Jharkhand Board JAC Class 9 Maths Solutions Chapter 8 Quadrilaterals Ex 8.1 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 8 Quadrilaterals Ex 8.1

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Question 1.

The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.

Answer:

Let the angles be 3x, 5x, 9x and 13x.

Sum of the interior angles of the quadrilateral = 360°

Now, 3x + 5x + 9x + 13x = 360°

⇒ 30x = 360°

⇒ x = 12°

Angles of the quadrilateral are:

3x = 3 × 12° = 36°

5x = 5 × 12° = 60°

9x = 9 × 12° = 108°

13x= 13 × 12°= 156°

Question 2.

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

Answer:

Given: AC = BD

To prove: ABCD is a rectangle.

Proof: In order to prove ABCD is a rectangle we need to prove that one of its interior angle is right angle.

In ΔABD and ΔBAD,

BA = BA (Common)

BC = AD (Opposite sides of a parallelogram are equal)

AC = BD (Given)

Therefore, ΔABC = ΔBAD by SSS congruence criterion.

⇒ ∠A = ∠B (by CPCT)

Also, ∠A + ∠B = 180° (Sum of the angles on the same side of the transversal)

⇒ 2∠A = 180°

⇒ ∠A = 90° = ∠B

Thus, ABCD is a rectangle.

Question 3.

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.

Answer:

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.

Given: OA = OC, OB = OD and ∠AOB = ∠BOC = ∠COD = ∠DOA = 90°

To prove: ABCD is parallelogram and AB = BC = CD = AD

Proof: In ΔAOB and ΔCOB,

OA = OC

∠AOB = ∠COB = 90°

OB = OB (Common)

Therefore, ΔAOB ≅ ΔBOC by SAS congruence criterion.

Thus, AB = BC (by CPCT)

Similarly, we can prove,

AB = BC = CD = AD Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.

Thus, ABCD is rhombus.

Question 4.

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

Answer:

Given: Let ABCD be a square and its diagonals AC and BD intersect each other at O.

To prove: AC = BD, AO = OC and ∠AOB = 90°

Proof: In ΔABC and ΔBAD,

AB = BA (Common)

∠ABC = ∠BAD = 90° (Each angle of square is of 90°)

BC = AD (All sides of square are equal)

Therefore, ΔABC ≅ ΔBAD by SAS congruence criterion.

Thus, AC = BD (by CPCT).

i.e., diagonals are equal.

Now, In ΔAOB and ΔCOD,

∠BAO = ∠DCO (Alternate interior angles)

∠AOB = ∠COD (Vertically opposite angles)

AB = CD (Given)

Therefore, ΔAOB = ΔCOD by AAS congruence criterion.

Thus, AO = CO and BO = DO by CPCT i.e.

Diagonals bisect each other.

Now, In ΔAOB and ΔCOB,

OB = OB (common)

AO = CO (Proved above)

AB = CB (Sides of the square)

Therefore, ΔAOB ≅ ΔCOB by SSS congruence criterion.

∴ ∠AOB = ∠COB (CPCT)

Also, ∠AOB + ∠COB = 180°

(Linear pair)

Thus, ∠AOB = ∠COB = 90° i.e.

Diagonals bisect each other at right angles.

Question 5.

Show that if the diagonals of a which diagonals AC and BD bisect each other at right angles at O.

Answer:

Given: Let ABCD be a quadrilateral in which diagonals AC and BD bisect each other at right angles at O.

To prove: Quadrilateral ABCD is a square.

Proof: In ΔAOB and ΔCOD,

AO = CO (Diagonals bisect each other)

∠AOB = ∠COD (Vertically opposite angles)

OB = OD (Diagonals bisect each other)

Therefore, ΔAOB ≅ ΔCOD by SAS congruence criterion.

Thus, AB = CD by CPCT. ……. (i)

Now, In ΔAOD and ΔCOD,

AO = CO (Diagonals bisect each other)

∠AOD = ∠COD = 90° (Diagonals bisect each other at right angle)

OD = OD (Common)

Therefore, ΔAOD ≅ ΔCOD by SAS congruence criterion.

