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MCQ Questions on Geometry for Competitive Exams

Prepare for competitive exams with our Geometry MCQ questions. Practice essential theorems and problem-solving techniques with questions tailored to exam patterns. Page-6 section-1

Each section contains 100 MCQs on Geometry .
51
In ΔABC, D and E are the points on AB and AC respectively such that AD × AC = AB × AE. If ∠ADE = ∠ACB + 30° and ∠ABC = 78°, then ∠A = ?
Discuss
Answer & Solution
Answer: Option D
Solution:
Geometry mcq question image
$$\frac{{{\text{AD}}}}{{{\text{AB}}}} = \frac{{{\text{AE}}}}{{{\text{AC}}}}$$
(It's mean ΔADE ∽ ΔABC)
∠B = ∠D, ∠C = ∠E
∠D = 78°
∠ADE = ∠ACB + 30°
78° = ∠ACB + 30°
∠ACB = 48°
∠A + ∠D + ∠E = 180°
∠A + 78 + 48 = 180°
∠A = 54°
52
In the given figure, if KI = IT and EK = ET, then ∠TEI = . . . . . . . . .
Geometry mcq question image
Discuss
Answer & Solution
Answer: Option C
No explanation is given for this question. Let's Discuss on Board
53
If in a triangle ABC, BE and CF are two medians perpendicular to each other and if AB = 19 cm and AC = 22 cm then the length of BC is
Discuss
Answer & Solution
Answer: Option D
Solution:
Given
Geometry mcq question image
AB = 19 cm, AC = 22 cm,
∵ BE ⊥ CF (Given), [Medians CF & BE are perpendicular to each other]
In this case
We know,
AB2 + AC2 = 5(BC)2
⇒ (19)2 + (22)2 = 5(BC)2
⇒ 361 + 484 = 5(BC)2
⇒ 845 = 5(BC)2
⇒ (BC)2 = 169
⇒ BC = 13 cm
54
In a ΔABC, ∠ABC = 2∠CAB, if the side BC is extended to D and ∠ACD = 126°, then ∠CAB is:
Discuss
Answer & Solution
Answer: Option C
Solution:
Geometry mcq question image
Let,
∠CAB = x
∠ABC = 2x
∠ACD = 3x = 126° (by exterior angle)
3x = 126°
x = 42°
∠CAB = 42°
55
Two circles of radii 5 cm and 3 cm touch externally, then the ratio in which the direct common tangent to the circles divides externally the line joining the centres of the circles is:
Discuss
Answer & Solution
Answer: Option A
No explanation is given for this question. Let's Discuss on Board
56
In the figure, a circle touches all the four sides of a quadrilateral ABCD whose sides AB = 6.5 cm, BC = 5.4 cm and CD = 5.3 cm. The length of AD is:
Geometry mcq question image
Discuss
Answer & Solution
Answer: Option B
Solution:
Geometry mcq question image
We know that AB + CD = BC + AD
6.5 + 5.3 = 5.4 + AD
11.8 = 5.4 + AD
AD = 11.8 - 5.4
AD = 6.4 cm
Note:-
Geometry mcq question image
AB + CD = (AB + PB) + (DR + RC) . . . . . . (i)
The length of tangent drawn from an external point to a circle are equal.
AB = CD, AS = BQ, CQ = RC, DR = DS
From equation (i)
AB + CD = AS + BQ + DS + CQ
= (AS + DS) + (BQ + BC)
$$\boxed{{\text{AB}} + {\text{CD}} = {\text{AD}} + {\text{BC}}}$$
If a circle is inscribed inside a quadrilateral then the sum of a pair opposite side equal to the sum of another pair of opposite sides.
i.e. AB + CD = AD + BC
57
PQRA is a rectangle, AP = 22 cm, PQ = 8 cm. ΔABC is a triangle whose vertices lie on the sides of PQRA such that BQ = 2 cm and QC = 16 cm. Then the length of the line joining the mid points of the sides AB and BC is.
Discuss
Answer & Solution
Answer: Option B
Solution:
Given that AP = 22 cm and PQ = 8 cm
Geometry mcq question image
Made a triangle such that B, is on side PQ and BQ = 2 cm
And C is on RQ such that QC = 16 cm
Because all the vertices are on sides of PQRA.
Now, PQRA is a rectangle so all the angle will be of 90°.
∠ARQ = 90°
and RC = RQ - CQ = 22 - 16 cm = 6 cm
In right angle ΔARC
AC2 = AR2 + RC2 = 82 + 62
$$\boxed{{\text{AC}} = 10}$$
Now in ΔABC, AC is 10 cm and M, N are the mid point of ΔABC.
So, MN = $$\frac{{10}}{2}$$ = 5 cm
58
In ΔABC, a line through A cuts the side BC at D such that BD : DC = 4 : 5. If the area of ΔABD = 60 cm2, then the area of ΔADC is:
Discuss
Answer & Solution
Answer: Option C
Solution:
Height will be same for both triangles
Geometry mcq question image
In triangles ADB, if base = 4x and area = 60 cm2
Area of ΔADB = 60 cm2
$$\frac{1}{2}$$ × base × height = 60
⇒ $$\frac{1}{2}$$ × 4x × height = 60
⇒ height = $$\frac{{60}}{{2x}}$$
⇒ height = $$\frac{{30}}{x}$$ cm
⇒ Therefore using height Area of ΔADC will be
= $$\frac{1}{2}$$ × base × height
= $$\frac{1}{2}$$ × 5x × $$\frac{{30}}{x}$$
= 75 cm2
59
If the perimeter of an isosceles right-angle triangle is 8($$\sqrt 2 $$ + 1) cm, then the length of the hypotenuse of the triangle is:
Discuss
Answer & Solution
Answer: Option A
Solution:
$$\eqalign{ & \left( {2 + \sqrt 2 } \right){\text{unit}} \to 8\left( {\sqrt 2 + 1} \right) \cr & 1\,{\text{unit}} \to \frac{{8\left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}} = 4\sqrt 2 \cr & {\text{AC}} = 4\sqrt 2 \times \sqrt 2 = 8{\text{ cm}} \cr} $$
60
ABCD is a cyclic trapezium whose sides AD and BC are parallel to each other; if ∠ABC = 75° then the measure of ∠BCD is:
Discuss
Answer & Solution
Answer: Option A
Solution:
Geometry mcq question image
According to figure
⇒ AD || BC
⇒ ∠ABC = 75°
Then
⇒ ∠ABC + ∠ADC = 180°
⇒ 75° + ∠ADC = 180°
⇒ ∠ADC = 180° - 75°
⇒ ∠ADC = 105°
⇒ As we know in a cyclic trapezium
∠ADC + ∠DCB = 180°
(AD || BC, corresponding angle)
⇒ 105° + ∠DCB = 180°
⇒ ∠DCB = 75°