CBSE Class 9 Mathematics

NCERT Solutions
CHAPTER 8

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

Ans. Let in quadrilateral ABCD, $\mathrm{\angle }A=3x,\mathrm{\angle }B=5x,\mathrm{\angle }C=9x,\mathrm{\angle }D=13x$

Since, sum of all the angles of a quadrilateral = 360o

$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C+\mathrm{\angle }D={360}^{o}$

Now $\mathrm{\angle }A=3x=3×{12}^{o}={36}^{o}$
$\mathrm{\angle }B=5x=5×{12}^{o}={60}^{o}$
$\mathrm{\angle }C=9x=9×{12}^{o}={108}^{o}$
And $\mathrm{\angle }D=13x=13×{12}^{o}={156}^{o}$
Hence angles of given quadrilateral are ${36}^{o},{60}^{o},{108}^{o}$and ${156}^{o}$

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

Ans. Given: ABCD is a parallelogram with diagonal AC = diagonal BD
To prove: ABCD is a rectangle.

Proo
f: In triangles ABC and ABD,

AB = AB [ Common ]

AC = BD [ Given ]

AD = BC [ Opposite sides of a $\parallel gm$ ]

$\mathrm{\Delta }ABC\cong \mathrm{\Delta }BAD$ [ By SSS congruency ]

$\mathrm{\angle }DAB=\mathrm{\angle }CBA$ [ By C.P.C.T.] ……….(i)

But $\mathrm{\angle }DAB+\mathrm{\angle }CBA={180}^{o}$

[ The sum of consecutve angles of a parallelogram is$={180}^{o}$] ……….(ii)

From eq. (i) and (ii),

$\mathrm{\angle }DAB=\mathrm{\angle }CBA={90}^{o}$

Simillarly, the other two angles are of ${90}^{o}$ each.

Hence ABCD is a rectangle.

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

Ans. Let ABCD is a quadrilateral.

Given:
The diagonals AC and BD bisect each other at right angle at point O.

OA = OC, OB = OD

And $\mathrm{\angle }AOB=\mathrm{\angle }BOC=\mathrm{\angle }COD=\mathrm{\angle }DOA={90}^{o}$

To prove: ABCD is a rhombus, i.e.we have to prove that AB = BC = CD = DA

Proof: In $\mathrm{\Delta }DOA$ and $\mathrm{\Delta }BOC$,

OA = OC [ Given ]

$\mathrm{\angle }AOD=\mathrm{\angle }BOC$ [ Given ]

OB = OD [ Given ]

$\mathrm{\Delta }AOD\cong \mathrm{\Delta }COB$ [ By SAS congruency ]

AD = BC [ By C.P.C.T. ] ……….(i)

Again, In $\mathrm{\Delta }AOB$ and $\mathrm{\Delta }COD$,

OA = OC [ Given ]

$\mathrm{\angle }AOB=\mathrm{\angle }COD$ [ Given ]

OB = OD [ Given ]

$\mathrm{\Delta }AOB\cong \mathrm{\Delta }COD$ [ By SAS congruency ]

AB = CD [ By C.P.C.T. ] ……….(ii)

Now In $\mathrm{\Delta }AOB$ and $\mathrm{\Delta }BOC$,

OA = OC [ Given ]

$\mathrm{\angle }AOB=\mathrm{\angle }BOC$ [ Given ]

OB = OB [ Common ]

AOB COB [ By SAS congruency ]

AB = BC [ By C.P.C.T. ] ……….(iii)

From eq. (i), (ii) and (iii),

AD = BC = CD = AB

And the diagonals of quadrilateral ABCD bisect each other at right angle.

Therefore, ABCD is a rhombus.

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

Ans. Given: ABCD is a square. AC and BD are its diagonals intersect each other at point O.

To prove:(I) AC = BD

(ii) OA = OC and OB = OD

(iii) AC  BD at point O.

Proof: In triangles ABC and BAD,

AB = AB [ Common ]

ABC = BAD =  [ Because ABCD is a square ]

BC = AD [ Sides of a square ]

ABC BAD [ By SAS congruency ]

AC = BD [ By C.P.C.T. ]

Thus, the diagonals of a square are equal. Hence proved (i).

Again in AOB and COD,

$\mathrm{\angle }ABO$ = $\mathrm{\angle }CDO$ [ Alternate angles ]

$\mathrm{\angle }AOB=\mathrm{\angle }COD$ [ Vertically opposite angles ]

AB = CD [ Sides of a square ]

$\mathrm{\Delta }AOB\cong \mathrm{\Delta }COD$ [By AAS congruency ]

OA = OC AND OB = OD [ By C.P.C.T. ]

Thus, diagonals of a square bisect each other. Hence proved (ii).

