# Quadrilaterals

Chapters

- ap-scert-9th-class-Maths-Lesson-1-Real-Numbers
- ap-scert-9th-class-Maths-Lesson-2-Polynomials-and-Factorisation
- ap-scert-9th-class-Maths-Lesson-3-The-Elements-of-Geometry
- ap-scert-9th-class-Maths-Lesson-4-Lines-and-Angles
- ap-scert-9th-class-Maths-Lesson-5-Co-Ordinate-Geometry
- ap-scert-9th-class-Maths-Lesson-6-Linear-Equation-in-Two-Variables
- ap-scert-9th-class-Maths-Lesson-7-Triangles
- ap-scert-9th-class-Maths-Lesson-8-Quadrilaterals
- ap-scert-9th-class-Maths-Lesson-9-Statistics
- ap-scert-9th-class-Maths-Lesson-10-Surface-Areas-and-Volumes
- ap-scert-9th-class-Maths-Lesson-11-Areas
- ap-scert-9th-class-Maths-Lesson-12-Circles
- ap-scert-9th-class-Maths-Lesson-13-Geometrical-Constructions
- ap-scert-9th-class-Maths-Lesson-14-Probability
- ap-scert-9th-class-Maths-Lesson-15-Proofs-in-Mathematics

- ap-scert-9th-class-Maths-Lesson-1-Real-Numbers
- ap-scert-9th-class-Maths-Lesson-2-Polynomials-and-Factorisation
- ap-scert-9th-class-Maths-Lesson-3-The-Elements-of-Geometry
- ap-scert-9th-class-Maths-Lesson-4-Lines-and-Angles
- ap-scert-9th-class-Maths-Lesson-5-Co-Ordinate-Geometry
- ap-scert-9th-class-Maths-Lesson-6-Linear-Equation-in-Two-Variables
- ap-scert-9th-class-Maths-Lesson-7-Triangles
- ap-scert-9th-class-Maths-Lesson-8-Quadrilaterals
- ap-scert-9th-class-Maths-Lesson-9-Statistics
- ap-scert-9th-class-Maths-Lesson-10-Surface-Areas-and-Volumes
- ap-scert-9th-class-Maths-Lesson-11-Areas
- ap-scert-9th-class-Maths-Lesson-12-Circles
- ap-scert-9th-class-Maths-Lesson-13-Geometrical-Constructions
- ap-scert-9th-class-Maths-Lesson-14-Probability
- ap-scert-9th-class-Maths-Lesson-15-Proofs-in-Mathematics

- Solutions
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Solutions

State whether the statements are true or false.

- i) Every parallelogram is a trapezium.
- ii) All parallelograms are quadrilaterals.
- iii) All trapeziums are parallelograms.
- iv) A square is a rhombus.
- v) Every rhombus is a square.
- vi) All parallelograms are rectangles.

Solution:

- i) True
- ii) True
- iii) False
- iv) True
- v) False
- vi) False

Complete the following table by writing YES if the property holds for the particular quadrilateral and NO if property does not holds.

Solution

ABCD is a trapezium in which AB ‖ CD. If AD = BC, show that ∠A = ∠B and ∠C = ∠D.

Solution:

Given that in □ABCD AB ‖ CD; AD = BC

Mark a point ‘E’ on AB such DC = AE.

Join E, C.

Now in AECD quadrilateral

AE // DC and AE = DC

∴ □AECD is a parallelogram.

∴ AD//EC

∠DAE = ∠CEB (corresponding angles) ……………..(1)

In ΔCEB; CE = CB (∵ CE = AD)

∴ ∠CEB = ∠CBE (angles opp. to equal sides) …………….. (2)

From (1) & (2)

∠DAE = ∠CBE

⇒ ∠A = ∠B

Also ∠D = ∠AEC (∵ Opp. angles of a parallelogram)

= ∠ECB + ∠CBE [ ∵ ∠AEC is ext. angle of ΔBCE] |

= ∠ECB + ∠CEB [ ∵∠CBE = ∠CEB]

= ∠ECB + ∠ECD [∵ ∠ECD = ∠CEB alt. int. angles]

= ∠BCD = ∠C

∴ ∠C = ∠D

The four angles of a quadrilateral are in the ratio of 1 : 2 : 3 : 4. Find the measure of each angle of the quadri-lateral.

