## Quadrilaterals

**Chapter 8**

**Exercise 8.1**

**EX 8.1 QUESTION 1.**

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

**Solution:**

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

∴ 3x + 5x + 9x + 13x = 360° [Angle sum property of a quadrilateral]

⇒ 30x = 360°

⇒ x = 360°/30 = 12°

∴First angle = 3x = 3 x 12° = 36°

Second angle = 5x = 5 x 12° = 60°

Third angle = 9x = 9 x 12° = 108°

Forth angle = 13x = 13 x 12° = 156°

**EX 8.1 QUESTION 2.**

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

**Solution:**

In ∆ABC and ∆DCB,

AC = DB [Given]

AB = DC [Opposite sides of a parallelogram are equal]

BC = CB [Common]

∴ ∆ABC ≅ ∆DCB [By SSS congruency]

⇒ ∠ABC = ∠DCB [By C.P.C.T.] …(1)

Now, AB || DC and BC is a transversal. [ ∵ ABCD is a parallelogram]

∴ ∠ABC + ∠DCB = 180° … (2) [Co-interior angles]

From (1) and (2), we have

∠ABC = ∠DCB = 90°

i.e., ABCD is a parallelogram having an angle equal to 90°.

∴ ABCD is a rectangle.

**EX 8.1 QUESTION 3.**

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

**Solution:**

∴ In ∆AOB and ∆AOD, we have

AO = AO [Common]

OB = OD [O is the mid-point of BD]

∠AOB = ∠AOD [Each 90]

∴ ∆AQB ≅ ∆AOD [By,SAS congruency

∴ AB = AD [By C.P.C.T.] ……..(1)

Similarly, AB = BC .. .(2)

BC = CD …..(3)

CD = DA ……(4)

∴ From (1), (2), (3) and (4), we have

AB = BC = CD = DA

Thus, the quadrilateral ABCD is a rhombus.

Alternatively: ABCD can be proved first a parallelogram then proving one pair of adjacent sides equal will result in rhombus.

**EX 8.1 QUESTION 4.**

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

**Solution:**

In ∆ABC and ∆BAD,

AB = BA [Common]

BC = AD [Sides of a square ABCD]

∠ABC = ∠BAD [Each angle is 90°]

∴ ∆ABC ≅ ∆BAD [By SAS congruency]

AC = BD [ C.P.C.T.]

(ii)In ∆AOB and ∆COD

∴ ∠OAB = ∠OCD [Alternate interior angles are equal]

AB = CD [Opposite side of a square]

∠OAB = ∠OCD [Alternate interior angles are equal]

∆ABC ≅ ∆BAD [ASA congruency]

AO = OC , BO = OD [C.P.C.T.]

(iii) In ∆AOB and ∆AOD,

OB = OD [Proved]

AB = AD [Sides of a square ABCD]

OA = OA [Common]

∴ ∆BAD ≅ ∆ABC [By SSS congruency]

⇒ ∠AOB = ∠AOD [By C.P.C.T.]

∠AOB + ∠AOD = 180° [ linear pair.]

⇒2∠AOB = 180°

⇒ ∠AOB = 180°/2 = 90°

The diagonals of a square are equal and bisect each other at right angles.

**EX 8.1 QUESTION 5.**

**Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.**

**Solution:**

Now, in ∆BAD and ∆ABC,

AD = BC [Each 90°]

BD = AC [Common]

∴ ∆BAD ≅ ∆ABC [SAS congruency]

⇒ ∠BAD = ∠ABC [ C.P.C.T.] …(1)

∠BAD + ∠ABC = 180° [Co – interior angles]

∠ABC = 180°/2 = 90°

**EX 8.1 QUESTION 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.**

**Solution:**

(i) ABCD is a parallelogram.

∠DAC = ∠BAC ———(i)

∴ ∠DAC = ∠BCA ——(ii) [Alternate interior angles are equal]

∠BAC = ∠ACD ——–(iii) [Alternate interior angles are equal]

From equation (I),(ii) and (iii)

∠ACD = ∠BCA———(iv)

⇒ AC bisects ∠C.

(ii) From equation (ii) and (iv)

∠ACD = ∠DAC

In ∆ADC,

∠ACD = ∠DAC

AD = DC [Angles opposite to equal sides of a triangle are equal]

Parallelogram whose adjacent sides are equal is a rhombus. Hence, ABCD is a rhombus.

**EX 8.1 QUESTION 7.**

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

**Solution:**

ABCD is a rhombus.