Now in triangles AOB and AOD,

AO = AO [ Common ]

AB = AD [ Sides of a square ]

OB = OD [ Diagonals of a square bisect each other ]

AOB AOD [ By SSS congruency ]

AOB = AOD [ By C.P.C.T. ]

But AOB + AOD = [ Linear pair ]

AOB = AOD =

OA  BD or AC  BD. Hence proved (iii).

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

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

Given: AC = BD ……….(i)

OA = OC ...........(ii)

And OB = OD ……….(iii)

To Prove: ABCD is a square. i.e all the sides are equal and all the angles are of ${90}^{o}$ each.

Proof: Now OA + OC = OB + OD

OC + OC = OB + OB [ Using (i), (ii) & (iii) ]

2OC = 2OB

OC = OB ……….(iv)

From eq. (i), (ii), (iii) and (iv), we get, OA = OB = OC = OD ……….(v)

Now in AOB and COD,

OA = OD [ proved ]

AOB = COD [ Vertically opposite angles ]

OB = OC [ proved ]

AOB COD [ By SAS congruency ]

AB = CD [ By C.P.C.T. ] ……….(vi)

Similarly, BOC AOD [ By SAS congruency ]

BC = AD [ By C.P.C.T. ] ……….(vii)

From eq. (vi) and (vii), it is concluded that ABCD is a parallelogram because opposite sides of a quadrilateral are equal.

AB = BA [ Common ]

BC = AD [ proved above ]

AC = BD [ Given ]

ABC BAD [ By SSS congruency ]

ABC = BAD [ By C.P.C.T. ] ……….(viii)

But ABC + BAD = [ ABCD is a parallelogram ] ……….(ix)

ABC + ABC =  [ Using eq. (viii) and (ix) ]

2ABC = ABC =

$\mathrm{\angle }ABC=\mathrm{\angle }BAD=\phantom{\rule{thickmathspace}{0ex}}{90}^{o}$ ……….(x)

But ABC = ADC [ Opposite angles of a parallelogram are equal. ]

From eq. (xi) and (xii), we get

Now in AOB and BOC,

OA = OC [ Given ]

AOB = BOC =  [ Given ]

OB = OB [ Common ]

AOB COB [ By SAS congruency ]

AB = BC ……….(xiv)

From eq. (vi), (vii) and (xiv), we get,

AB = BC = CD = AD ……….(xv)

Now, from eq. (xiii) and (xv), we have a quadrilateral whose sides are equal make an angle of  with each other.

ABCD is a square. Hence Proved

6. Diagonal AC of a parallelogram ABCD bisects A (See figure). Show that:

(i) It bisects C also.

(ii) ABCD is a rhombus.

Ans. Diagonal AC bisects A of the parallelogram ABCD.

(i) Since AB  DC and AC intersects them.

1 = 3 [ Alternate angles ] ……….(i)

Similarly 2 = 4 ……….(ii)

But 1 = 2 [ Given ] ……….(iii)

3 = [ Using eq. (i), (ii) and (iii) ]

Thus AC bisects C.

(ii) 2 = 3 = 4 = 1 [ Proved above ]

Thus, 2 = 3

AD = CD [ Sides opposite to equal angles ] ..............(iv)

But as ABCD is a parallelogram, therfore AB = CD and BC = AD

So, we get AB = CD = AD = BC [ Using (iv) ]

Hence ABCD is a rhombus.

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

Ans. ABCD is a rhombus. Therefore, AB = BC = CD = AD

Let O be the point of intersection of diagonals.

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$\mathrm{\angle }1=\mathrm{\angle }3$ [ Angles opposite to equal sides are equal ]

$\mathrm{\angle }1=\mathrm{\angle }4$ [ Alternate interior angles ]

$\mathrm{\angle }3=\mathrm{\angle }4$

Thus, diagonal AC bisects $\mathrm{\angle }C$

Also, $\mathrm{\angle }3=\mathrm{\angle }2$ [ Alternate interior angles ]

So, we get ,

$\mathrm{\angle }1=\mathrm{\angle }2$

Thus AC bisects $\mathrm{\angle }A$ also.

Similarly,

$\mathrm{\angle }6=\mathrm{\angle }8$ [ Angles opposite to equal sides are equal ]

$\mathrm{\angle }6=\mathrm{\angle }7$ [ Alternate interior angles ]

$\mathrm{\angle }7=\mathrm{\angle }8$

Thus, diagonal BD bisects $\mathrm{\angle }B$ .

Again, $\mathrm{\angle }8=\mathrm{\angle }5$ [ Alternate interior angles ]

So, we get ,

$\mathrm{\angle }6=\mathrm{\angle }5$

Thus, BD bisects $\mathrm{\angle }D$ also.

