# Class IX, NCERT (CBSE) Mathematics | Quadrilaterals, Introduction (Chapter 8.1)

Class IX, Mathematics (Quadrilaterals)
NCERT (CBSE) Textbook Exercise 8.1 Solution
(Page 146 - 147)
Question 1: The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral.
Ans:
Let us assume that the common ratio between the angles of quadrilateral be x. Therefore, the angles will be 3x, 5x, 9x, and 13x respectively.
As the sum of all interior angles of a quadrilateral is 360Âº, we can write
3x + 5x + 9x + 13x = 360Âº
or, 30x = 360Âº
or, x = 12Âº
Hence, the angles of the quadrilateral are
3x = 3 × 12 = 36Âº
5x = 5 × 12 = 60Âº
9x = 9 × 12 = 108Âº
13x = 13 × 12 = 156Âº
Question 2: If the diagonals of a parallelogram are equal, then show that it is a rectangle.
Ans:
Let ABCD be a parallelogram. in order to show that ABCD is a rectangle, we have to prove that one of its interior angles of the ABCD is 90Âº.
Considering Î”ABC and Î”DCB,
AB = DC (Opposite sides of a parallelogram are equal)
BC = BC (Common) and AC = DB (Given)
∴ Î”ABC ≅ Î”DCB (By SSS Congruence rule)
∠ABC = ∠DCB (as the sum of the measures of angles on the same side of transversal is 180Âº).
or, ∠ABC + ∠DCB = 180Âº (AB || CD)
or, ∠ABC + ∠ABC = 180Âº
or, 2∠ABC = 180Âº
or, ∠ABC = 90Âº
Since ABCD is a parallelogram and one of its interior angles is 90Âº so 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.
Ans:
Let us consider ABCD a quadrilateral, whose diagonals AC and BD bisect each other at right angle i.e., OA = OC, OB = OD, and
∠AOB = ∠BOC = ∠COD = ∠AOD = 90Âº.
For ABCD to be a rhombus, we have to prove that ABCD is a parallelogram and all the sides of ABCD are equal.
Now considering Î”AOD and Î”COD,
OA = OC (Diagonals bisect each other)
∠AOD = ∠COD (Given)
OD = OD (Common)
Î”AOD ≅ Î”COD (By SAS congruence rule)
∴ AD = CD . . . . . . (1)
In the same way, it can be proved that
AD = AB and CD = BC . . . . . . (2)
From equations (1) and (2), we get
AB = BC = CD = AD
Since opposite sides of quadrilateral ABCD are equal, so ABCD is a parallelogram. Also since all sides of parallelogram ABCD are equal, so ABCD is a rhombus.
Question 4: Show that the diagonals of a square are equal and bisect each other at right angles.
Ans:
Let us consider a square ABCD whose diagonals AC and BD intersect each other at a point O. To prove that the diagonals of a square are equal and bisect each other at right angles, we have to prove AC = BD, OA = OC, OB = OD, and ∠AOB = 90Âº.
Considering Î”ABC and Î”DCB,
AB = DC (Sides of a square are equal)
∠ABC = ∠DCB (All interior angles are of 90 ) BC = CB (Common side)
∴ Î”ABC ≅ Î”DCB (By SAS congruency)
∴ AC = DB (By CPCT)
Hence, the diagonals of the square ABCD are equal in length.
Now considering Î”AOB and Î”COD,
∠AOB = ∠COD (Vertically opposite angles)
∠ABO = ∠CDO (Alternate interior angles)
AB = CD (Sides of a square are equal)
∴ Î”AOB ≅ Î”COD (By AAS congruence rule)
∴ AO = CO and OB = OD (By CPCT)
Hence, the diagonals of ABCD (square) bisect each other.
Similarly in Î”AOB and Î”COB,
As we had proved that diagonals bisect each other, therefore,
AO = CO; AB = CB (Sides of a square are equal); BO = BO (Common)
