Geometry MCQ Quiz - Objective Question with Answer for Geometry - Download Free PDF
Last updated on May 13, 2025
Latest Geometry MCQ Objective Questions
Geometry Question 1:
In ΔABC, ∠C = 90° and CD ⊥ AB, also ∠A = 65°, then ∠CBA is equal to
Answer (Detailed Solution Below)
Geometry Question 1 Detailed Solution
Given:
∠C = 90°
CD ⊥ AB
∠A = 65°
Concept used:
The sum of the angles of a triangle is 180°.
Calculation:
In ΔABC,
∠BAC + ∠CBA + ∠ACB = 180°
⇒ 65° + 90° + ∠CBA = 180°
⇒ ∠CBA = 25°
Important Points
We can arrive at the solution based on the perpendicular as well. But it is more time-consuming. Since ∠A and ∠C are given, it is wise to directly apply them in the concept and get the solution.
Geometry Question 2:
The area of the figure ABCD formed by joining A(-1, 1) B, (5, 1), C(5, 6) and D(-1, 6) in order is a
Answer (Detailed Solution Below)
Geometry Question 2 Detailed Solution
Given:
A(-1, 1) B, (5, 1), C(5, 6) and D(-1, 6)
Formula used:
Area of the quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD
Area of the triangle having vertices (x1, y1), (x2, y2) and (x3, y3)
= \(\dfrac{1}{2}\) |x1(y2-y3) +x2(y3-y1) +x3(y1-y2) |
Calculations:
Area of ∆ABC = \(\dfrac{1}{2}\) | − 1[1 − 6)] + (5) (6 - 1) + 5 [1 - 1)] |
= \(\dfrac{1}{2}\) | 5 + 25 |
= 15 sq. units
Area of ∆ACD = \(\dfrac{1}{2}\) |− 1(6 – 6) + 5(6 – 1) + (-1)( 1 − 6) |
= \(\dfrac{1}{2}\) | 25 + 5 |
= 15 sq. units
Area of the quadrilateral ABCD = 15 + 15
= 30 sq. units
∴ the area of the quadrilateral is 30 sq. units.
∴ The answer is 30 sq.units.
Geometry Question 3:
If the points P(-2, 1), Q(α, β) and R(4, -1) are collinear and α - β = -3, then the value of (α + β) is:
Answer (Detailed Solution Below)
Geometry Question 3 Detailed Solution
Given:
Points: P(-2, 1), Q(α, β), R(4, -1)
α - β = -3
Formula used:
For collinear points, slope of PQ = slope of QR
Slope formula: (y2 - y1) / (x2 - x1)
Calculation:
Slope of PQ = (β - 1) / (α + 2)
Slope of QR = (-1 - β) / (4 - α)
Since P, Q, R are collinear:
⇒ (β - 1) / (α + 2) = (-1 - β) / (4 - α)
Cross-multiplying:
⇒ (β - 1)(4 - α) = (-1 - β)(α + 2)
⇒ 4β - β α - 4 + α = -α -2 - β α - 2β
⇒ 4β + α + α + 2β = -2 + 4
⇒ 3β + α = 1
Given that α - β = -3:
From α - β = -3 ⇒ α = β - 3
Substitute in 3β + α = 1:
⇒ 3β + (β - 3) = 1
⇒ 4β - 3 = 1
⇒ β = 1
So, α = 1 - 3 = -2
α + β = -2 + 1 = -1
∴ The correct answer is -1.
Geometry Question 4:
If a regular polygon has 10 sides, then the measure of its interior angle is greater than the measure of its exterior angle by how many degrees?
Answer (Detailed Solution Below)
Geometry Question 4 Detailed Solution
Concept used:
If the number of sides of a regular polygon be n, then
The interior angle = \( \frac{(n - 2) \times 180^\circ }{n}\)
Calculation:
Let the required difference be x.
No. of sides of a regular polygon = 10
According to the question
\(\Rightarrow \frac{{\left( {n - 2} \right) \times 180^\circ }}{n} - \frac{{360^\circ }}{n} = x\)
\(\Rightarrow \frac{{\left( {10 - 2} \right) \times 180^\circ }}{{10}} - \frac{{360^\circ }}{{10}} = x\)
\(\Rightarrow \frac{{8 \times 180^\circ }}{{10}} - 36^\circ = x\)
⇒ 144 - 36° = x
∴ x = 108°Geometry Question 5:
Find the ratio of the measure of the angle of a regular pentagon to the measure of the angle of a regular octagon.
Answer (Detailed Solution Below)
Geometry Question 5 Detailed Solution
Given:
A regular pentagon has 5 sides, and a regular octagon has 8 sides.
