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**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 = 360Â°

Now,

3x + 5x + 9x + 13x = 360Â°

â‡’ 30x = 360Â°

â‡’ x = 12Â°

Angles of the quadrilateral are:

3x = 3Ã—12Â° = 36Â°

5x = 5Ã—12Â° = 60Â°

9x = 9Ã—12Â° = 108Â°

13x = 13Ã—12Â° = 156Â°**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 Î”ABC and Î”BAD,

BC = BA (Common)

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

AC = BD (Given)

Therefore, Î”ABC â‰… Î”BAD by SSS congruence condition.

âˆ A = âˆ B (by CPCT)

also,

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

â‡’ 2âˆ A = 180Â°

â‡’ âˆ A = 90Â°

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 âˆ AOB = âˆ BOC = âˆ OCD = âˆ ODA = 90Â°

To show,

ABCD is parallelogram and AB = BC = CD = AD

Proof,

In Î”AOB and Î”COB,

OA = OC (Given)

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

OB = OB (Common)

Therefore, Î”AOB â‰… Î”COB 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 âˆ AOB = 90Â°

Proof,

In Î”ABC and Î”BAD,

BC = BA (Common)

âˆ ABC = âˆ BAD = 90Â°

AC = AD (Given)

Therefore, Î”ABC â‰… Î”BAD by SAS congruence condition.

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

Now,

In Î”AOB and Î”COD,

âˆ BAO = âˆ DCO (Alternate interior angles)

âˆ AOB = âˆ COD (Vertically opposite)

AB = CD (Given)

Therefore, Î”AOB â‰… Î”COD by AAS congruence condition.

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

Now,

In Î”AOB and Î”COB,

OB = OB (Given)

AO = CO (diagonals are bisected)

AB = CB (Sides of the square)

Therefore, Î”AOB â‰… Î”COB by SSS congruence condition.

also, âˆ AOB = âˆ COB

âˆ AOB + âˆ COB = 180Â° (Linear pair)

Thus, âˆ AOB = âˆ COB = 90Â° (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 Î”AOB and Î”COD,

AO = CO (Diagonals bisect each other)

âˆ AOB = âˆ COD (Vertically opposite)

OB = OD (Diagonals bisect each other)

Therefore, Î”AOB â‰… Î”COD by SAS congruence condition.

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

also,

âˆ OAB = âˆ OCD (Alternate interior angles)

â‡’ AB || CD

Now,

In Î”AOD and Î”COD,

AO = CO (Diagonals bisect each other)

âˆ AOD = âˆ COD (Vertically opposite)

OD = OD (Common)

Therefore, Î”AOD â‰… Î”COD by SAS congruence condition.

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

also,

AD = BC and AD = CD

â‡’ AD = BC = CD = AB --- (ii)

also, âˆ ADC = âˆ BCD by CPCT.

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

â‡’ 2âˆ ADC = 180Â°

â‡’ âˆ ADC = 90Â° --- (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 âˆ A (see Fig. 8.19). Show that (i) it bisects âˆ C also, (ii) ABCD is a rhombus.**

**Answer**

(i)

In Î”ADC and Î”CBA,

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

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

AC = CA (Common)

Therefore, Î”ADC â‰… Î”CBA by SSS congruence condition.

Thus,

âˆ ACD = âˆ CAB by CPCT

and âˆ CAB = âˆ CAD (Given)

â‡’ âˆ ACD = âˆ BCA

Thus, AC bisects âˆ C also.

(ii) âˆ ACD = âˆ CAD (Proved)

â‡’ 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 âˆ A as well as âˆ C and diagonal BD bisects âˆ B as well as âˆ D.**

**Answer **

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

Proof,

AD = CD (Sides of a rhombus)

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

also, AB || CD

â‡’ âˆ DAC = âˆ BCA (Alternate interior angles)

â‡’ âˆ DCA = âˆ BCA

Therefore, AC bisects âˆ C.

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

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

**8. ABCD is a rectangle in which diagonal AC bisects âˆ A as well as âˆ C. Show that: (i) ABCD is a square (ii) diagonal BD bisects âˆ B as well as âˆ D.**

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

â‡’ 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 Î”BCD,

BC = CD

â‡’ âˆ CDB = âˆ CBD (Angles opposite to equal sides are equal)

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

â‡’ âˆ CBD = âˆ ABD

Thus, BD bisects âˆ B

Now,

âˆ CBD = âˆ ADB

â‡’ âˆ CDB = âˆ ADB

Thus, BD bisects âˆ 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) Î”APD â‰… Î”CQB (ii) AP = CQ (iii) Î”AQB â‰… Î”CPD (iv) AQ = CP (v) APCQ is a parallelogram **

**Answer**

(i) In Î”APD and Î”CQB,

DP = BQ (Given)

âˆ ADP = âˆ CBQ (Alternate interior angles)

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

Thus, Î”APD â‰… Î”CQB by SAS congruence condition.

(ii) AP = CQ by CPCT as Î”APD â‰… Î”CQB.

(iii) In Î”AQB and Î”CPD,

BQ = DP (Given)

âˆ ABQ = âˆ CDP (Alternate interior angles)

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

Thus, Î”AQB â‰… Î”CPD by SAS congruence condition.

(iv) AQ = CP by CPCT as Î”AQB â‰… Î”CPD.

(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) Î”APB â‰… Î”CQD (ii) AP = CQ **

**Answer**

(i) In Î”APB and Î”CQD,

âˆ ABP = âˆ CDQ (Alternate interior angles)

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

AB = CD (ABCD is a parallelogram)

Thus, Î”APB â‰… Î”CQD by AAS congruence condition.

(ii) AP = CQ by CPCT as Î”APB â‰… Î”CQD.

**11. In Î”ABC and Î”DEF, 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) Î”ABC â‰… Î”DEF. **

**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.

â‡’ 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 Î”ABC and Î”DEF,

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

Thus, Î”ABC â‰… Î”DEF by SSS congruence condition.

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

**AD = BC (see Fig. 8.23). Show that (i) âˆ A = âˆ B (ii) âˆ C = âˆ D (iii) Î”ABC â‰… Î”BAD (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

â‡’ âˆ CBE = âˆ CEB

also,

âˆ A + âˆ CBE = 180Â° (Angles on the same side of transversal and âˆ CBE = âˆ CEB)

âˆ B + âˆ CBE = 180Â° (Linear pair)

â‡’ âˆ A = âˆ B

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

â‡’ âˆ A + âˆ D = âˆ A + âˆ C (âˆ A = âˆ B)

â‡’ âˆ D = âˆ C

(iii) In Î”ABC and Î”BAD,

AB = AB (Common)

âˆ DBA = âˆ CBA

AD = BC (Given)

Thus, Î”ABC â‰… Î”BAD by SAS congruence condition.

(iv) Diagonal AC = diagonal BD by CPCT as Î”ABC â‰… Î”BA.