MAK LABS — Std 9 Maths Part 2 Chapter 5 Quadrilaterals — Practice Sets 5.1 to 5.5 & Problem Set 5 Solutions

1Diagonals of a parallelogram WXYZ intersect each other at point O. If ∠XYZ = 135° then what is the measure of ∠XWZ and ∠YZW? If l(OY) = 5 cm then l(WY) = ?

∠XWZ = ∠XYZ ...... opposite angles of a parallelogram
∴ ∠XWZ = 135°

∠YZW + ∠XYZ = 180° ...... adjacent angles of a parallelogram are supplementary
∴ ∠YZW = 180° − 135° = 45°

Diagonals of a parallelogram bisect each other.
∴ WY = 2 × OY = 2 × 5 = 10 cm

∴ ∠XWZ = 135°, ∠YZW = 45°, l(WY) = 10 cm

2In a parallelogram ABCD, if ∠A = (3x + 12)°, ∠B = (2x − 32)° then find the value of x and then find the measures of ∠C and ∠D.

∠A and ∠B are adjacent angles, so they are supplementary.

(3x + 12) + (2x − 32) = 180

5x − 20 = 180 ∴ 5x = 200 ∴ x = 40

∠A = 3(40) + 12 = 132° and ∠B = 2(40) − 32 = 48°

∠C = ∠A = 132° and ∠D = ∠B = 48° ...... opposite angles of a parallelogram

∴ x = 40, ∠C = 132°, ∠D = 48°

3Perimeter of a parallelogram is 150 cm. One of its sides is greater than the other side by 25 cm. Find the lengths of all sides.

Let the smaller side be x cm. Then the adjacent side is (x + 25) cm.

Opposite sides of a parallelogram are equal, so perimeter = 2[x + (x + 25)].

2(2x + 25) = 150 ∴ 2x + 25 = 75 ∴ 2x = 50 ∴ x = 25

x + 25 = 50

∴ The sides are 25 cm, 50 cm, 25 cm and 50 cm

4If the ratio of measures of two adjacent angles of a parallelogram is 1 : 2, find the measures of all angles of the parallelogram.

Let the adjacent angles be x° and 2x°. Adjacent angles of a parallelogram are supplementary.

x + 2x = 180 ∴ 3x = 180 ∴ x = 60

So the adjacent angles are 60° and 120°. Opposite angles are equal.

∴ The angles are 60°, 120°, 60° and 120°

5*Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO = 12 and AB = 13 then show that □ABCD is a rhombus.

Given: □ABCD is a parallelogram; diagonals meet at O; AO = 5, BO = 12, AB = 13.

To prove: □ABCD is a rhombus.

AO² + BO² = 5² + 12² = 25 + 144 = 169 and AB² = 13² = 169

∴ AO² + BO² = AB²
∴ ΔAOB is a right angled triangle and ∠AOB = 90° ...... converse of Pythagoras theorem

∴ ∠AOD = 180° − 90° = 90° ...... angles in a linear pair

In ΔAOB and ΔAOD,
seg AO ≅ seg AO ...... common side
∠AOB ≅ ∠AOD ...... each is a right angle
seg BO ≅ seg DO ...... diagonals of a parallelogram bisect each other

∴ ΔAOB ≅ ΔAOD ...... S-A-S test
∴ AB = AD ...... c.s.c.t.

In parallelogram ABCD, AB = CD and AD = BC ...... opposite sides
∴ AB = BC = CD = DA

∴ □ABCD is a rhombus.

6In the figure 5.12, □PQRS and □ABCR are two parallelograms. If ∠P = 110° then find the measures of all angles of □ABCR.

Fig. 5.12

∠R = ∠P = 110° ...... opposite angles of parallelogram PQRS

In parallelogram ABCR, ∠B = ∠R = 110° ...... opposite angles

∠A + ∠R = 180° ...... adjacent angles are supplementary
∴ ∠A = 180° − 110° = 70°

∠C = ∠A = 70° ...... opposite angles

∴ ∠A = 70°, ∠B = 110°, ∠C = 70°, ∠R = 110°

7In figure 5.13 □ABCD is a parallelogram. Point E is on the ray AB such that BE = AB then prove that line ED bisects seg BC at point F.

Fig. 5.13

Given: □ABCD is a parallelogram; A − B − E and BE = AB; seg ED meets seg BC at F.

To prove: BF = CF

AB = DC ...... opposite sides of a parallelogram
But AB = BE ...... given ∴ DC = BE ...... (I)

Side DC ∥ side AB, so seg DC ∥ seg BE.