The measure of the interior angle of a regular polygon is given by:
Interior angle = [(n - 2) × 180] / n, where n is the number of sides.
Formula used:
Ratio = Measure of the angle of the pentagon / Measure of the angle of the octagon
Calculations:
Step 1: Calculate the measure of the angle of the pentagon:
Angle = [(5 - 2) × 180] / 5
Angle = (3 × 180) / 5
Angle = 540 / 5
Angle = 108°
Step 2: Calculate the measure of the angle of the octagon:
Angle = [(8 - 2) × 180] / 8
Angle = (6 × 180) / 8
Angle = 1080 / 8
Angle = 135°
Step 3: Find the ratio:
Ratio = 108 / 135
Ratio = 4 / 5
The ratio of the measure of the angle of a regular pentagon to the measure of the angle of a regular octagon is 4:5.
Top Geometry MCQ Objective Questions
The area of the triangle whose vertices are given by the coordinates (1, 2), (-4, -3) and (4, 1) is:
Answer (Detailed Solution Below)
Geometry Question 6 Detailed Solution
Download Solution PDFGiven:-
Vertices of triangle = (1,2), (-4,-3), (4,1)
Formula Used:
Area of triangle = ½ [x1 (y2 - y3) + x2 (y3 - y1) + x3 (y1 - y2)]
whose vertices are (x1, y1), (x2, y2) and (x3, y3)
Calculation:
⇒ Area of triangle = (1/2) × [1(-3 – 1) + (-4) (1 – 2) + 4{2 – (-3)}]
= (1/2) × {(-4) + 4 + 20}
= 20/2
= 10 sq. units
In the triangle ABC, AB = 12 cm and AC = 10 cm, and ∠BAC = 60°. What is the value of the length of the side BC?
Answer (Detailed Solution Below)
Geometry Question 7 Detailed Solution
Download Solution PDFGiven:
In the triangle, ABC, AB = 12 cm and AC = 10 cm, and ∠BAC = 60°.
Concept used:
According to the law of cosine, if a, b, and c are three sides of a triangle ΔABC and ∠A is the angle between AC and AB then, a2 = b2 + c2 - 2bc × cos∠A
Calculation:
According to the concept,
BC2 = AB2 + AC2 - 2 × AB × AC × cos60°
⇒ BC2 = 122 + 102 - 2 × 12 × 10 × 1/2
⇒ BC2 = 124
⇒ BC ≈ 11.13
∴ The measure of BC is 11.13 cm.
A circle touches all four sides of a quadrilateral PQRS. If PQ = 11 cm. QR = 12 cm and PS = 8 cm. then what is the length of RS ?
Answer (Detailed Solution Below)
Geometry Question 8 Detailed Solution
Download Solution PDFGiven :
A circle touches all four sides of a quadrilateral PQRS. If PQ = 11 cm. QR = 12 cm and PS = 8 cm
Calculations :
If a circle touches all four sides of quadrilateral PQRS then,
PQ + RS = SP + RQ
So,
⇒ 11 + RS = 8 + 12
⇒ RS = 20 - 11
⇒ RS = 9
∴ The correct choice is option 3.
AB and CD are two parallel chords of a circle of radius 13 cm such that AB = 10 cm and CD = 24cm. Find the distance between them(Both the chord are on the same side)
Answer (Detailed Solution Below)
Geometry Question 9 Detailed Solution
Download Solution PDFGiven∶
AB ∥ CD, and
AB = 10cm, CD = 24 cm
Radii OA and OC = 13 cm
Formula Used∶
Perpendicular from the centre to the chord, bisects the chord.
Pythagoras theorem.
Calculation∶
Draw OP perpendicular on AB and CD, and
AB ∥ CD, So, the points O, Q, P are collinear.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
AP = 1/2 AB = 1/2 × 10 = 5cm
CQ = 1/2 CD = 1/2 × 24 = 12 cm
Join OA and OC
Then, OA = OC = 13 cm
From the right ΔOPA, we have
OP2 = OA2 - AP2 [Pythagoras theorem]
⇒ OP2 = 132- 52
⇒ OP2 = 169 - 25 = 144
⇒ OP = 12cm
From the right ΔOQC, we have
OQ2 = OC2- CQ2 [Pythagoras theorem]
⇒ OQ2 = 132 - 122
⇒ OQ2 = 169 - 144 = 25
⇒ OQ = 5
So, PQ = OP - OQ = 12 -5 = 7 cm
∴ The distance between the chord is of 7 cm.
