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All questions of Quadrilaterals and Polygons for SSC CGL Exam
Directions: Kindly study the following questions carefully and choose the right answer:In a trapezium, the two non-parallel sides are equal in length, each being of 5 cm. The parallel sides are at a distance of 3 cm apart. If the smaller side of the parallel sides is of length 2 cm, then the sum of the diagonals of the trapezium is- a)10√5 cm
- b)6√5 cm
- c)3√5 cm
- d)5√5 cm
Correct answer is option 'B'. Can you explain this answer?
Directions: Kindly study the following questions carefully and choose the right answer:
In a trapezium, the two non-parallel sides are equal in length, each being of 5 cm. The parallel sides are at a distance of 3 cm apart. If the smaller side of the parallel sides is of length 2 cm, then the sum of the diagonals of the trapezium is
a)
10√5 cm
b)
6√5 cm
c)
3√5 cm
d)
5√5 cm
 | EduRev SSC CGL answered |
In ΔBCF,
By the pythagoras theorem,
BF2 = BC2 – CF2
(BF)2 = (5)2 – (3)2 ⇒ BF = 4 cm
∴ AB = 2 + 4 + 4 = 10 cm
Now, in ΔACF,
Similarly, BD = 45 cm
Hence, option B is correct.
By the pythagoras theorem,
BF2 = BC2 – CF2
(BF)2 = (5)2 – (3)2 ⇒ BF = 4 cm
∴ AB = 2 + 4 + 4 = 10 cm
Now, in ΔACF,
Similarly, BD = 45 cm
Hence, option B is correct.
Directions: Kindly study the following Questions carefully and choose the right answer:Ratio of the number of sides of two regular polygons is 5 : 6 and the ratio of their each interior angle is 24 : 25. Then the number of sides of these two polygons are- a)20, 24
- b)15, 18
- c)10, 12
- d)5, 6
Correct answer is option 'C'. Can you explain this answer?
Directions: Kindly study the following Questions carefully and choose the right answer:
Ratio of the number of sides of two regular polygons is 5 : 6 and the ratio of their each interior angle is 24 : 25. Then the number of sides of these two polygons are
a)
20, 24
b)
15, 18
c)
10, 12
d)
5, 6
 | Bayshore Academy answered |
Let the number of sides be 5x and 6x respectively. Then,
75x – 30 = 72x – 24
3x = 6
x = 2
∴ Number of sides = 5x = 5 × 2 = 10 and 6x = 6 × 2 = 12
Hence, option C is correct.
Directions: Kindly study the following questions carefully and choose the right answer:If ABCD be a cyclic quadrilateral in which ∠A = 4x°, ∠B = 7x°, ∠C = 5y°, ∠D = y°, then x : y is- a)3 : 4
- b)4 : 3
- c)5 : 4
- d)4 : 5
Correct answer is option 'B'. Can you explain this answer?
Directions: Kindly study the following questions carefully and choose the right answer:
If ABCD be a cyclic quadrilateral in which ∠A = 4x°, ∠B = 7x°, ∠C = 5y°, ∠D = y°, then x : y is
a)
3 : 4
b)
4 : 3
c)
5 : 4
d)
4 : 5
 | Target Study Academy answered |
The sum of opposite angles of a concyclic quadrilateral is 180°.
∴ ∠A + ∠C = 180°
4x + 5y = 180° ...(i)
∠B + ∠D = 180°
7x + y = 180° ...(ii)
By equation (ii) × 5 – (i), we get
∴ ∠A + ∠C = 180°
4x + 5y = 180° ...(i)
∠B + ∠D = 180°
7x + y = 180° ...(ii)
By equation (ii) × 5 – (i), we get
From equation (ii),
Hence, option B is correct.
Directions: Kindly study the following Questions carefully and choose the right answer:If the sum of the interior angles of a regular polygon be 1080°, the number of sides of the polygon is- a)6
- b)8
- c)10
- d)12
Correct answer is option 'B'. Can you explain this answer?
Directions: Kindly study the following Questions carefully and choose the right answer:
If the sum of the interior angles of a regular polygon be 1080°, the number of sides of the polygon is
a)
6
b)
8
c)
10
d)
12
 | Ssc Cgl answered |
We know that,
Sum of the interior angles of a regular polygon of n sides = (2n – 4) × 90°
∴ (2n – 4) × 90° = 1080°
2n – 4 = 12
2n = 16 ⇒ n = 8
Hence, option B is correct.
Sum of the interior angles of a regular polygon of n sides = (2n – 4) × 90°
∴ (2n – 4) × 90° = 1080°
2n – 4 = 12
2n = 16 ⇒ n = 8
Hence, option B is correct.
Directions: Kindly study the following Questions carefully and choose the right answer:A quadrilateral ABCD is inscribed in a circle. If AB is parallel to CD and AC = BD, then the quadrilateral must be a- a)parallelogram
- b)rhombus
- c)trapezium
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Directions: Kindly study the following Questions carefully and choose the right answer:
A quadrilateral ABCD is inscribed in a circle. If AB is parallel to CD and AC = BD, then the quadrilateral must be a
a)
parallelogram
b)
rhombus
c)
trapezium
d)
None of these
 | G.K Academy answered |
The quadrilateral must be a trapezium because a quadrilateral where only one pair of opposite sides are parallel (in the case AB || CD) is a trapezium.
