The document NCERT Solutions Chapter 8 - Quadrilaterals (I), Class 9, Maths Class 9 Notes | EduRev is a part of Class 9 category.

All you need of Class 9 at this link: Class 9

**Exercise 8.1****1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.**

**Answer**

Let x be the common ratio between the angles.

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

Given,

AC = BD

To show,

To show ABCD is a rectangle we have to prove that one of its interior angle is right angled.

Proof,

In ΔABC and ΔBAD,

BC = BA (Common)

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

AC = BD (Given)

Therefore, ΔABC ≅ ΔBAD by SSS congruence condition.

∠A = ∠B (by CPCT)

also,

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

⇒ 2∠A = 180°

⇒ ∠A = 90°

Thus ABCD is a rectangle.**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 = ∠OCD = ∠ODA = 90°

To show,

ABCD is parallelogram and AB = BC = CD = AD

Proof,

In ΔAOB and ΔCOB,

OA = OC (Given)

∠AOB = ∠COB (Opposite sides of a parallelogram are equal)

OB = OB (Common)

Therefore, ΔAOB ≅ ΔCOB by SAS congruence condition.

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 as it is a parallelogram whose diagonals intersect at right angle.**4. Show that the diagonals of a square are equal and bisect each other at right angles. Answer**

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

To show,

AC = BD, AO = OC and ∠AOB = 90°

Proof,

In ΔABC and ΔBAD,

BC = BA (Common)

∠ABC = ∠BAD = 90°

AC = AD (Given)

Therefore, ΔABC ≅ ΔBAD by SAS congruence condition.

Thus, AC = BD by CPCT. Therefore, diagonals are equal.

Now,

In ΔAOB and ΔCOD,

∠BAO = ∠DCO (Alternate interior angles)

∠AOB = ∠COD (Vertically opposite)

AB = CD (Given)

Therefore, ΔAOB ≅ ΔCOD by AAS congruence condition.

Thus, AO = CO by CPCT. (Diagonal bisect each other.)

Now,

In ΔAOB and ΔCOB,

OB = OB (Given)

AO = CO (diagonals are bisected)

AB = CB (Sides of the square)

Therefore, ΔAOB ≅ ΔCOB by SSS congruence condition.

also, ∠AOB = ∠COB

∠AOB + ∠COB = 180° (Linear pair)

Thus, ∠AOB = ∠COB = 90° (Diagonals bisect each other at right angles)**5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square. Answer**

Given,

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

To prove,

Quadrilateral ABCD is a square.

Proof,

In ΔAOB and ΔCOD,

AO = CO (Diagonals bisect each other)

∠AOB = ∠COD (Vertically opposite)

OB = OD (Diagonals bisect each other)

Therefore, ΔAOB ≅ ΔCOD by SAS congruence condition.

Thus, AB = CD by CPCT. --- (i)

also,

∠OAB = ∠OCD (Alternate interior angles)

⇒ AB || CD

Now,

In ΔAOD and ΔCOD,

AO = CO (Diagonals bisect each other)

∠AOD = ∠COD (Vertically opposite)

OD = OD (Common)

Therefore, ΔAOD ≅ ΔCOD by SAS congruence condition.

Thus, AD = CD by CPCT. --- (ii)

also,

AD = BC and AD = CD

⇒ AD = BC = CD = AB --- (ii)

also, ∠ADC = ∠BCD by CPCT.

and ∠ADC + ∠BCD = 180° (co-interior angles)

⇒ 2∠ADC = 180°

⇒ ∠ADC = 90° --- (iii)

One of the interior ang is right angle.

Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.**6. Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus.**

**Answer**

(i)

In ΔADC and ΔCBA,

AD = CB (Opposite sides of a ||gm)

DC = BA (Opposite sides of a ||gm)

AC = CA (Common)

Therefore, ΔADC ≅ ΔCBA by SSS congruence condition.

Thus,

∠ACD = ∠CAB by CPCT

and ∠CAB = ∠CAD (Given)

⇒ ∠ACD = ∠BCA

Thus, AC bisects ∠C also.

(ii) ∠ACD = ∠CAD (Proved)

⇒ AD = CD (Opposite sides of equal angles of a triangle are equal)

Also, AB = BC = CD = DA (Opposite sides of a ||gm)

Thus, ABCD is a rhombus.

**7. ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.**

**Answer **

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

Proof,

AD = CD (Sides of a rhombus)

∠DAC = ∠DCA (Angles opposite of equal sides of a triangle are equal.)

also, AB || CD

⇒ ∠DAC = ∠BCA (Alternate interior angles)

⇒ ∠DCA = ∠BCA

Therefore, AC bisects ∠C.

Similarly, we can prove that diagonal AC bisects ∠A.

Also, by preceding above method we can prove that diagonal BD bisects ∠B as well as ∠D.

**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.**

(i)∠DAC = ∠DCA (AC bisects ∠A as well as ∠C)

⇒ AD = CD (Sides opposite to equal angles of a triangle are equal)

also, CD = AB (Opposite sides of a rectangle)

Therefore, AB = BC = CD = AD

Thus, ABCD is a square.

(ii) In ΔBCD,

BC = CD

⇒ ∠CDB = ∠CBD (Angles opposite to equal sides are equal)

also, ∠CDB = ∠ABD (Alternate interior angles)

⇒ ∠CBD = ∠ABD

Thus, BD bisects ∠B

Now,

∠CBD = ∠ADB

⇒ ∠CDB = ∠ADB

Thus, BD bisects ∠D

**9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). 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 condition.

(ii) AP = CQ by CPCT as ΔAPD ≅ ΔCQB.

(iii) In ΔAQB and ΔCPD,

BQ = DP (Given)

∠ABQ = ∠CDP (Alternate interior angles)

AB = BCCD (Opposite sides of a ||gm)

Thus, ΔAQB ≅ ΔCPD by SAS congruence condition.

(iv) AQ = CP by CPCT as ΔAQB ≅ ΔCPD.

(v) From (ii) and (iv), it is clear that APCQ has equal opposite sides also it has equal opposite angles. Thus, APCQ is a ||gm.

**10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see Fig. 8.21). Show that (i) ΔAPB ≅ ΔCQD (ii) AP = CQ **

**Answer**

(i) In ΔAPB and ΔCQD,

∠ABP = ∠CDQ (Alternate interior angles)

∠APB = ∠CQD (equal to right angles as AP and CQ are perpendiculars)

AB = CD (ABCD is a parallelogram)

Thus, ΔAPB ≅ ΔCQD by AAS congruence condition.

(ii) AP = CQ by CPCT as ΔAPB ≅ ΔCQD.

**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. 8.22). 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 two 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, it is a parallelogram.

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

(vi) In ΔABC and ΔDEF,

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

Thus, ΔABC ≅ ΔDEF by SSS congruence condition.

**12. ABCD is a trapezium in which AB || CD and**

**AD = BC (see Fig. 8.23). 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.

(i) CE = AD (Opposite sides of a parallelogram)

AD = BC (Given)

Therefor, BC = CE

⇒ ∠CBE = ∠CEB

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 ΔABC and ΔBAD,

AB = AB (Common)

∠DBA = ∠CBA

AD = BC (Given)

Thus, ΔABC ≅ ΔBAD by SAS congruence condition.

(iv) Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!