circle, triangle qox and triangle po'x are similar.So po'/qo=o'x/ox=>o'x=45/4
- 10 years agoHelpfull: Yes(0) No(0)
- rest part
discrimant is less than 0
root will be imgnry left them
put x=1 and 6 in orignal eqn
log6 (x^2-6x+9)^1/2 +log6(x^2-8x+16)^1/2=1
noumber sln=0 - 10 years agoHelpfull: Yes(0) No(5)
- OP/OX=O'Q/O'X
THEN O'X=9*OX/7
THEN O'X=9*(OO'-O'X)/7
THEN 16*O'X/7=9*20/7
TEHN O'X=35/4 IS THE ANSWER - 9 years agoHelpfull: Yes(0) No(4)
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