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Quadrilaterals

    Exercise 8.1

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9th class > Quadrilaterals > Exercise 8.1

Exercise 8.1

1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Instructions

Assuming ABCD is a parallelogram where AC = BD.
Considering ∆ABC and ∆DCB: AC = (given), AB = (opposite sides of a parallelogram are ) and BC = ( side)
Thus, ∆ABC ≅ ∆ using the congruency rule.
By CPCT ⇒ ∠ABC = ∠ (1)
Since, ABCD is a : AB || and is a transversal.
∴ ∠ABC + ∠ = ° … (2) (Co-interior angles)
From (1) and (2), we have: ∠ABC = ∠DCB = ° i.e., ABCD is a having an angle equal to 90°.
So, ABCD is a rectangle.

2. Show that the diagonals of a square are equal and bisect each other at right angles.

Instructions

(i) In ∆ABC and ∆BAD: AB = ( side), BC = (sides of a ABCD) and ∠ABC = ∠ (each angle is °)
Thus, ∆ABC ≅ ∆ (by congruency). Thus, AC = (by CPCT) (1)
(ii) AD || (a square is also a ) and is a transversal. Thus, ∠1 = ∠ and ∠2 = ∠ ( angles are )
In ∆OAD and ∆OCB, we have: AD = (sides of a square ABCD), ∠1 = ∠ (proved) and ∠2 = ∠ (proved).
Thus, ∆OAD ≅ ∆ (Using congruency rule) ⇒ OA = and OD = i.e., the diagonals AC and BD bisect each other at O. (2)
(iii) In ∆OBA and ∆ODA, we have: OB = (proved), BA = (sides of a square ABCD) and OA = ( side)
Thus, by congruency rule: ∆OBA ≅ ∆. By CPCT: ∠AOB = ∠ (3)
As ∠AOB and ∠AOD form a linear pair i.e. ∠AOB + ∠AOD = ° which gives us: ∠AOB = ∠AOD = ° ⇒ AC ⊥ BD (4)
From (1), (2) and (4), we get AC and BD are and bisect each other at angles.

3. Diagonal AC of a parallelogram ABCD bisects ∠ A. Show that:

(i) it bisects ∠ C also,

(ii) ABCD is a rhombus.

Instructions

We have a parallelogram ABCD in which diagonal AC bisects ∠A ⇒ ∠DAC = ∠
(i) Since, ABCD is a parallelogram: AB || and is a transversal. Thus, ∠1 = ∠ (1) (as alternate interior angles are )
Also, BC || and is a transversal. Thus, ∠2 = ∠ (2) (as alternate interior angles are )
Also, ∠1 = ∠ (3) as AC bisects ∠ From (1), (2) and (3), we have: ∠3 = ∠4 ⇒ AC bisects ∠C.
(ii) In ∆ABC, we have: ∠1 = ∠ (from (2) and (3)) which means BC = (4) (as sides opposite to equal angles of a ∆ are )
Similarly, AD = (5). Since, we know ABCD is a ,
thus, AB = (6)
From (4), (5) and (6), we have AB = BC = CD = DA. Thus, ABCD is a .

4. ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C. Show that:

(i) ABCD is a square

(ii) diagonal BD bisects ∠ B as well as ∠ D.

Solution:

Instructions

We have a rectangle ABCD such that AC bisects ∠A as well as ∠C. i.e., ∠1 = ∠ and ∠2 = ∠ (1)
(i) Since, every rectangle is a .ABCD is a parallelogram ⇒ AB || and is a transversal i.e. ∠2 = ∠ (2) (as Alternate interior angles are )
From (1) and (2), we have: ∠3 = ∠. In ∆ABC, ∠3 = ∠4 ⇒ AB = (as sides opposite to equal angles of a triangle are equal)
Similarly, CD = . So, ABCD is a rectangle having sides equal ⇒ ABCD is a .
(ii) Since, ABCD is a square and diagonals of a square the angles. So, bisects ∠B as well as ∠D.

5. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that:

(i) ∆ APD ≅ ∆ CQB

(ii) AP = CQ

(iii) ∆ AQB ≅ ∆CPD

(iv) AQ = CP

(v) APCQ is a parallelogram

Instructions

Prove the above statements

  • Consider the triangles ΔAPD and ΔCQB: DP =
  • ∠ADP = ∠ (Alternate interior angles)
  • AD = (Opposite sides of a parallelogram)
  • By congruency rule, ΔAPD ≅ Δ
  • By CPCT, AP =
  • Now, considering the triangles ΔAQB and ΔCPD:
  • BQ = and ∠ABQ = ∠ (Alternate interior angles)
  • Also, = CD (Opposite sides of a parallelogram)
  • Thus, by congruency, ΔAQB ≅ Δ
  • By CPCT, AQ =
  • For quadrilateral APCQ, we have proved that the opposite sides are equal: AQ = CP and AP = CQ
  • Thus, APCQ is a parallelogram.

6. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that:

(i) ∆ APB ≅ ∆ CQD

(ii) AP = CQ

Instructions

Prove the above statements

  • Considering the triangles ΔAPB and ΔCQD: ∠ABP = ∠ (Alternate interior angles)
  • ∠APB = ∠ (= 90° as AP and CQ are perpendiculars)
  • AB = (opposite sides of parallelogram)
  • By congruency rule: ΔAPB ≅ Δ
  • By CPCT: AP =
  • Hence, proved.

7. ABCD is a trapezium in which AB || CD and AD = BC. Show that:

(i) ∠ A = ∠ B

(ii) ∠ C = ∠ D

(iii) ∆ ABC ≅ ∆ BAD

(iv) diagonal AC = diagonal BD

Before starting the proof, let's construct a line through C parallel to DA intersecting AB produced at E (as shown in the below figure). We obtain a parallelogram AECD.

Instructions

Now, prove the above statements

  • Considering the triangle BCE: CE = (Opposite sides of a parallelogram) (i)
  • From (i) and (ii), BC = i.e. the triangle BCE is an triangle.
  • This means ∠CBE = ∠ (Angles opposite to equal sides)
  • We also have: ∠A + ∠CEB = ° (Angles on the same side of transversal)
  • We can re-write this as: ∠A + ∠ = 180°
  • We also have: ∠B + ∠CBE = ° (Linear pair)
  • Thus, ∠A = ∠
  • ∠A + ∠ = ∠ + ∠C = ° (Angles on the same side of transversal)
  • Since ∠A = ∠B: ∠D = ∠
  • Considering the triangles ΔABC and ΔBAD: AB = (common side)
  • ∠ DAB = ∠ (Already proved) and AD = (Given)
  • ΔABC ≅ Δ by congruency
  • By CPCT: AC =
  • Hence, proved.