In ΔDFC and ΔEFB,
∠FDC ≅ ∠FEB ...... alternate angles, transversal DE
seg DC ≅ seg EB ...... from (I)
∠DCF ≅ ∠EBF ...... alternate angles, transversal BC

∴ ΔDFC ≅ ΔEFB ...... A-S-A test

∴ seg CF ≅ seg BF ...... c.s.c.t.

∴ Line ED bisects seg BC at point F.

1In figure 5.22, □ABCD is a parallelogram, P and Q are midpoints of side AB and DC respectively, then prove □APCQ is a parallelogram.

Fig. 5.22

Given: □ABCD is a parallelogram; P and Q are midpoints of sides AB and DC.

To prove: □APCQ is a parallelogram.

AB = DC ...... opposite sides of a parallelogram
∴ ½AB = ½DC
∴ AP = QC ...... (I) P and Q are midpoints

Side AB ∥ side DC ...... opposite sides of a parallelogram
∴ seg AP ∥ seg QC ...... (II) A − P − B, D − Q − C

From (I) and (II), one pair of opposite sides of □APCQ is parallel and congruent.

∴ □APCQ is a parallelogram ...... test for parallelogram

2Using opposite angles test for parallelogram, prove that every rectangle is a parallelogram.

Given: □ABCD is a rectangle.

To prove: □ABCD is a parallelogram.

∠A = ∠B = ∠C = ∠D = 90° ...... angles of a rectangle

∴ ∠A ≅ ∠C and ∠B ≅ ∠D

Both pairs of opposite angles of □ABCD are congruent.

∴ □ABCD is a parallelogram ...... opposite angles test

3In figure 5.23, G is the point of concurrence of medians of ΔDEF. Take point H on ray DG such that D − G − H and DG = GH, then prove that □GEHF is a parallelogram.

Fig. 5.23 — median DK meets EF at its midpoint K

Given: G is the centroid of ΔDEF; D − G − H and DG = GH.

To prove: □GEHF is a parallelogram.

Let ray DG meet side EF at K. Then seg DK is a median, so K is the midpoint of EF.
∴ EK = KF ...... (I)

The centroid divides each median in the ratio 2 : 1.
∴ DG = 2 GK ∴ GK = ½ DG

But DG = GH ...... given ∴ GK = ½ GH
∴ K is the midpoint of seg GH, that is, GK = KH ...... (II)

Seg GH and seg EF are the diagonals of □GEHF. From (I) and (II), they bisect each other at K.

∴ □GEHF is a parallelogram ...... diagonals test

4Prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle. (Figure 5.24)

Bisectors of ∠A and ∠B meet at P, ∠B and ∠C at Q, ∠C and ∠D at R, ∠D and ∠A at S

Given: □ABCD is a parallelogram; the bisectors of its angles form □PQRS as shown.

To prove: □PQRS is a rectangle.

∠A + ∠B = 180° ...... adjacent angles of a parallelogram
∴ ½∠A + ½∠B = 90°
∴ ∠PAB + ∠PBA = 90° ...... rays AP and BP are bisectors

In ΔAPB, ∠APB = 180° − (∠PAB + ∠PBA) = 180° − 90° = 90°
∴ ∠SPQ = 90° ...... (I) ∠SPQ and ∠APB are the same angle

Similarly, from ΔBQC, ∠BQC = 90°
∴ ∠PQR = ∠BQC = 90° ...... (II) vertically opposite angles

From ΔCRD, ∠CRD = 90°
∴ ∠QRS = 90° ...... (III) ∠QRS and ∠CRD are the same angle

From ΔASD, ∠ASD = 90°
∴ ∠PSR = ∠ASD = 90° ...... (IV) vertically opposite angles

From (I) to (IV), all angles of □PQRS are right angles.

∴ □PQRS is a rectangle.

The position of P, Q, R, S depends on the shape of the parallelogram. In the figure, P and R lie near sides DC and AB respectively; the proof works the same way for any shape.

5In figure 5.25, if points P, Q, R, S are on the sides of parallelogram such that AP = BQ = CR = DS then prove that □PQRS is a parallelogram.

Fig. 5.25

Given: □ABCD is a parallelogram; AP = BQ = CR = DS.

To prove: □PQRS is a parallelogram.

AB = CD ...... opposite sides
∴ AB − AP = CD − CR ∴ PB = RD ...... (I)

AD = BC ...... opposite sides
∴ AD − DS = BC − BQ ∴ AS = CQ ...... (II)

In ΔAPS and ΔCRQ,
AP = CR ...... given
∠A ≅ ∠C ...... opposite angles of a parallelogram
AS = CQ ...... from (II)
∴ ΔAPS ≅ ΔCRQ ...... S-A-S test
∴ PS = RQ ...... c.s.c.t.

