Latest Elitmus Aptitude Question SOLUTION: PR is a tangent to a circle at point P.Q is another point on the circle such that PQ is the diameter and RQ cuts the circle at point M. If the radius of the circle is 4 units and P

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(#M40008869) ELITMUS Numerical Ability question Keep an EYE

PR is a tangent to a circle at point P.Q is another point on the circle such that PQ is the diameter and RQ cuts the circle at point M. If the radius of the circle is 4 units and PR=6 units then find the ratio of the perimeter of triangle PMR to the triangle PQR
a)11/20
b)3/5
c)13/20
d)18/25

Asked In Elitmus amrit bhagat (12 years ago)

Unsolved Read Solution (10)

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Sourabh Sar  (12 years ago)

The ans is (b) 3/5

Sol:
I hope the geometry of this questn is clr.

Nw since PR is the tangent,Hence angle QPR is 90.
=>Triangle PQR is rit-angled triangle.
=>hyp QR=10. (Pythagoras Theorem)........(1)

Also PM is perpendicular to QR (Since triangle PQR is made inside circle with diameter as hyp
hence PMQ is rit-angled...itz a property of circle)
=>three triangles PQR,PMQ & PMR are similar to each other.
(For similarity of such type of triangles u can refer any 10th std book).

NOw in that pic consider the triangles PQR & PMR.
PQR~PMR
hence PR/RM=QR/PR.
=>PR^2=RM*QR
putting the values & from(1)
=>RM=18/5....(2)

Also
PQR~PQM
=>PQ/QM=QR/PQ
=>PQ^2=QM*QR
putting the values & from(1)
=>QM=32/5.....(3)

Also,
PQM~PMR
=>PM/QM=RM/PM
=>PM^2=QM*RM
from(2) & (3)...
PM=24/5......(4)

Hence peimeter of triangle PMR=PM+RM+PR
Putting the obtained values perimeter=72/5

Also perimeter of triangle PQR=PQ+QR+PR=24.

Hence th ratio PMR/PQR=72/(5*24)=3/5

:-))

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AB  (9 years ago)

mr ashu dont be oversmart

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Ashu  (12 years ago)

b)3/5 is the ratio of the perimeter of triangle PMR to the triangle PQR.

Is this solution Helpfull? Yes (8) | No (12)  

Tuhin Kole  (8 years ago)

Answer


Since PR is tangent, angle RPQ = 90°. Hence QPR is a right angled triangle.

Given that PR = 6 units ----(1)
PQ = Diameter of the circle = 2×4 = 8 units ----(2)

Using Pythagorean theorem, QR2 = PR2 + PQ2 = 62+82 = 100
=> QR = 10 units ----(3)

Angle inscribed in a semicircle is always 90° (Thales' theorem)
=> angle PMQ is 90°

sin PRM = sin PRQ
PM/PR=PQ/RQ
PM/6=8/10
PM = 48/10 = 4.8 units ----(3)

tan PRM = tan PRQ
PM/RM=PQ/PR
4.8/RM=86
RM = 3.6 units ----(4)

Perimeter of triangle PMR = PM + RM + PR = 4.8 + 3.6 + 6 = 14.4 units
Perimeter of triangle PQR = PQ + QR + PR = 8 + 10 + 6 = 24 units

Perimeter of triangle PMRPerimeter of triangle PQR=14.4/24=3/5

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amrit bhagat  (12 years ago)

explian the ans plz

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SACHIN AGGARWAL  (10 years ago)

the ans is (c)
13/20

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amit kumar  (8 years ago)

Since PR is tangent, angle RPQ = 90°. Hence QPR is a right angled triangle.

Given that PR = 6 units ----(1)
PQ = Diameter of the circle = 2×4 = 8 units ----(2)

Using Pythagorean theorem, QR2 = PR2 + PQ2 = 62+82 = 100
=> QR = 10 units ----(3)

Angle inscribed in a semicircle is always 90° (Thales' theorem)
=> angle PMQ is 90°

sin PRM = sin PRQ
PMPR=PQRQPM6=810
PM = 4810 = 4.8 units ----(3)

tan PRM = tan PRQ
PMRM=PQPR4.8RM=86
RM = 3.6 units ----(4)

Perimeter of triangle PMR = PM + RM + PR = 4.8 + 3.6 + 6 = 14.4 units
Perimeter of triangle PQR = PQ + QR + PR = 8 + 10 + 6 = 24 units

Perimeter of triangle PMR Perimeter of triangle PQR=14.424=144/240=3/5

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Gourab Bhowmik  (7 years ago)

Since PR is tangent, angle RPQ = 90°. Hence QPR is a right angled triangle.

Given that PR = 6 units ----(1)
PQ = Diameter of the circle = 2×4 = 8 units ----(2)

Using Pythagorean theorem, QR2 = PR2 + PQ2 = 62+82 = 100
=> QR = 10 units ----(3)

Angle inscribed in a semicircle is always 90° (Thales' theorem)
=> angle PMQ is 90°

sin PRM = sin PRQ
PM
PR
=
PQ
RQ
PM
6
=
8
10
PMPR=PQRQPM6=810
PM =
48
10
4810 = 4.8 units ----(3)

tan PRM = tan PRQ
PM
RM
=
PQ
PR
4.8
RM
=
8
6
PMRM=PQPR4.8RM=86
RM = 3.6 units ----(4)

Perimeter of triangle PMR = PM + RM + PR = 4.8 + 3.6 + 6 = 14.4 units
Perimeter of triangle PQR = PQ + QR + PR = 8 + 10 + 6 = 24 units

Perimeter of triangle PMR
Perimeter of triangle PQR
=
14.4
24
=
144
240
=
3
5
Perimeter of triangle PMRPerimeter of triangle PQR=14.424=144240=35

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Aman  (6 years ago)

PQ=8(diameter of the circle)
PR=6
so, QR=10 (pythagorean theorem)

angle QMP=90 degree (angle subtanding diameter of a circle)
angle QMP=angle RMP=90 degree

assume QM=a and MP=b
MR=10-a

we get equations, a^2 + b^2=8^2......(i)
and, b^2 + (10-a)^2=6^2......(ii)

solving these two equations, we get, a=6.4 and b=4.8

so, perimeter(triangle PMR)/ perimeter(triangle PQR)=(b+10-a+6)/(8+10+6)=14.4/20=18/25......option (d)

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RAGA  (4 years ago)

Ans:13/20
PQ-8
PR-6
QR-10 by pythagores theorm
let PM be k
PQR right 3gle
PM- x
MR- 10-x
PMR also right 3gle as formed by diameter
64=k^2+x^2
36=(10-x)^2+k^2

solving for x gives
x=(64/10)
then k=(48/10)
perimeter of PMR=6+(48/10)+(48/10)
perimetr of PQR =6+8+10
ratio
13/20

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