∴ Î”AOB ≅ Î”COB (By SSS congruency)
∴ ∠AOB = ∠COB (By CPCT)
∠AOB + ∠COB = 180Âº (Linear pair)
or, 2∠AOB = 180Âº
or, ∠AOB = 90Âº
Hence proved that the diagonals of a square bisect each other at right angles.
Question 5: Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Ans:
Consider a quadrilateral ABCD whose diagonals AC and BD intersect each other at O. It is given that the diagonals of ABCD are equal and bisect each other at right angles. Therefore, AC = BD, OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90Âº. To prove ABCD is a square, we have to show that ABCD is a parallelogram where AB = BC = CD = AD, and one of its interior angles is 90Âº.
Consider Î”AOB and Î”COD,
AO = CO (Diagonals bisect each other)
OB = OD (Diagonals bisect each other)
∠AOB = ∠COD (Vertically opposite angles)
∴ Î”AOB ≅ Î”COD (SAS congruence rule)
∴ AB = CD (By CPCT)  . . . . . . . . . . . . (1)
Also, ∠OAB = ∠OCD (By CPCT)
However, these angles are alternate interior angles of  AB and CD and alternate interior angles are equal to each other only when the two lines are parallel.
∴ AB || CD  . . . . . . . . . . . . . . . (2)
From equations (1) and (2), we can say that
ABCD is a parallelogram.
Consider Î”AOD and Î”COD,
AO = CO (Diagonals bisect each other)
∠AOD = ∠COD (Given that each is 90Âº)
OD = OD (Common)
∴ Î”AOD ≅ Î”COD (SAS congruence rule)
∴ AD = DC  . . . . . . . . . . . . . (3)
But, AD = BC and AB = CD (Opposite sides of parallelogram ABCD)
∴ AB = BC = CD = DA
Therefore, all the sides of quadrilateral ABCD are equal to each other.
Consider Î”ADC and Î”BCD,
AC = BD (Given)
DC = CD (Common)
∴ Î”ADC ≅ Î”BCD (SSS Congruence rule)
∴ ∠ADC = ∠BCD (By CPCT)
However, ∠ADC + ∠BCD = 180° (Co-interior angles)
or, 2∠ADC = 180°
or, ∠ADC = 90°
One of the interior angles of quadrilateral ABCD is a right angle.
Therefore ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angles is 90Âº. Hence, ABCD is a square.
Question 6: Diagonal AC of a parallelogram ABCD bisects ∠A (see the given figure). Show that
(i) It bisects ∠C also,
(ii) ABCD is a rhombus.
Ans:
(i) ABCD is a parallelogram.
DAC = BCA (Alternate interior angles)  . . . . . . (1)
And, BAC = DCA (Alternate interior angles)  . . . . . . . . (2)
Given that AC bisects A.
so, DAC = BAC  . . . . . . . . (3)
From the above equations (1), (2), and (3), we get
DAC = BCA = BAC = DCA  . . . . . . . . . . (4)
or,  DCA = BCA
Hence, AC bisects C.
(ii)
From equation (4), we obtain
DAC = DCA
or, DA = DC (Side opposite to equal angles are equal)
DA = BC and AB = CD (Opposite sides of a parallelogram)
AB = BC = CD = DA
Hence, ABCD is a rhombus.
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.
Ans:
Let us join AC.
In Î”ABC,
BC = AB (as sides of a rhombus are equal to each other)
1 = 2 (Angles opposite to equal sides of a triangle are equal)
However, 1 = 3 (Alternate interior angles for parallel lines AB and CD)
or, 2 = 3
Therefore, AC bisects C.
2 = 4 (Alternate interior angles for || lines BC and DA)
or, 1 = 4
Therefore, AC bisects A.
In the same way it can be proved that BD bisects B and D as well.
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.
Ans:

(i) As per the given conditions ABCD is a rectangle. whose ∠A = ∠C

CD = DA (Sides opposite to equal angles are also equal)
However, DA = BC and AB = CD (Opposite sides of a rectangle are equal)
AB = BC = CD = DA
Therefore, ABCD is a rectangle whose all sides are equal.
Hence, ABCD is a square.
(ii) Let us join BD and in Î”BCD,
BC = CD (Sides of a square are equal to each other)
CDB = CBD (Angles opposite to equal sides are equal)
However, CDB = ABD (Alternate interior angles for AB || CD)
CBD = ABD
or, BD bisects B.
Also, CBD = ADB (Alternate interior angles for BC || AD)
CDB = ABD
Hence, BD bisects D.
Question 9: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see the given figure). Show that:
(i) Î”APD ≅ Î”CQB
(ii) AP = CQ
(iii) Î”AQB Î”CPD
(iv) AQ = CP
(v) APCQ is a parallelogram

Ans:
(i) Consider Î”APD and Î”CQB,
ADP = CBQ (Alternate interior angles for BC || AD)
AD = CB (Opposite sides of parallelogram ABCD)
DP = BQ (Given)
Î”APD Î”CQB (Using SAS congruence rule)
(ii) As we had observed that Î”APD Î”CQB,
AP = CQ (CPCT)
(iii) Consider Î”AQB and Î”CPD,
ABQ = CDP (Alternate interior angles for AB || CD)
AB = CD (Opposite sides of parallelogram ABCD)
BQ = DP (Given)
Therefore, Î”AQB Î”CPD (Using SAS congruence rule)
(iv) As we had observed that Î”AQB Î”CPD,
AQ = CP (CPCT)
(v) From the results obtained in (ii) and (iv) above,
AQ = CP and
AP = CQ
Since opposite sides in quadrilateral APCQ are equal to each other so, APCQ is a parallelogram.
Question 10: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD. Show that
(i) Î”APB Î”CQD
(ii) AP = CQ
Ans:
(i) In Î”APB and Î”CQD,
APB = CQD (Each 90°)
AB = CD (Opposite sides of parallelogram ABCD)
ABP = CDQ (Alternate interior angles for AB || CD)
Î”APB Î”CQD (By AAS congruency)
(ii) By using the above result
Î”APB Î”CQD, we obtain
AP = CQ (By CPCT)
Question 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 the given figure). 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 ≅ Î”DEF.
Ans:
(i) Given that AB = DE and AB || DE.
If two opposite sides of a quadrilateral are equal and parallel to each other, then it will be a parallelogram.
Therefore, quadrilateral ABED is a parallelogram.
(ii) Also, BC = EF and BC || EF
Therefore, quadrilateral BCEF is a parallelogram.
(iii) As ABED and BEFC are parallelograms, therefore
AD = BE and AD || BE
(Opposite sides of a parallelogram are equal and parallel)
And, BE = CF and BE || CF
(Opposite sides of a parallelogram are equal and parallel)
AD = CF and AD || CF
(iv) We had already observed that one pair of opposite sides (AD and CF) of quadrilateral ACFD are equal and parallel to each other, therefore, it is a parallelogram.
(v) As ACFD is a parallelogram, therefore, the pair of opposite sides will be equal and parallel to each other.
i.e. AC || DF and AC = DF
(vi) Consider Î”ABC and Î”DEF,
AB = DE (Given)
BC = EF (Given)
AC = DF (ACFD is a parallelogram)
Î”ABC Î”DEF (By SSS congruence rule)
Question 12: ABCD is a trapezium in which AB || CD and AD = BC (see the given figure). Show that

(i) A = B (ii) C = D
(iii) Î”ABC Î”BAD (iv) diagonal AC = diagonal BD
Ans:
First extend AB. Now draw a line through C, which is parallel to AD, intersecting AE at point E. It is clear that AECD is a parallelogram.
(i) AD = CE (Opposite sides of parallelogram AECD)
However, AD = BC (Given)
Therefore, BC = CE
CEB = CBE (Angle opposite to equal sides are also equal)
Considering the parallel lines AD and CE then AE which is a  transversal line for them.
We have,
A + CEB = 180Âº (Angles on the same side of transversal)
A + CBE = 180Âº (Using the relationCEB = CBE)  . . . . . . . (1)
However, B + CBE = 180Âº (Linear pair angles)  . . . . . . . . (2)
From the equations (1) and (2), we obtain
A = B
(ii) AB || CD
A + D = 180Âº (Angles on the same side of the transversal)
Also, C + B = 180° (Angles on the same side of the transversal)
so, ∠A + D = C + B
However, A = B [Using the result obtained in (i)]
or, C = D
(iii) Consider Î”ABC and Î”BAD,
AB = BA (Common side) BC = AD (Given)
B = A (already proved )
Î”ABC Î”BAD (by SAS congruence rule)
Therefore, AC = BD (By CPCT)

1. plzz give the solutions of chapter9 area of parallelogram amd area of triagle.........plzzzzzzz

2. This site is the best site i have ever seen ..... its wonderful that the site provide us the information and we can avoid the teachers in liu Thanks

3. This site is very wonderful site as its provides the information and avoids the student to go the the useless teachers in liu

4. please give solution of chapter 9

5. Very helpfulll and easy 2 understand you saved my day.. i had to complete these questions and show to me tuition teacher.. dont think that i straight away started copyin it ..i first tried to understand it then did it myself.. cheers !

6. plzzzzzzzzzzzzzzzz post ex.8.2 of maths of std.9...plzzz hv to submit on 22nd..please plzzzzzzzzzzzzzzzzzzzzzzzzz

7. Dis site is little bit helpful bt certain required materials r nt given....so plz add full information...hopfully wish dat u will listen dis request....!

8. We thankfully acknowledge the overwhelming response and appreciation that we are getting from our readers everyday. Rest assure, you will get full information, answers of all type of questions, sample papers, and everything in this site and at http://extrascore.blogspot.com FREE of COST!!
Keep visiting and happy searching.

9. Please Put 8.2 exercise also please i need it very urgently :)..=P

10. Sir,I am not getting answer for all the chapter so pls tell me how i can answer for chapter

11. thankyou very much this was sooo usefulll to me , thanks a lot :)

12. very good site

13. cool but like this students can copy everything.... :I

14. it was help full...........
thanks!!!!!!

15. Fantastic I like it lam happy but plzz... post solution for 8.2 from this chapter

1. This comment has been removed by a blog administrator.

16. I tried to visit your blog but i am not able to read it as I am not invited for the same what next?

17. BORINGGGGGG................ ;-(