MCQ: Quadrilaterals and Polygons - 1 - SSC CGL MCQ

Since, they are symmetrically on horizontal plane.
∴ AC = BD
∴ AE = BF = x
Now, AB = (a – b) + 2x
i.e. a + b = a – b +2x ⇒ 2b = 2x ⇒ b = x
Now in ΔACE,

Hence, option D is correct.

In rectangle PQRS,
PQ || RS
∴ ∠RPQ = ∠ PRS ...(i)
( ∵ vertically opposite angles)
Now in ΔPQR,

⇒ ∠QPR = 30°
∴ ∠PRS = 30° [ From the equation (i) ]
Hence, option C is correct.

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.

Area of rectangle lies between 40 cm² and 45 cm²
Now, one side = 5 cm
Since, area can't be less than 40 cm²
∴ Other side can't be less than = 40/5 = 8 cm
Since, area can't be greater than 45 cm².
∴ Other side can't be greater than =45/5 = 9 cm
∴ Minimum value of diagonal 

∴ Maximum value of diagonal 
So, diagonal lies between 9 cm and 11 cm.
Hence, option B is correct. 

Area of ΔAPS = Area of ΔDSR
∵ AS = SD and AP = DR
∴ ar (ΔABC) = 4 ar (ΔBPQ).
Hence, option B is correct.

∠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.

∠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.

∠ABC + ∠ADC = 180°
∠CBE = 50°
∴ ∠ABC = 180° – ∠CBE = 180° – 50° = 130°
∴ ∠ADC = 180° – ∠ABC = 180° – 130° = 50°
Hence, option C is correct.

The angle subtended at the centre by an arc is twice to that of angle subtended at the circumference.

∴ ∠CAD =1/2∠COD = 70°
∴ ∠BAD = ∠BAC + ∠CAD = 70° + 40° = 110°
∴ ∠BCD = 180° – ∠BAD = 180° – 110° = 70°
Hence, option A is correct.

Let the number of sides of a regular polygon be n. Then,
According to question,
Exterior angle : Interior angle = 1 : 17

n – 2 = 34
n = 36
Hence, option C is correct.

The sum of opposite angles of a concyclic quadrilateral = 180°
∴ ∠A + ∠C = ∠B + ∠D = 180°
Hence, option A is correct.

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

From equation (ii),

Hence, option B 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, option A is correct.

∠BAC = 55°
∠ACB = 90°
[∵ Angle of semi-circle]
In ΔABC, we know that
∠ABC = 180° – 90° – 55° = 35°
∴ ∠ABC + ∠ADC = 180°
∠ADC = 180° – ∠ABC = 180° – 35° = 145°
Hence, option C is correcrt.

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

Hence, option C is correct.