10th Grade > Mathematics
CIRCLES MCQs
:
B
The tangent at any point of a circle is perpendicular to the radius through the point of contact.
∴∠OPC=90∘
Given, OP = PC.
So, △OPC is an isosceles right angled triangle.⇒∠PCO=∠POC
∠PCO+∠POC+∠OPC=180∘(Angle sum property of a triangle)
∠PCO+∠POC+90∘=180∘
∠PCO+∠POC=90∘
Hence, ∠POC=∠OCP=45∘
:
D
OMN = 90∘
Radius and tangent are perpendicular at point of contact remaining part of ∠OMN is ∠OMP=90∘−60∘=30∘
OP = OM = Radius
Hence, ∠OMP = ∠OPM = 30∘
Therefore, ∠MOP = 120∘.
:
Join the centre of the circle and the vertices of the triangle. Observe that the sides of the triangle become the tangents to the circle. Hence, the radii of the circle as shown in the question become the heights of the smaller triangles.
According to formula
12 (r) (sum of sides) = area of triangle
12 (r) ( 3+ 4 + 5) = 12 × 3 × 4
r = 1 cm
:
C
Given that ∠APB=120∘
Also, we know that if two tangents are drawn from an external point to a circle, then the line joining the external point and the centre of the circle bisects the angle between the tangents.
⟹∠APO=∠OPB=60∘
Thus, cos ∠OPA=cos 60∘=APOP
⟹12 = APOP
Thus, OP=2AP
:
B
Given that AO = AB=OB.
Since all sides are equal, △AOB is equilateral, and hence equiangular. Also, each angle of the triangle equals 60∘.
i.e.,∠AOB = 60∘
⟹∠ACB=12AOB
(∵ Angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. )
⟹∠ACB=60∘2=30∘
:
A
We know that the tangents drawn from an external point to a circle are equal.
∴ AD = AF, . . . (i) [tangents from A]
BD = BE, . . . (ii) [tangents from B]
CE = CF . . . (iii) [tangents from C]
Now, AB = AC [given]
⇒ AD + BD = AF + CF
⇒ BD = CF
⇒ BE = CE [using (ii) and (iii)]
⇒ E bisects BC.
:
A
A parallelogram ABCD circumscribes a circle with centre O.
We know that the lengths of tangents drawn from an exterior point to a circle are equal.
∴ AP = AS, . . . (i) [tangents from A]
BP = BQ, . . . (ii) [tangents from B]
CR = CQ, . . . (iii) [tangents from C]
DR = DS . . . (iv) [tangents from D]
Adding (i), (ii), (iii) and (iv), we get,
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC . . . (v)
But we know that in a parallelogram opposite sides are equal.
∴ AD = BC and AB = CD, putting these in (v) we get
2 AB = 2 AD
AB = AD = CD = BC.
Hence, ABCD is a rhombus.