Olympiad Test: Quadrilaterals - 2 - Class 8 MCQ

∵ O is the bisector of both the diagonals.
∴ PQRS is a parallelogram.
∵ Diagonals of the parallelogram are equal. 
∴ PQRS may be a square or rectangle.

∵ AB is parallel to BC, and, AB is transversal.
∴ ∠A + ∠B = 180°
⇒ ∠B = 180° - ∠A = 180° - 110° = 70°

In ∆LAM, 
∠ALM + 90° + ∠M = 180°
⇒∠M = 65°
∴x=∠M= 65°  (∵KLMN is a parallelogram).
In ∆BLK,
∠BLK + 65° + 90° = 180°
⇒ ∠M = 180° - 90° = 90°
⇒∠BLK = 25°
sum of all angles of parallelogram = 360
∠K + ∠M + ∠N + ∠L = 360°
65° + 65° ∠N + ∠N = 360°
130°+ 2∠N = 360°
2∠N = 360° -130°
2∠N = 230°
∠N = 115°

∠L = 25° + 25°+ y°
115° = 50+ y°
y°= 115° - 50°
y° = 65°

∠OBC = ∠ODA = 35°
(Alternate Interior ∠S).

ABCD is a parallelogram and P,Q,R,S are the midpoints of AB, BC, CD, DA respectively. Consider, AC as a diagonal of ABCD.
Now, According to midpoint theorem,
PQ =1/2 AC and PQ is parallel to AC,
and SR = 1/2 AC and PQ is parallel to AC.
∴ PQ = RS and PQ || RS || AC.
∴ PQRS will be a parallelogram. 

∵ AB is a common side in both square and rhombus.
∴ AB = BC = CD = DA = FA = FE = EB
In ∆EBC,
∠BEC + ∠C + ∠B = 180° [∠C = ∠BEC]
[∵ EB = BC]

[∠B = 90° + 60° = 150°]
∵ AE is the diagonal of square. ∴
∴∠AEF = 45°
Now,
∵∠ E is the angle of a square, 
∴∠E = 90°
⇒ 45° + 15° + a = 90°
⇒ a = 30°

From ∆EDC,
∠EDC = 180° - 2 × 30° = 120°
∠EDG = ∠EDC - ∠D = 120° - 90° = 30°
Also, 

5K × 13 K = 195
⇒ K = 3
∴  Sides of the rectangle are 15 cm, 39 cm 
∴ Perimeter = 2 (15 + 39)
= 2 × 54
= 108 cm.

∵ AE = EB.
∴ ∠BAE = ∠ABE = x.
∵ ΔDO = OB, and , AO = OC
In ΔABE,
∠DAE + ∠ABE = ∠AEF
⇒ x + x = 50°
⇒ 2x = 50°
⇒ x = 25°

In ΔAPS, 

∴ Perimeter (PQRS) = 4 × 4 cm
= 16 cm