Solution:

Given that, the ratio of angles of a quad-rilateral = 1 : 2 : 3 : 4

Sum of the terms of the ratio

= 1 +2 + 3 + 4= 10

Sum of the four interior angles of a quadrilateral = 360°

∴ The measure of first angle

ABCD is a rectangle, AC is diagonal. Find the angles of ΔACD. Give reasons.

Solution:

Given that □ABCD is a rectangle;

AC is its diagonal.

In ΔACD; ∠D = 90° [ ∵ ∠D is also angle of the rectangle]

∠A + ∠C = 90° [ ∵ ∠D = 90° ⇒ ∠A + ∠C = 180°-90° = 90°]

(i.e,,) ∠D right angle and

∠A, ∠C are complementary angles.

In the given figure ABCD is a parallelogram. ABEF is a rectangle. Show that ΔAFD ≅ ΔBEC

Solution:

Given that □ABCD is a parallelogram.

□ABEF is a rectangle.

In ΔAFD and ΔBEC

AF = BE ( ∵ opp. sides of rectangle □ABEF)

AD = BC (∵ opp. sides of //gm □ABCD)

DF = CE (∵ AB = DC = DE + EC , AB = EF = DE + DF)

∴ ΔAFD ≅ ΔBEC (SSS congruence)

Show that the diagonals of a rhombus divide it into four congruent triangles.

Solution:

□ABCD is a rhombus.

Let AC and BD meet at O.

In ΔAOB and ΔCOD

∠OAB = ∠OCD (alt.int. angles)

AB = CD (def. of rhombus)

∠OBA = ∠ODC ........(1) (alt. int. angles)

∴ ΔAOB ≅ ΔCOD (ASA congruence)

Thus AO = OC (CPCT)

Also ΔAOD ≅ ΔCOD ......(2)

[ ∵ AO = OC; AD = CD; OD = OD SSS congruence]

Similarly we can prove

ΔAOD ≅ ΔCOB ...... (3)

From (1), (2) and (3) we have

ΔAOB ≅ ΔBOC ≅ ΔCOD ≅ ΔAOD

∴ Diagonals of a rhombus divide it into four congruent triangles.

In a quadrilateral ABCD, the bisector of ∠C and ∠D intersect at O. Prove that ∠COD = 1/2 (∠A + ∠B) .

(OR)

In a quadrilateral ABCD, the bisectors of ∠A and ∠B are intersects at ‘O’ then prove that ∠AOB = 1/2 (∠C + ∠D)

Solution:

In a quadrilateral □ABCD

∠A + ∠B + ∠C + ∠D = 360°

(angle sum property)

∠C + ∠D = 360° - (∠A + ∠B)

1/2 (∠C + ∠D) = 180 - 1/2 (∠A + ∠B) ...... (1)

(∵ dividing both sides by 2) .

But in ΔCOD

1/2∠C + 1/2 ∠D + ∠COD = 180°

1/2∠C + 1/2 ∠D = 180° - ∠COD

∴12(∠C +∠D) = 180° -∠COD.....(2)

From (1) and (2);

180° - ∠COD = 180° - 1/2 (∠A + ∠B)

∴ ∠COD = 1/2 (∠A + ∠B)

Hence proved.

The opposite angles of a parallelogram are (3x - 2)° and (x + 48)°. Find the measure of each angle of the parallelogram.