Hence Proved.

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 both B as well as D.

Ans. ABCD is a rectangle.

Therefore AB = DC ……….(i)

Also A = B = C = D =

1 = 2 and 3 = 4

[AC bisects A and C (given)]

AC = AC [Common]

ABC ADC [ By ASA congruency ]

From eq. (i) and (ii), AB = BC = CD = AD

Hence ABCD is a square.

(ii) In ABD and CBD

AB = BC [ Since ABCD is a square ]

AD = DC [ Since ABCD is a square ]

BD = BD [ Common ]

ABD CBD [ By SSS congruency ]

ABD = CBD [ By C.P.C.T. ] ……….(iii)

And ADB = CDB [ By C.P.C.T. ] ……….(iv)

From eq. (iii) and (iv), it is clear that diagonal BD bisects both B and D.

Hence, Proved.

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

(i) APD CQB

(ii) AP = CQ

(iii) AQB CPD

(iv) AQ = CP

(v) APCQ is a parallelogram.

Ans. (i) In APD and CQB,

DP = BQ [ Given ]

ADP = QBC [ Alternate angles (ADBC and BD is transversal) ]

AD = CB [ Opposite sides of parallelogram ]

APD CQB [ By SAS congruency ]

(ii) Since APD CQB

AP = CQ [ By C.P.C.T. ]

(iii) In AQB and CPD,

BQ = DP [ Given]

ABQ = PDC [ Alternate angles (ABCD and BD is transversal)]

AB = CD [ Opposite sides of parallelogram]

AQB CPD [ By SAS congruency]

(iv) Since AQB CPD

AQ = CP [ By C.P.C.T.]

AP = CQ [proved in part (i)]

AQ = CP [proved in part (iv)]

Since opposite sides of quadrilateral APCQ are equal.

Hence APCQ is a parallelogram.

10. ABCD is a parallelogram and AP and CQ are the perpendiculars from vertices A and C on its diagonal BD (See figure). Show that:

(i) APB CQD

(ii) AP = CQ

Ans. Given: ABCD is a parallelogram. AP  BD and CQ  BD

To prove: (i) APB CQD (ii) AP = CQ

Proof: (i) In APB and CQD,

1 = 2 [ Alternate interior angles ]

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

APB = CQD =

APB CQD [ By AAS Congruency ]

(ii) Since APB CQD

AP = CQ [ By C.P.C.T. ]

11. In ABC and 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 figure). Show that:

(i) Quadrilateral ABED is a parallelogram.

(ii) Quadrilateral BEFC is a parallelogram.

(iv) Quadrilateral ACFD is a parallelogram.

(v) AC = DF

(vi) ABC DEF

Ans. (i) AB = DE [ Given ]

And AB  DE [ Given ]

ABED is a parallelogram.

(ii) BC = EF [ Given ]

And BC  EF [ Given ]

BEFC is a parallelogram.

(iii) As ABED is a parallelogram.

Also BEFC is a parallelogram.

CF  BE and CF = BE ……….(ii)

From (i) and (ii), we get

ACFD is a parallelogram.

(v) As ACFD is a parallelogram.

AC = DF

(vi) In ABC and DEF,

AB = DE [ Given ]

BC = EF [ Given ]

AC = DF [ Proved in (iii) ]

ABC  DEF [ By SSS congruency ]

12. ABCD is a trapezium in which AB  CD and AD = BC (See figure). Show that:

(i) A = B

(ii) C = D

(iv) Diagonal AC = Diagonal BD

Ans. Given: ABCD is a trapezium.

AB  CD and AD = BC

To prove
: (i) A = B

(ii) C = D

(iv) Diagonal AC = Diagonal BD

Construction: Draw CE  AD and extend AB to intersect CE at E.

Proof(i) As AECD is a parallelogram.

[ By construction, CE is parallel to AD and AE is parallel to CD ]

BC = EC

3 = 4 [ Angles opposite to equal sides are equal ]

Now 1 + 4 =  [ Consecutive Interior angles of a parallelogram ]

And 2 + 3 =  [ Linear pair ]

1 + 4 = 2 + 3

1 = 2 [ 3 = 4, so gets cancelled with each other ] ............... (i)

A = B

(ii) 3 = C [ Alternate interior angles ]

And D = 4 [ Opposite angles of a parallelogram ]

But 3 = 4 [ BCE is an isosceles triangle ]

C = D

AB = AB [ Common ]

1 = 2 [ Proved , see eqn (i) ]

AD = BC [ Given ]

ABC BAD [ By SAS congruency ]