Quantitative Aptitude
MENSURATION MCQs
Regular Polygons, Triangles, Circles
Total Questions : 254
| Page 3 of 26 pages
Answer: Option D. -> AB, PQ
:
D
The diameter of a circle is a line segment which passes through the center of the circle and whose endpoints lie on the circle.
AB and PQ are the diameters of the circle.
:
D
The diameter of a circle is a line segment which passes through the center of the circle and whose endpoints lie on the circle.
AB and PQ are the diameters of the circle.
Answer: Option B. -> AB
:
B
The radius of a circle is a line segment connecting the center of the circle to any point on the circumferenceof the circle.
So, OA, OC, OD and OB are the radii of the given circle.
However, AB is the diameter.
:
B
The radius of a circle is a line segment connecting the center of the circle to any point on the circumferenceof the circle.
So, OA, OC, OD and OB are the radii of the given circle.
However, AB is the diameter.
Answer: Option C. -> Chord
:
C
A chord of a circle is a line segment whose endpoints lie on the circle.
The diameter of a circle is the chord which passes through the center of the circle.
The line segment AB is a chord as it does not pass through the center of the circle.
:
C
A chord of a circle is a line segment whose endpoints lie on the circle.
The diameter of a circle is the chord which passes through the center of the circle.
The line segment AB is a chord as it does not pass through the center of the circle.
Answer: Option A. -> 36 cm
:
A
Given that
The radius of thecircle = 18 cm
The diameter of thecircle = radius × 2
On substituting the values we get:
The diameter of the circle = 18 × 2 = 36 cm
:
A
Given that
The radius of thecircle = 18 cm
The diameter of thecircle = radius × 2
On substituting the values we get:
The diameter of the circle = 18 × 2 = 36 cm
Answer: Option B. -> D = 2r
:
B
Diameter = 2 × radius (or) twice the radius
∴ D = 2r is the correct relation between them.
:
B
Diameter = 2 × radius (or) twice the radius
∴ D = 2r is the correct relation between them.
Answer: Option B. -> radius
:
B
The radius of a circle is a line segment joining the center of the circle to any point on the circumferenceof the circle.
The line segment OA is the radius of the circle.
:
B
The radius of a circle is a line segment joining the center of the circle to any point on the circumferenceof the circle.
The line segment OA is the radius of the circle.
Answer: Option C. -> 20∘
:
C
ADB is a straight line. Bylinear pair axiom,
∠ADP + ∠PDB =180∘
100 + ∠PDB =180∘
∠PDB =80∘
Similarly ∠QED =80∘
We have, ∠DPE = 90∘(angle subtended by a diameter)
In △DPE,
∠DPE + ∠PED + ∠EDP = 180
[Angle sum property of a triangle]
∠PED = 10∘
Similarly ∠QDE =10∘
In △DRE,
∠DRE + ∠RDE + ∠RED = 180
[Angle sum property of a triangle]
∠DRE = 160∘
PRE is a straight line.Bylinear pair axiom,
∠PRD + ∠DRE =180∘
∠PRD + 160 =180∘
∠PRD =20∘
:
C
ADB is a straight line. Bylinear pair axiom,
∠ADP + ∠PDB =180∘
100 + ∠PDB =180∘
∠PDB =80∘
Similarly ∠QED =80∘
We have, ∠DPE = 90∘(angle subtended by a diameter)
In △DPE,
∠DPE + ∠PED + ∠EDP = 180
[Angle sum property of a triangle]
∠PED = 10∘
Similarly ∠QDE =10∘
In △DRE,
∠DRE + ∠RDE + ∠RED = 180
[Angle sum property of a triangle]
∠DRE = 160∘
PRE is a straight line.Bylinear pair axiom,
∠PRD + ∠DRE =180∘
∠PRD + 160 =180∘
∠PRD =20∘
Answer: Option B. -> 45∘
:
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,△OPCis an isosceles rightangled triangle.⇒∠PCO=∠POC
∠PCO+∠POC+∠OPC=180∘(Angle sum property of a triangle)
∠PCO+∠POC+90∘=180∘
∠PCO+∠POC=90∘
Hence,∠POC=∠OCP=45∘
:
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,△OPCis an isosceles rightangled triangle.⇒∠PCO=∠POC
∠PCO+∠POC+∠OPC=180∘(Angle sum property of a triangle)
∠PCO+∠POC+90∘=180∘
∠PCO+∠POC=90∘
Hence,∠POC=∠OCP=45∘