Solution:

Given that the opposite angles of a parallelogram are (3x - 2)° and (x + 48)°

Thus 3x – 2 = x + 48

(∵ opp. angles of a //gm are equal)

3x - x = 48 + 2

2x = 50

x = 50/2 = 25°

∴ The given angles are (3 x 25 - 2)° and (25 + 48) °

= (75 - 2)° and 73° = 73° and 73°

We know the consecutive angles are supplementary.

∴ The other two angles are (180°-73°) and (180°-73°)

= 107° and 107°

∴ The four angles are 73°, 107°, 73° and 107°.

Find the measure of all the angles of a parallelogram, if one angle is 24° less than the twice of the smallest angle.

Solution:

Let the smallest angle = x

Then its consecutive angle = 180 - x°

By problem (180 - x)° = (2x- 24)°

(∵ opp. angles are equal)

180 + 24 = 2x + x

3x = 204

x =204/3 = 68°

∴ The angles are

68°; (2 x 68 - 24)°; 68°; (2 x 68 - 24)°

= 68°, 112°, 68°, 112°

In the given figure ABCD is a paral-lelogram and E is the mid point of the side BC. If DE and AB are produced to meet at F, show that AF = 2AB.

Solution

Given that □ABCD is a parallelogram.

E is the midpoint of BC.

Let G be the midpoint of AD.

Join G, E.

Now in ΔAFD, GE is the line joining the midpoints G, E of two sides AD and FD.

∴GE // AF and GE = 1/2 AF

But GE = AB [ ∵ ABEG is a parallelo¬gram and AB, GE forms a pair of opp. sides]

12 = AB ⇒ AF = 2AB

Hence Proved.

In the given figure ABCD is a paral¬lelogram. P, Q are the midpoints of sides AB and DC

Solution

□ABCD is a parallelogram.

P, Q are the mid points of AB and CD.

Join Q, P.

Now AB = CD (Opp. sides of a //gm)

12AB = 12CD

PB = QC

Also PB // QC.

Now in □PBCQ;

PB = QC; PB//QC

Hence □PBCQ is a parallelogram.

ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle QAC and CD//BA as shown in the figure. Show that i) ∠DAC = ∠BCA

ii) ABCD is a parallelogram.

Solution:

Given that AABC is isosceles; AB = AC

AD is bisector of ∠QAC

i) In ΔABC, AB = AC ⇒ ∠B = ∠ACB

(angles opp. to equal sides)

Also ∠QAC = ∠B + ∠ACB

∠QAC = ∠BCA + ∠BCA

(∵∠BCA = ∠B)

⇒ 1/2∠QAC = 1/2 [2 ∠BCA]

⇒ ∠DAC = ∠BCA [ ∵ AD is bisector of ∠QAC]

ii) From (i) ∠DAC = ∠BCA

But these forms a pair of alt. int. angles for the pair of lines AD and BC; AC as a transversal.

∴ AD//BC

In □ABCD ; AB // DC; BC // AD

□ABCD is a parallelogram.

ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C on diagonal BD (see figure). Show that 1) ΔAPB ≅ ΔCQD ii) AP = CQ.

Solution:

Given that □ABCD is a parallelogram.

BD is a diagonal.

AP ⊥ BD and CQ ⊥ BD

i) In ΔAPB and ΔCQD

AB = CD ( ∵ Opp. sides of //gm ABCD)

∠APB = ∠CQD (each 90°)

∠PBA = ∠QDC (alt. int. angles for the lines AB and DC)

∴ ΔAPB ≅ ΔCQD (AAS congruence)

ii) From (1) ΔAPB ≅ ΔCQD

⇒ AP = CQ (CPCT)

In Δs ABC and Δs DEF, AB = DC and 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) ABED is a parallelogram
- ii) BCFE is a parallelogram
- iii) AC = DF
- iv) ΔABC = ΔDEF

Solution

Given that in ΔABC and ΔDEF

AB = DE and AB // DE

BC = EF and BC//EF.

i) In □ABED AB//ED and AB = ED

Hence □ABED is a parallelogram.

ii) In □BCFE; BC = EF and BC//EF

Hence □BCFE is a parallelogram.

iii) ACFD is a parallelogram (In a paral-lelogram opposite sides are equal).

