Geometry MCQ Quiz - Objective Question with Answer for Geometry - Download Free PDF

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.

 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.

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.

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

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.

Given:-

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

Given:

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.

Given :

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.

Given∶

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.

Concept:

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

Given:

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°.

Given:

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

Given:

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.

Concept:

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°

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