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

Given:

In Δ ABC, AB = 12 cm, BC = 16 cm, AC = 20 cm

Formula used:

Area of the triangle (Δ) = \(\sqrt{s(s-a)(s-b)(s-c)}\)

Where s = semi-perimeter = \(\frac{a+b+c}{2}\)

Radius (r) of the inscribed circle = \(\frac{\Delta}{s}\)

Calculations:

a = 12 cm, b = 16 cm, c = 20 cm

s = \(\frac{12+16+20}{2}\) = 24 cm

Area (Δ) = \(\sqrt{24(24-12)(24-16)(24-20)}\)

⇒ Area (Δ) = \(\sqrt{24×12×8×4}\)

⇒ Area (Δ) = \(\sqrt{9216}\)

⇒ Area (Δ) = 96 cm2

Radius (r) = \(\frac{96}{24}\)

⇒ Radius (r) = 4 cm

∴ The correct answer is option (2).

Given:

Three circles touch each other externally when the distance between their centres are 4 cm

and 5 cm and 6 cm.

Calculation:

Let O, P, Q be the centres of the three circles and they touch each other at M, N and S points.

OP = 4 cm, PQ = 5 cm, QO = 6 cm

Let OM = OS = r

⇒ MP = PN = 4 – r

⇒ SQ = NQ = QO – OS = 6 – r

⇒ NQ + PN = PQ = 5

⇒ 6 – r + 4 – r = 5

⇒ 2r = 5

⇒ r = 5/2

⇒ OM = 5/2

⇒ MP = 4 – 5/2 = 3/2

⇒ NQ = 6 – 5/2 = 7/2

⇒ OM + MP + NQ = 15/2 = 7.5

∴ Total radius of three circles is 7.5 cm.

Given:

The sum of their areas =  74 πsq cm

Distance between their centers = 12 cm.

Formula Used:

Area of circle = πr²

Calculation:

Let assume that radius of circle 1 = x

So, radius of circle 2 = 12 - x

Area of circle 1 = π(x)²

Area of circle 2 = π(12 - x)²

According to question ⇒ π(x)² + π(12 - x)² = 74π

⇒ x² + 144 - 24x + x² = 74 

⇒ 2x² - 24x + 70 = 0

⇒ x² - 12x + 35 = 0

⇒ (x - 7)(x - 5) = 0

⇒ x = 7 ⇒ x = 5 

∴ The radius of smaller circle is 5 cm

Given:

Vertices are (3, 5), (- 2, 0) and (6, 4)

Formula used:

Area of triangle = 1/2[x₁(y₂ – y₃) + x₂(y₃ – y₁) + x₃(y₁ – y₂)]

Calculations:

 Area of triangle = 1/2[3(0 – 4) - 2(4 – 5) + 6(5 – 0)]

⇒ 1/2[- 12 + 2 + 30]

⇒ 1/2 × 20

⇒ 10

⇒ Area of triangle = 10 unit2

∴ The area of the triangle is 10 unit2

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:-

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

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°

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.

Given:

ABC is a right-angled triangle. A circle is inscribed in it.

The length of the two sides containing the right angle are 10 cm and 24 cm

Calculations:

Hypotenuse² = 10² + 24²    (Pythagoras theorem)

Hypotenuse = √676 = 26

Radius of the circle (incircle) inside a triangle = ( Sum of sides containing right angle – Hypotenuse)/2

⇒ (10 + 24 - 26)/2

⇒ 8/2

⇒ 4

∴ The correct choice is option 4.