So, AC = DF.

iv) Consider ΔABC = ΔDEF

AB = DE (given);

AC = DF (proved)

BC = EF (given)

∴ ΔABC ≅ ΔDEF (SSS congruency rule).

ABCD is a parallelogram. AC and BD are the diagonals intersect at ‘O’. P and Q are the points of trisection of the diagonal BD. Prove that CQ//AP and also AC bisects PQ.

Solution:

Given □ABCD is a parallelogram;

BD is a diagonal.

P, Q are the points of trisection of BD.

In ΔAPB and ΔCQD

AB = CD (... Opp. sides of //gm ABCD)

BP = DQ (given)

∠ABP = ∠CDQ (alt. int. angles for the lines AB//DC, BD as a transversal)

ΔAPB = ΔCQD (SAS congruence)

Similarly in ΔAQD and ΔCPB

AD = BC (opp. sides of //gm ABCD)

DQ = BP (given)

∠ADQ = ∠CBP (all int. angles for the lines AD//BC, BD as a transversal)

ΔAQD ≅ ΔCPB

Now in □APCQ

AP = CQ (CPCT of AAPB, ACQD)

AQ = CP (CPCT of AAQD and ACPB)

∴ □APCQ is a parallelogram.

∴ CQ//AP (opp. sides of//gm APCQ)

Also AC bisects PQ. [ ∵ diagonals of //gm APCQ]

ABCD is a square. E, F, G and H are the mid points of AB, BC, CD and DA respectively. Such that AE = BF = CG = DH. Prove that EFGH is a square.

Solution:

Given that ABCD is a square.

E, F, G, H are the mid points of AB, BC, CD and DA.

Also AE = BF = CG = DH

In ΔABC; E, F are the mid points of sides AB and BC.

∴ EF//AC and EF = 1/2 AC

Similarly GH//AC and GH = AC

GF//BD and GF = 1/2 BD

HE//BD and HE = 1/2 BD

But AC = BD (∵ diagonals of a square)

∴ EF = FG = GH = HE

Hence EFGH is a rhombus.

Also AC ⊥ BD

(∵ diagonals of a rhombus)

∴ In //gm OIEJ [ ∵ 0I // EJ; IE // OJ]

We have ∠IOJ = ∠E

[ ∵ Opp. angles of a //gm]

∴ ∠E - 90°

Hence in quad. EFGH; all sides are equal and one angle is 90°.

∴ EFGH is a square.

ABC is a triangle. D is a point on AB such that AD =1/4 AB and E is a point on AC such that AE =1/4 AC. If DE = 2 cm find BC.

Solution:

Given that D and E are points on AB and AC.

Such that AD = 1/4 AB and AE = 1/4 AC

Let X, Y be midpoints of AB and AC.

Joint D, E and X, Y.

Now in ΔAXY; D, E are the midpoints of sides AX and AY.

∴ DE // XY and DE = 1/2 XY

⇒ 2 cm = 1/2 XY

⇒ XY = 2 x 2 = 4cm

Also in ΔABC; X, Y are the midpoints of AB and AC.

∴ XY//BC and XY = 1/2 BC

4 cm = 1/2 BC

⇒ BC = 4 x 2 = 8 cm

ABCD is a quadrilateral. E, F, G and H are the midpoints of AB, BC, CD and DA respectively. Prove that EFGH is a parallelogram.

Solution:

Given that E, F, G and H are the midpoints of the sides of quad. ABCD.

In ΔABC; E, F are the midpoints of the sides AB and BC.

∴ EF//AC and EF = 1/2 AC

Also in ΔACD; HG // AC

and HG = 1/2 AC

∴ EF // HG and EF = HG

Now in □EFGH; EF = HG and EF // HG

∴ □EFGH is a parallelogram.

Show that the figure formed by joining the midpoints of sides of a rhom¬bus successively is a rectangle.