Thus, AD = CD by CPCT

Similarly, ΔCOD ≅ ΔBOC and BC = CD (CPCT)

⇒ AD = BC = CD = AB ……… (ii)

Also, as diagonals bisect each other, so, ABCD is a parallelogram.

Consider ΔACD and ΔBDC

AD = BC (Proved above)

CD = CD (Common)

AC = BD (Given)

⇒ ΔACD ≅ ΔBDC (by SSS congruence criterion)

⇒ ∠ADC = ∠BCD (CPCT)

As ABCD is a parallelogram so,

AD || BC

⇒ ∠ADC + ∠BDC = 180° (Angles on the same side of transversal)

∴ ∠ADC – ∠BCD = 90°.

So, ABCD is a parallelogram in which all sides are equal and one angle is 90°

⇒ ABCD is a square.

Question 6.

Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig.). Show that

(i) it bisects ∠C also,

(ii) ABCD is a rhombus.

Answer:

(i) As ABCD is a parallelogram

⇒ AB || CD and AD || BC

⇒ ∠BAC = ∠ACD …….(i)

(Alternate interior angles)

∠CAD = ∠ACB ……. (ii)

(Alternate interior angles)

Also, ∠CAD = ABAC …….. (iii)

(As AC bisects ∠A)

Therefore, ∠BCA=∠DCA [By (i), (ii), (iii)]

⇒ AC bisects ∠C.

(ii) As ABCD is a parallelogram

⇒ AD = BC and AB = CD …(iv)

Also, as ∠CAD = ∠ACB (Alternate interior angles)

and ∠ACB = ∠ACD (Proved)

⇒ ∠CAD = ∠ACD

⇒ AD = CD …(v)

(sides opposites to equal angles are equal)

Similarly, AB = BC …(vi)

Therefore, AB = BC = CD = AD (By (iv), (v) (vi))

So, ABCD is a parallelogram in which all sides are equal.

⇒ ABCD is a rhombus.

Question 7.

ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and

Answer:

Let ABCD is a rhombus and AC and BD be its diagonals.

Proof: In ΔACD – AD = CD (Sides of a rhombus)

∠DAC = ∠DCA

(Angles opposite to equal sides of a triangle are equal)

Also, AB || CD (Opposite sides of rhombus are parallel)

⇒ ∠DAC = ∠BCA (Alternate interior angles)

⇒ ∠DCA = ∠BCA

Therefore, AC bisects ∠C.

In ∆ABC,

AB = BC (Sides of a rhombus)

⇒ ∠BAC = ∠BCA (Angles opposite to equal sides are equal)

Also, AD || BC (Opposite sides of rhombus are parallel)

⇒ ∠DAC = ∠BCA (Alternate interior angles)

∴ ∠BAC = ∠DAC

⇒ AC bisects ∠A

Similarly we can prove that diagonal BD bisects ∠B as well as ∠D.

Question 8.

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.

Answer:

(i) AB || CD and AD || BC (∵ opposite sides of rectangle are parallel)

⇒ ∠1 = ∠4 and∠2=∠3 (Alternate interior angles)

Also, as AC bisects ∠A and ∠C

⇒ ∠1 = ∠2 and ∠3 = ∠4

Therefore ∠2 = ∠4

So, in ΔACD,

AD = CD (sides opposite to equal angles are equal)

Also, AD = BC and AB = CD (∵ opposite sides of rectangle are equal)

∴ AB = BC = CD = AD

⇒ ABCD is a rectangle in which all sides are equal

⇒ ABCD is a square.

(ii) In ΔABD,

AB = AD (sides of a square)

⇒ ∠5 = ∠7 (Angles opposite to equal sides are equal)

Also, as AB || CD

⇒ ∠5 = ∠8 (Alternate interior angles)

∴ ∠7 = ∠8

⇒ BD bisects ∠D

Similarly, BD bisects ∠B.

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Question 9.

