Math Olympiad Test: Quadrilaterals- 2 - Class 9 MCQ

Let the angles of the quadrilateral be x°, 2x°, 3x° and 4x°
∵ Sum of the angles of quadrilateral = 360°
⇒ x + 2x + 3x + 4x = 360°
⇒10x = 360°
⇒ x = 36°
∵ Measure of largest angle
= 4 x x = 1 = 4 x 36° = 144°

Perimeter of parallelogram = 2(a + b) = 24 cm
⇒ a + b = 12 cm
Given a = 8 cm
then 8 + b = 12 cm
⇒ b = 4 cm

∵ ABCD is a square
∴ ∠D = 90° and AD = DC = AB = BC
In ΔADC
AD = DC
∴ ∠CAD = ∠ACD, and
∠D + ∠ACD + ∠CAD = 180°
⇒ 90° + 2 ∠ACD = 180°
⇒ ∠ACD = 90°​​​​​​​/2 = 45°

The figure formed by joining the midpoints of consecutive side of a quadrilateral is a parallelogram.

Let the angles be 2x, 4x, 5x and 7x respectively.
∴ Difference between largest and smallest angle = 7x - 2x = 5x
∵ Sum of all angles of a quadrilateral = 360°
⇒ 2x + 4x + 5x + 7x = 360°
⇒ 18x = 360°
⇒ x = 20°
∴ Required difference
= 7x - 2x = 5x
= 5 x 20° = 100°

Let the angle be x°
∴ Its adjacent angle = (180 - x)°
A/Q,

⇒ 3x° = 360° - 2x°
⇒ 5x = 360°
⇒ x = 72°

Rhombus, parallelogram, and kite have their opposite angles equal (i.e. in the ratio 1 : 1). Therefore, the quadrilateral PQRS is a trapezium.

P, Q, R and S are the mid-points of BA, BC, CD and DA respectively.

∴ AP = AS = PB = BQ = QC

∴ In ΔAPS
AP = AS

Similarly
Similarly ∠ASP = ∠APS =∠BPQ = ∠BQP = ∠CQR
= ∠CRQ = ∠DSR = ∠DRS = 45°
Now ∠P + ∠ASP + ∠APS = 180°
⇒ ∠P = 90°
Similarly,
∠P = ∠Q = ∠R = ∠S = 90°
∴ PQRS is a parallelogram having each of its angles = 90°
Now,
using midpoint theorem in ΔABC and ΔACD
SR = PQ = 1/2 AC, and in ΔABD and ΔBDC

∴ SP = PQ = QR = RS
∴ PQRS is a square

In ΔABR

DP ∥ BR, and P is the mid-point of side AB
∴ Using mid-point theorem (converse) Point P is the mid-point of AB and is parallel to BR
∴ 
D will be the mid-point of side AR
∴ 
⇒ AR = 2BC

∠DAC = ∠​​​​​​​DCA = 90°​​​​​​​/2 = 45°

∵ The opposite angles of a parallelogram are equal.
∴ (3x - 2)° = (50 - x)°
⇒  4x = 52°
⇒ x = 13°
∴ (3x - 2)° = 37°
∴ (50 - x)° = 37°

Let the angle be x°
∴ Adjacent angle = (180 - x)°
⇒ 
⇒ 4x = 180°
⇒ x = 45°

∵ Diagonals of a rectangle bisect each other and are also equal in length.
∴ In ΔPOS,
OP = OS
⇒ ∠OPS = ∠OSP
(angles opposite to equal sides are equal)
Also,
∠POS + ∠OSP + ∠OPS = 180°
⇒ 2∠OPS = 180° - ∠POS
= 180° - 64° (∵ ∠POS + ∠QOR) {vertically opposite∠s}
⇒ ∠POS​​​​​​​ = 116°​​​​​​​/2 = 58°

Let the measure of 4th angle be x° 
∴ 98° + 92° + 70° + x° = 180° × 2 = 360°
⇒ x° = 100°

∵ Diagonals of a rhombus bisect each other at 90°
In ΔAOD
AD = 15 cm, AC = 12 cm,




∴ BD = 2OD
= 2 × 9 = 18 cm