The ratio of the measures of each interior angle of a regular octagon to that of the regular dodecagon is:
Answer (Detailed Solution Below)
Geometry Question 10 Detailed Solution
Download Solution PDFConcept:
Octagon has eight sides.
Dodecagon has twelve sides.
Formula:
Interior angle of polygon = [(n – 2) × 180°] /n
Calculation:
Interior angle of octagon = [(8 – 2)/8] × 180° = 1080°/8 = 135°
Interior angle of dodecagon = [(12 – 2)/12] × 180° = 1800°/12 = 150°
∴ The ratio of the measures of the interior angles for octagon and dodecagon is 9 : 10
Two circles touch each other externally at P. AB is a direct common tangent to the two circles, A and B are points of contact, and ∠PAB = 40° . The measure of ∠ABP is:
Answer (Detailed Solution Below)
Geometry Question 11 Detailed Solution
Download Solution PDFGiven:
Two circles touch each other externally at P.
AB is a direct common tangent to the two circles, A and B are points of contact, and ∠PAB = 40°.
Concept used:
If two circles touch each other externally at some point and a direct common tangent is drawn to both circles, the angle subtended by the direct common tangent at the point where two circles touch each other is 90°.
Calculation:
According to the concept, ∠APB = 90°
Considering ΔAPB,
∠ABP
⇒ 90° - ∠PAB
⇒ 90° - 40° = 50°
∴ The measure of ∠ABP is 50°.
Two common tangents AC and BD touch two equal circles each of radius 7 cm, at points A, C, B and D, respectively, as shown in the figure. If the length of BD is 48 cm, what is the length of AC?
Answer (Detailed Solution Below)
Geometry Question 12 Detailed Solution
Download Solution PDFGiven:
Radius of each circle = 7 cm
BD = transverse common tangent between two circles = 48 cm
Concept used:
Length of direct transverse tangents = √(Square of the distance between the circle - Square of sum radius of the circles)
Length of the direct common tangents =√(Square of the distance between the circle - Square of the difference between the radius of circles)
Calculation:
AC = Length of the direct common tangents
BD = Length of direct transverse tangents
Let, the distance between two circles = x cm
So, BD = √[x2 - (7 + 7)2]
⇒ 48 = √(x2 - 142)
⇒ 482 = x2 - 196 [Squaring on both sides]
⇒ 2304 = x2 - 196
⇒ x2 = 2304 + 196 = 2500
⇒ x = √2500 = 50 cm
Also, AC = √[502 - (7 - 7)2]
⇒ AC = √(2500 - 0) = √2500 = 50 cm
∴ The length of BD is 48 cm, length of AC is 50 cm
The complementary angle of supplementary angle of 130°
Answer (Detailed Solution Below)
Geometry Question 13 Detailed Solution
Download Solution PDFGiven:
One of the supplement angles is 130°.
Concept used:
For supplementary angle: The sum of two angles is 180°.
For complementary angle: The sum of two angles is 90°.
Calculation:
The supplement angle of 130° = 180° - 130° = 50°
The complement angle of 50° = 90° - 50° = 40°
∴ The complement angle of the supplement angle of 130° is 40°
Mistake PointsPlease note that first, we have to find the supplementary angle of 130° & after that, we will find the complementary angle of the resultant value.
To draw a pair of tangents to a circle which are inclined to each other at an angle of 75°, it is required to draw tangents at the end points of those two radii of the circle, the angle between whom is
Answer (Detailed Solution Below)
Geometry Question 14 Detailed Solution
Download Solution PDFConcept:
Radius is perpendicular to the tangent at the point of contact
Sum of all the angles of a Quadrilateral = 360°
Calculation:
PA and PB are tangents drawn from an external point P to the circle.
∠OAP = ∠OBP = 90° (Radius is perpendicular to the tangent at the point of contact)
Now, In quadrilateral OAPB,
∠APB + ∠OAP + ∠AOB + ∠OBP = 360°
75° + 90° + ∠AOB + 90° = 360°
∠AOB = 105°
Thus, the angle between the two radii, OA and OB is 105°
Two circles touch each other externally at point X. PQ is a simple common tangent to both the circles touching the circles at point P and point Q. If the radii of the circles are R and r, then find PQ2.
Answer (Detailed Solution Below)
Geometry Question 15 Detailed Solution
Download Solution PDF
We know,
Length of direct common tangent = √[d2 - (R - r)2]
where d is the distance between the centers and R and r are the radii of the circles.
PQ = √[(R + r)2 - (R - r)2]
⇒ PQ = √[R2 + r2 + 2Rr - (R2 + r2 - 2Rr)]
⇒ PQ = √4Rr
⇒ PQ2 = 4Rr