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

(i) ΔAPD ≅ ΔCQB

(ii) AP = CQ

(iii) ΔAQB ≅ ΔCPD

(iv) AQ = CP

(v) APCQ is a parallelogram

Answer:

(i) In ΔAPD and ΔCQB,

DP = BQ (Given)

∠ADP = ∠CBQ (Alternate interior angles)

AD = BC (Opposite sides of a ||^{gm})

Thus, ΔAPD ≅ ΔCQB

(by SAS congruence criterion)

(ii) AP = CQ by CPCT as ΔAPD ≅ ΔCQB.

(iii) In ΔAQB and ΔCPD,

BQ = DP (Given)

∠ABQ = ∠CDP (Alternate interior angles)

AB = CD (Opposite sides of a ||^{gm})

Thus, ΔAQB ≅ ΔCPD by SAS congruence criterion.

(iv) AQ = CP by CPCT as ΔAQB ≅ ΔCPD.

(v) From (ii) and (iv), we get opposite sides of quadrilateral APCQ are equal. Thus, APCQ is a ||^{gm}.

Question 10.

ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig). Show that

(i) ΔAPB ≅ ΔCQD

(ii) AP = CQ

Answer:

(i) In ΔAPB and ΔCQD,

∠ABP = ∠CDQ (Alternate interior angles)

∠APB = ∠CQD = 90° (∵ AP and CQ are perpendiculars)

AB = CD (Opposite sides of a is a parallelogram)

Thus, ΔAPB ≅ ΔCQD (by AAS congruence criterion)

(ii) AP = CQ by CPCT as ΔAPB ≅ ΔCQD.

Question 11.

In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see Fig).

Show that

(i) quadrilateral ABED is a parallelogram

(ii) quadrilateral BEFC is a parallelogram

(iii) AD || CF and AD = CF

(iv) quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) ΔABC ≅ ΔDEF.

Answer:

(i) AB = DE and AB || DE (Given)

Thus, quadrilateral ABED is a parallelogram because a pair of opposite sides of a quadrilateral are equal and parallel to each other.

(ii) Again BC = EF and BC || EF.

Thus, quadrilateral BEFC is a parallelogram.

(iii) Since ABED and BEFC are parallelograms.

⇒ AD = BE and BE = CF

(Opposite sides of a parallelogram are equal)

Thus, AD = CF.

Also, AD || BE and BE || CF

(Opposite sides of a parallelogram are parallel)

Thus, AD || CF

(iv) AD and CF are opposite sides of quadrilateral ACFD which are equal and parallel to each other.

Thus, ACFD is a parallelogram.

(v) AC || DF and AC = DF because ACFD is a parallelogram.

(vi) In ∆ABC and ∆DEF,

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram ACFD)

Thus, ∆ABC ≅ ∆DEF (by SSS congruence criterion)

Question 12.

ABCD is a trapezium in which AB || CD and AD = BC.

Show that

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) ΔABC ≅ ΔBAD

(iv) diagonal AC = diagonal BD

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Answer:

Construction: Draw a line through C parallel to DA intersecting AB produced at E.

Proof:

(i) In quadrilateral ∆DCE

AE || CD (as AB || CD)

AD || CE (By construction)

∴ ∆DCE is a parallelogram as opposite sides are parallel

CE = AD (Opposite sides of a parallelogram)

AD = BC (Given)

Therefore, BC = CE

⇒ ∠CBE = ∠CEB

(Angles opposite to equal sides)

also, ∠A + ∠CBE = 180°

(Angles on the same side of transversal and ∠CBE = ∠CEB)

∠B + ∠CBE = 180° (Linear pair)

⇒ ∠A = ∠B

(ii) ∠A + ∠D = ∠B+ ∠C= 180° (Angles on the same side of transversal)

⇒ ∠A + ∠D = ∠A + ∠C (∵ ∠A = ∠B)

⇒ ∠D = ∠C

(iii) In ∆ABC and ∆BAD,

AB = AB (Common)

∠DAB = ∠CBA (Proved)

AD = BC (Given)

Thus, ∆ABC ≅ ∆BAD by SAS congruence criterion.

(iv) Diagonal AC = diagonal BD by CPCT as ∆ABC ≅ ∆BAD.