Thursday, December 1, 2016

CBSE Class 9 - Maths - Quadrilaterals (Questions and Answers)

Quadrilaterals 

CBSE Class 9 - Maths - Quadrilaterals (Questions and Answers)


Class 9 NCERT Exemplar Problems and Solutions

Question 1: If angles A,B,C and D of a quadrilateral ABCD, taken in order, are in the ratio 3:7:6:4 then ABCD is a trapezium. Is this statement true? Give reason for your answer.

Solution: Since, the angles are in the ratio 3:7:6:4,

Let these angles be 3x,7x,6x and 4x.

∵Since sum of angles of a quadrilateral is 360°,

∴3x+7x+6x+4x=360°

⇒ 20x=360°

⇒x=18°

∴The angles are:
   ∠A=3×18°=54°,
  ∠B=7×18°=126°,
  ∠C=6×18°=108° and
 ∠D=4×18°=72°

Since ∠A+∠B=54°+126°=180°. Thus, the sum of co-interior angles is 180°formed by lines AD,BC and transversal AB, ∴, AD∥BC.

So,ABCD is a trapezium.

Hence, the given statement is true.




Question 2: Diagonals of a quadrilateral ABCD bisect each other. If∠A=45°, determine ∠B.

Solution: Since the diagonals AC and BD of quadrilateral ABCD bisect each other, ABCD is a parallelogram.

⇒ AD∥BC and AB is a transversal,

⇒ ∠A+∠B=180°(sum of co-interior angle=180°)

⇒ 45°+∠B=180°

⇒ ∠B=135°.



Question 3: Three angles of a quadrilateral ABCD are equal. Is it a parallelogram? Why or why not?

Solution: It may or may not be a parallelogram; because we have ∠A=∠B=∠C=80°, then ∠D=360°-3×80°=120°,

⇒ ∠B≠ ∠D (opposite angles are not equal).


Question 4: Diagonals AC and BD of a parallelogram ABCD intersect each other at O=3cm and OD=2cm, determine the length of AC and BD.

Solution: Since the diagonals of a parallelogram bisect each other,

       AC=2 OA=(2×3)cm= 6cm and

      BD=2 OD=(2×2) cm=4cm.


Question 5: The angle between two attitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60°.Find the angles of the parallelogram.

Solution: Let ABCD be parallelogram such that DM⊥AB, DN⊥BC and ∠MDN=60°.

In quadrilateral DMBN,

    ∠MDN+∠M+∠N+∠B=360°

⇒60°+90°+90°+∠B=360°

⇒∠B=120°.

AD∥BC and AB is a transversal,

∠A+∠B=180°

⇒∠A+120°=180°⇒∠A=60°.

∠C=∠A and ∠D=∠B

⇒∠C=60° and ∠D=120°.

Thus, the angles of∥gm ABCD are 60°,120°,60°,120° .


Question 6: ABCD is a rhombus in which altitude from D to side AB bisects AB. Find the angles of the rhombus. 

Solution:
Given: ABCD is a rhombus in which DM⊥AB such that M is mid-point of AB.

Construction: Join BD.

Proof:
    In △DAM and △DBM, ∠AMD=∠BMD (each=90°)

    AM=BM                        (M is mid-point of AB)

     DM=DM (common)

    ∴ △DAM ≌ △DBM          (SAS axiom of congruency)

⇒ AD=BD

Also, AD=AB         (∵ABCD is a rhombus)


⇒ AD=BD=AB

⇒ △ABD is an equilateral triangle

⇒ ∠A=60°

Since, AD∥BC and AB is a transversal,

∴ ∠A+∠B=180°       (sum of co-interior ∠s is 180°)

⇒ 60°+∠B=180°

⇒∠B=120°

  ∠C=∠A and∠D=∠B          (opp.∠s in a ∥gm are equal)

⇒ ∠C=60° and ∠D=120°

 Hence, the angles of the rhombus are 60°,120°,60°,120°.

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