Hence, option C is correct.
Hence, option C is correct.
Directions: Kindly study the following Questions carefully and choose the right answer:Consider the following statements
I. Let ABCD be a parallelogram which is not a rectangle. Then, 2(AB2 + BC2) ≠ AC2 + BD2
II. If ABCD is a rhombus with AB = 4 cm, then AC2 + BD2 = n3 some positive integer n.
Which of the above statements is/are correct?- a)only I
- b)Only II
- c)Both I and II
- d)Neither I nor II
Correct answer is option 'B'. Can you explain this answer?
Directions: Kindly study the following Questions carefully and choose the right answer:
Consider the following statements
I. Let ABCD be a parallelogram which is not a rectangle. Then, 2(AB2 + BC2) ≠ AC2 + BD2
II. If ABCD is a rhombus with AB = 4 cm, then AC2 + BD2 = n3 some positive integer n.
Which of the above statements is/are correct?
I. Let ABCD be a parallelogram which is not a rectangle. Then, 2(AB2 + BC2) ≠ AC2 + BD2
II. If ABCD is a rhombus with AB = 4 cm, then AC2 + BD2 = n3 some positive integer n.
Which of the above statements is/are correct?
a)
only I
b)
Only II
c)
Both I and II
d)
Neither I nor II
| | G.K Academy answered |
I. ABCD is a parallelogram, then
AC2 + BD2 = 2(AB2 + BC2)
II. ABCD is a rhombus and diagonals AC and BD bisect each other.
∴ AO = OC and OB = OD
In ΔAOB, AB2 = AO2 + OB2
AC2 + BD2 = 2(AB2 + BC2)
II. ABCD is a rhombus and diagonals AC and BD bisect each other.
∴ AO = OC and OB = OD
In ΔAOB, AB2 = AO2 + OB2
Hence, option B is correct.
Directions: Kindly study the following questions carefully and choose the right answer:Let ABCD be a parallelogram. Let P, Q, R and S be the mid-points of sides AB,
BC, CD and DA respectively. Consider the following statements.
I. Area of triangle APS < Area of triangle DSR, if BD < AC.
II. Area of triangle ABC = 4 (Area of triangle BPQ).
Select the correct answer using the codes given below.- a)Only I
- b)Only II
- c)Both I and II
- d)Neither I nor II
Correct answer is option 'B'. Can you explain this answer?
Directions: Kindly study the following questions carefully and choose the right answer:
Let ABCD be a parallelogram. Let P, Q, R and S be the mid-points of sides AB,
BC, CD and DA respectively. Consider the following statements.
I. Area of triangle APS < Area of triangle DSR, if BD < AC.
II. Area of triangle ABC = 4 (Area of triangle BPQ).
Select the correct answer using the codes given below.
BC, CD and DA respectively. Consider the following statements.
I. Area of triangle APS < Area of triangle DSR, if BD < AC.
II. Area of triangle ABC = 4 (Area of triangle BPQ).
Select the correct answer using the codes given below.
a)
Only I
b)
Only II
c)
Both I and II
d)
Neither I nor II
| | Ssc Cgl answered |
Area of ΔAPS = Area of ΔDSR
∵ AS = SD and AP = DR
∴ ar (ΔABC) = 4 ar (ΔBPQ).
Hence, option B is correct.
∵ AS = SD and AP = DR
∴ ar (ΔABC) = 4 ar (ΔBPQ).
Hence, option B is correct.
Directions: Kindly study the following Questions carefully and choose the right answer:The area of a rhombus with side of 13 cm and one diagonal 10 cm will be- a)140 cm`
- b)130 cm2
- c)120 cm2
- d)110 cm2
Correct answer is option 'C'. Can you explain this answer?
Directions: Kindly study the following Questions carefully and choose the right answer:
The area of a rhombus with side of 13 cm and one diagonal 10 cm will be
a)
140 cm`
b)
130 cm2
c)
120 cm2
d)
110 cm2
 | Iq Funda answered |
As we know that diagonals of a rhombus bisect each other at right angles.
Therefore, applying the Pythagoras theorem taking triangle ΔOCD into consideration, we get
Therefore, applying the Pythagoras theorem taking triangle ΔOCD into consideration, we get
Therefore, Diagonal (d2) = 12 + 12 = 24 cm and Diagonal (d1) = 10 cm
= 120 square cm.
Hence, option C is correct.
