MCQ: Quadrilaterals and Polygons - 2 - SSC CGL MCQ

In ΔAFD and Δ BFE,
∠AFD = ∠BFE (∵ vertically opposite angles)
and ∠ADC = ∠ABC (alternate angles)
∴ ΔAFD – ΔBFE

Hence, option D is correct.

We know tangents drawn to a circle from same external point are
equal
AM = AQ = x (let)
∴ MB = 6 – x = BN
QD = 7 – x = DP
PC = y (let) = CN
Now, CD = DP + PC = 5
⇒ 7 – x + y = 5
⇒ y – x = – 2
BC = CN + BN
= y + 6 – x = y – x + 6 = – 2 + 6 = 4
Hence, optjon A is correct.

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.

Given, ∠ADC = 70°
∠ADC + ∠ABC = 180°
∠ABC = 180° – ∠ADC = 180° – 70° = 110°
∴ ∠PBC = 180° – ∠ABC = 180° – 110° = 70°
And, ∠DAB = 60°
∠DAB + ∠DCB = 180°
∠DCB = 180° – ∠DAB = 180° – 60° = 120°
∴ ∠PCB = 180° – ∠DCB = 180° – 120° = 60°
∴ ∠PBC + ∠PCB = 70° + 60° = 130°
Hence, option A is correct.

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

∴ ∠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.

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.

Let the number of sides of a regular polygon be n.

Hence, option B is correct.

If two chords AC and BD of a circle intersect inside or outside the circle when produced at a point P, then
AP . PC = BP . DP
[∵ AC and BD are diagonals of cyclic quadrilateral and let these are chords of a circle and intersect at point P]
Hence, optjon B is correct.

Let the number of sides be 5x and 4x respectively.

∴ Number of sides = 5x = 5 × 3 = 15 and 4x = 4 × 3 = 12.
Hence, option A is correct.

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, Diagonal (d₂) = 12 + 12 = 24 cm and Diagonal (d₁) = 10 cm

= 120 square cm.

Hence, option C is correct.

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.

Given, ∠ADC = 70° and ∠BAD = 95°
∠ADC + ∠ABC = 180°
∠ABC = 180° – ∠ADC = 180° – 70° = 110°
∴ ∠CBE = 180° – ∠ABC = 180° – 110° = 70°
BC = BE (given)
∴ ∠BEC = ∠BCE
In ΔBCE, we know that
∠BCE + ∠BEC + ∠CBE = 180°
2∠BCE = 180° – ∠CBE = 180° – 70° = 110°

∠BCE = 55° = ∠BEC
∠BAD + ∠BCD = 180°
∠BCD = 180° – ∠BAD = 180° – 95° = 85°
∴ ∠DCE = ∠BCD + ∠BCE = 85° + 55° = 140°
Hence, option A is correct.

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

Hence, option B is correct.

If the number of sides of a polygon be n. Then,

Hence, option C is correct.

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.