⇒ AB = BC = CD = DA

Also, AB || CD and AD || BC

Now, CD = AD ⇒ ∠1 = ∠2 …….(1) [ ∵ Angles opposite to equal sides of a triangle are equal]

Also, AD || BC and AC is the transversal. [ ∵ Every rhombus is a parallelogram]

⇒ ∠1 = ∠3 …(2) [ ∵ Alternate interior angles are equal]

From (1) and (2),

∠2 = ∠3 …(3)

Since AB || DC and AC is transversal.

∴ ∠2 = ∠4 …(4) [ ∵ Alternate interior angles are equal] From (1) and (4),

∠1 = ∠4

∴ AC bisects ∠C as well as ∠A.

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

**EX 8.1 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.**

**Solution:**

**EX 8.1 QUESTION 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**

**Solution:**

(i) In ∆APD and ∆CQB

DP = BQ

∴ ∠ADB = ∠CBD [Alternate interior angles are equal]

AD = BC [Opposite sides of a parallelogram are equal]

∴ ∆APD ≅ ∆CQB [SAS congruency]

(ii) ∆APD ≅ ∆CQB [Proved]

⇒ AP = CQ [C.P.C.T.]

(iii) In ∆AQB and ∆CPD,

QB = DP [Given]

∠ABQ = ∠CDP [Alternate interior angles are equal]

AB = CD [ Opposite sides of a parallelogram are equal]

∴ ∆AQB = ∆CPD [SAS congruency]

(iv) ∆AQB = ∆CPD [Proved]

⇒ AQ = CP [ C.P.C.T.]

(v) In a quadrilateral ∆PCQ,

Opposite sides are equal. [Proved]

∴ ∆PCQ is a parallelogram.

**EX 8.1 QUESTION 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**

**Solution:**

(i) In ∆APB and ∆CQD,

∠APB = ∠CQD [Each 90°]

AB = CD [ ∵ Opposite sides of a parallelogram are equal]

∠ABP = ∠CDQ [Alternate angles are equal ]

∴ ∆APB = ∆CQD [AAS congruency]

(ii) ∆APB ≅ ∆CQD [Proved]

⇒ AP = CQ [ C.P.C.T.]

**EX 8.1 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. 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.**

**Solution:**

(i) In ABED

AB = DE [Given]

AB || DE [Given]

∴ ABED is a parallelogram.

(ii)In BEFC BC = EF [Given]

BC = EF [Given]

BC || EF [Given]

∴ BEFC is a parallelogram.

(iii) In ABED

∴ AD || BE and AD = BE …(i)[Opposite sides of a parallelogram are equal and parallel]

In BEFC

BE || CF and BE = CF …(ii) [Opposite sides of a parallelogram are equal and parallel]

From (i) and (ii), we have

AD || CF and AD = CF

(iv) In ACFD

AD || CF and AD = CF [Proved]

∴Quadrilateral ACFD is a parallelogram.

(v)In ACFD [Proved]

AC =DF [Opposite sides of a parallelogram are equal]

(vi) In ∆ABC and ∆DFF, we have

AB = DE [Given]

BC = EF [Given]

AC = DE [Proved]

∆ABC ≅ ∆DFF [SSS congruency]

**EX 8.1 QUESTION 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**

**diagonal AC = diagonal BD**

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

**Solution:**

(i) Produce AB to E and draw CF || AD.. .(1)

∵ AB || DC

⇒ AE || DC Also AD || CE

∴ AECD is a parallelogram.

⇒ AD = CE …(1) [Opposite sides of the parallelogram are equal]

AD = BC …(2) [Given]

BC = CE [From equation (1) and (2)]

Now, in ∆BCE, we have BC = CF

⇒ ∠CEB = ∠CBE …(3) [Angles opposite to equal sides of a triangle are equal]

∠ABC + ∠CBE = 180° … (4) [Linear pair]

∠A + ∠CEB = 180° …(5) [Co-interior angles of a parallelogram ADCE]

From (4) and (5),

∠ABC + ∠CBE = ∠A + ∠CEB

⇒ ∠ABC = ∠A [From (3)]

⇒ ∠B = ∠A …(6)

(ii) ABCD is a trapezium in which AB || CD

∴ ∠A + ∠D = 180° …(7) [Co-interior angles]

∠B + ∠C = 180° … (8)

From (7) and (8), we get

∠A + ∠D = ∠B + ∠C

⇒ ∠C = ∠D [From (6)]

(iii) In ∆ABC and ∆BAD, we have

AB = BA [Common]

BC = AD [Given]

∠ABC = ∠BAD [Proved]

∴ ∆ABC = ∆BAD [By SAS congruency]

(iv) Since, ∆ABC = ∆BAD [Proved]

⇒ AC = BD [By C.P.C.T.]