Directions: Kindly study the following Questions carefully and choose the right answer:ABCD is a parallelogram. E is a point on BC such that BE : EC = m : n. If AE and DB intersect in F, then what is the ratio of the area of ΔFEB to the area of ΔAFD?- a)m/n
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Directions: Kindly study the following Questions carefully and choose the right answer:
ABCD is a parallelogram. E is a point on BC such that BE : EC = m : n. If AE and DB intersect in F, then what is the ratio of the area of ΔFEB to the area of ΔAFD?
a)
m/n
b)
c)
d)
| | G.K Academy answered |
In ΔAFD and Δ BFE,
∠AFD = ∠BFE (∵ vertically opposite angles)
and ∠ADC = ∠ABC (alternate angles)
∴ ΔAFD – ΔBFE
∠AFD = ∠BFE (∵ vertically opposite angles)
and ∠ADC = ∠ABC (alternate angles)
∴ ΔAFD – ΔBFE
Hence, option D is correct.
Directions: Kindly study the following Questions carefully and choose the right answer:ABCD is a quadrilateral such that BC = BA and CD > AD. Which one of the following is correct?- a)∠BAD = ∠BCD
- b)∠BAD < ∠BCD
- c)∠BAD > ∠BCD
- d)2∠BAD = ∠BCD
Correct answer is option 'C'. Can you explain this answer?
Directions: Kindly study the following Questions carefully and choose the right answer:
ABCD is a quadrilateral such that BC = BA and CD > AD. Which one of the following is correct?
a)
∠BAD = ∠BCD
b)
∠BAD < ∠BCD
c)
∠BAD > ∠BCD
d)
2∠BAD = ∠BCD
| | EduRev SSC CGL answered |
Join AC.
Now, in ΔABC
∵ AB = BC
∠BAC = ∠BCA ....(i) ( ∵ angles opposite to equal
side)
In ΔADC,
CD > AD
∠DAC > ∠DCA
(Since in a triangle opposite to greater side is bigger than the angle opposite to smaller side)
On adding Eq. (i) and (ii), we get
∠BAD > ∠BCD
Hence, option C is correct.
Now, in ΔABC
∵ AB = BC
∠BAC = ∠BCA ....(i) ( ∵ angles opposite to equal
side)
In ΔADC,
CD > AD
∠DAC > ∠DCA
(Since in a triangle opposite to greater side is bigger than the angle opposite to smaller side)
On adding Eq. (i) and (ii), we get
∠BAD > ∠BCD
Hence, option C is correct.
Directions: Kindly study the following Questions carefully and choose the right answer:ABCD is a cyclic trapezium such that AD || BC, if ∠ABC = 70°, then the value of ∠BCD is :- a)60°
- b)70°
- c)40°
- d)80°
Correct answer is option 'B'. Can you explain this answer?
Directions: Kindly study the following Questions carefully and choose the right answer:
ABCD is a cyclic trapezium such that AD || BC, if ∠ABC = 70°, then the value of ∠BCD is :
a)
60°
b)
70°
c)
40°
d)
80°
 | T.S Academy answered |
∠ABC + ∠CDA = 180°
∠CDA = 180° – ∠ABC = 180° – 70° = 110°
We know that,
∠BCD + ∠CDA = 180°
∴ ∠BCD = 180° – ∠CDA = 180° – 110° = 70°
Hence, opton B is correct.
∠CDA = 180° – ∠ABC = 180° – 70° = 110°
We know that,
∠BCD + ∠CDA = 180°
∴ ∠BCD = 180° – ∠CDA = 180° – 110° = 70°
Hence, opton B is correct.
Directions: Kindly study the following Questions carefully and choose the right answer:A cyclic quadrilateral ABCD is such that AB = BC, AD = DC, AC ⊥ BD. ∠CAD = Θ. Then the angle ∠ABC =- a)Θ
- b)
- c)2Θ
- d)3Θ
Correct answer is option 'C'. Can you explain this answer?
Directions: Kindly study the following Questions carefully and choose the right answer:
A cyclic quadrilateral ABCD is such that AB = BC, AD = DC, AC ⊥ BD. ∠CAD = Θ. Then the angle ∠ABC =
a)
Θ
b)
c)
2Θ
d)
3Θ
| | EduRev SSC CGL answered |
In ΔABC,
Given, AB = BC
∠BCA = ∠BAC
And, AD = DC
∠CAD = ∠ACD
∴ ∠DAB = ∠DCB
∴ ∠DAB + ∠DCB = 180° ⇒ 2∠DAB = 180°
∴ ∠DAB = 90° = ∠DCB
In ΔADE, we know that
Given, AB = BC
∠BCA = ∠BAC
And, AD = DC
∠CAD = ∠ACD
∴ ∠DAB = ∠DCB
∴ ∠DAB + ∠DCB = 180° ⇒ 2∠DAB = 180°
∴ ∠DAB = 90° = ∠DCB
In ΔADE, we know that
∴ ∠ADE = 180° – 90° – Θ = 90° – Θ
∴ ∠ADC = ∠ADE + ∠CDE = 2(90° – Θ)
∠ABC + ∠ADC = 180°
∠ABC = 180° – ∠ADC = 180° – 2(90° – Θ) = 2Θ
Hence, option C is correct.
∴ ∠ADC = ∠ADE + ∠CDE = 2(90° – Θ)
∠ABC + ∠ADC = 180°
∠ABC = 180° – ∠ADC = 180° – 2(90° – Θ) = 2Θ
Hence, option C is correct.
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