Solution:

Let □ABCD be a rhombus.

P, Q, R and S be the midpoints of sides of □ABCD

In ΔABC,

P, Q are the midpoints of AB and BC.

∴ PQ//AC and PQ = 1/2 AC .........(1)

Also in ΔADC, ,

S, R are the midpoints of AD and CD.

∴ SR//AC and SR = 1/2 AC ......(2)

From (1) and (2);

PQ // SR and PQ = SR

Similarly QR // PS and QR = PS

∴ □PQRS is a parallelogram.

As the diagonals of a rhombus bisect at right angles.

∠AOB - 90°

∴ ∠P = ∠AOB = 90°

[opp. angles of //gm PYOX] Hence □PQRS is a rectangle as both pairs of opp. sides are equal and parallel, one angle being 90°.

In a parallelogram ABCD, E and F are the midpoints of the sides AB and DC respectively. Show that the line segments AF and EC trisect the diagonal BD.

Solution:

□ABCD is a parallelogram. E and F are the mid points of AB and CD.

∴ AE = 12AB and CF = 12CD

Thus AE = CF [∵ AB - CD]

Now in □AECF, AE = CF and AE ‖CF

Thus □AECF is a parallelogram.

Now in ΔEQB and ΔFDP

EB = FD [Half of equal sides of a //gm]

∠EBQ = ∠FDP[alt. int.angles of EB//FD]

∠QEB = ∠PFD

[∵∠QED = ∠QCF = ∠PFD]

∴ ΔEQB ≅ ΔFPD [A.S.A. congruence]

∴ BQ = DP [CPCT] ...... (1)

Now in ΔDQC; PF // QC and F is the midpoint of DC.

Hence P must be the midpoint of DQ

Thus DP = PQ ....... (2)

From (1) and (2), DP = PQ = QB

Hence AF and CE trisect the diagonal BD.

Show that the line segments joining the mid points of the opposite sides of a quadrilateral and bisect each other.

Solution:

Let ABCD be a quadrilateral.

P, Q, R, S are the midpoints of sides of □ABCD.

Join (P, Q), (Q, R), (R, S) and (S, P).

In ΔABC; P, Q are the midpoints of AB and BC.

∴ PQ // AC and PQ = 1/2 AC .....(1)

Also from ΔADC

S, R are the midpoints of AD and CD

SR // AC and SR = 1/2 AC .......(2)

∴ From (1) & (2)

PQ = SR and PQ //SR

∴ □PQRS is a parallelogram.

Now PR and QS are the diagonals of □ PQRS.

∴ PR and QS bisect each other.

ABC is a triangle right angled at’C’. A line through the midpoint M of hypotenuse AB and parallel to BC intersects AC at D. Show that

- i) D is the midpoint of AC
- ii) MD ⊥ AC
- iii) CM = MA=1/2 AB

Solution

Given that in ΔABC; ∠C = 90°

M is the midpoint of AB.

i) If ‘D’ is the midpoints of AC.

The proof is trivial.

Let us suppose D is not the mid point of AC.

Then there exists D’ such that AD’ = D’C

Then D’M is a line parallel to BC through M.

Also DM is a line parallel to BC through M.

There exist two lines parallel to same line through a point M.

This is a contradiction.

There exists only one line parallel to a given line through a point not on the line.

∴ D’ must coincides with D

∴ D is the midpoint of AC

ii) From (i) DM // BC

Thus ∠ADM = ∠ACB = 90°

[corresponding angles]

⇒ MD ⊥ AC

iii) In ΔADM and ΔCDM

AD = CD [ ∵ D is midpoint from (i)]

∠ADM = ∠MDC (∵ 90° each)

DM = DM (Common side)

∴ ∠ADM = ∠CDM (SAS congruence)

⇒ CM = MA (CPCT)

CM = 1/2 AB (∵ M is the midpoint of AB)

∴ CM = MA = 1/2 AB

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