Chapter 8: Quadrilaterals

1 : Diagonals of a parallelogram ABCD intersect at O. If ∠BOC = 90° and ∠BDC = 50° , then ∠OAB is
(A)90°              (B)50°          (C)40°        (D)10°
Solution: (C) 40°

[Hints: Here, ∠BOC=∠AOB=90°

And  ∠BDC=∠OBA=50°  

In ∆AOB , we have

 ∠AOB+∠OAB+∠OBA=180°

 ⇒90°+∠OAB+50°=180°

 ⇒∠OAB+140°=180°

⇒∠OAB=180°-140° 

⇒∠OAB=40° ]

1.Three angles of a quadrilateral are 75° , 90° and 75° . The fourth angle is
(A) 90°       (B) 95°       (C) 105°           (D) 120°
Solution:  (D) 120° .

[ let x be fourth angle of the quadrilateral .

 ∴   75°+90°+75°+x=360°

 ⇒x+240°=360°

⇒x=360°-240° 

⇒x=120° ]

2. A diagonal of a rectangle is inclined to one side of the rectangle at 25°. The acute angle between the diagonals is
  (A) 55°           (B) 50°                (C) 40°              (D) 25°
Solution:  (A) 50°  .

[ Hints: Here, ∠BAC=25° and ∠ABC=90°  

     In ∆ABC , we have

  ∠BAC+∠ACB+∠ABC=180°

 ⇒25°+∠ACB+90°=180°

 ⇒∠ACB+115°=180°

⇒∠ACB=65°   ]

3. ABCD is a rhombus such that ∠ACB = 40°. Then ∠ADB is
(A) 40°     (B) 45°       (C) 50°          (D) 60°
Solution:

  [ Hints: Here, ∠ACB=∠CAD=40° and ∠∠∠∠∠∠∠ ]

4.The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rectangle, if
(A) PQRS is a rectangle
(B) PQRS is a parallelogram
(C) diagonals of PQRS are perpendicular
(D) diagonals of PQRS are equal.
Solution:  (C) diagonals of PQRS are perpendicular .

5.The quadrilateral formed by joining the mid-points of the sides of a quadrilateral PQRS, taken in order, is a rhombus, if
(A) PQRS is a rhombus
(B) PQRS is a parallelogram
(C) diagonals of PQRS are perpendicular
(D) diagonals of PQRS are equal.

Solution: (D) diagonals of PQRS are equal.

6.If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3:7:6:4, then ABCD is a
(A) rhombus            (B) parallelogram
(C) trapezium          (D) kite
Solution: (C) trapezium  .