In ΔPBQ and ΔRDS,
PB = RD ...... from (I)
∠B ≅ ∠D ...... opposite angles
BQ = DS ...... given
∴ ΔPBQ ≅ ΔRDS ...... S-A-S test
∴ PQ = RS ...... c.s.c.t.

Both pairs of opposite sides of □PQRS are congruent.

∴ □PQRS is a parallelogram ...... opposite sides test

1Diagonals of a rectangle ABCD intersect at point O. If AC = 8 cm then find the length of BO and if ∠CAD = 35° then find the measure of ∠ACB.

Diagonals of a rectangle are congruent ∴ BD = AC = 8 cm

Diagonals bisect each other ∴ BO = ½ BD = ½ × 8 = 4 cm

Side AD ∥ side BC and seg AC is a transversal.
∴ ∠ACB = ∠CAD = 35° ...... alternate angles

∴ BO = 4 cm, ∠ACB = 35°

2In a rhombus PQRS if PQ = 7.5 then find the length of QR. If ∠QPS = 75° then find the measure of ∠PQR and ∠SRQ.

All sides of a rhombus are congruent ∴ QR = PQ = 7.5

∠QPS + ∠PQR = 180° ...... adjacent angles of a parallelogram
∴ ∠PQR = 180° − 75° = 105°

∠SRQ = ∠QPS = 75° ...... opposite angles

∴ QR = 7.5, ∠PQR = 105°, ∠SRQ = 75°

3Diagonals of a square IJKL intersect at point M. Find the measures of ∠IMJ, ∠JIK and ∠LJK.

Diagonals of a square are perpendicular bisectors of each other ∴ ∠IMJ = 90°

Diagonals of a square bisect its angles, and each angle of the square is 90°.
∴ ∠JIK = ½ × 90° = 45° and ∠LJK = ½ × 90° = 45°

∴ ∠IMJ = 90°, ∠JIK = 45°, ∠LJK = 45°

4Diagonals of a rhombus are 20 cm and 21 cm respectively, then find the side of rhombus and its perimeter.

Rhombus PQRS with diagonals 21 cm and 20 cm meeting at O

Diagonals of a rhombus are perpendicular bisectors of each other.
∴ PO = ½ × 21 = 10.5 cm, SO = ½ × 20 = 10 cm, ∠POS = 90°

In ΔPOS, PS² = PO² + SO² ...... Pythagoras theorem
= 10.5² + 10² = 110.25 + 100 = 210.25

∴ PS = 14.5 cm

Perimeter = 4 × 14.5 = 58 cm

∴ Side = 14.5 cm, perimeter = 58 cm

5State with reasons whether the following statements are ‘true’ or ‘false’.

Every parallelogram is a rhombus — False. A parallelogram need not have all sides equal.
Every rhombus is a rectangle — False. A rhombus need not have right angles.
Every rectangle is a parallelogram — True. Both pairs of opposite sides of a rectangle are parallel.
Every square is a rectangle — True. All angles of a square are right angles.
Every square is a rhombus — True. All sides of a square are congruent.
Every parallelogram is a rectangle — False. The angles of a parallelogram need not be right angles.

1In □IJKL, side IJ ∥ side KL, ∠I = 108°, ∠K = 53° then find the measures of ∠J and ∠L.

Side IJ ∥ side KL and side IL is a transversal.
∴ ∠I + ∠L = 180° ...... interior angles
∴ ∠L = 180° − 108° = 72°

Side JK is a transversal.
∴ ∠J + ∠K = 180° ∴ ∠J = 180° − 53° = 127°

∴ ∠J = 127°, ∠L = 72°

2In □ABCD, side BC ∥ side AD, side AB ≅ side DC. If ∠A = 72° then find the measures of ∠B and ∠D.

Side BC ∥ side AD and side AB is a transversal.
∴ ∠A + ∠B = 180° ...... interior angles
∴ ∠B = 180° − 72° = 108°

□ABCD is an isosceles trapezium (AB ≅ DC), so its base angles are congruent.
∴ ∠D = ∠A = 72°

∴ ∠B = 108°, ∠D = 72°

3In □ABCD, side BC < side AD (Figure 5.32), side BC ∥ side AD and if side BA ≅ side CD then prove that ∠ABC ≅ ∠DCB.

Fig. 5.32 with construction

Given: Side BC ∥ side AD, BC < AD, side BA ≅ side CD

To prove: ∠ABC ≅ ∠DCB

Construction: Draw a segment through C parallel to side BA, meeting side AD at T.

Seg BC ∥ seg AT ...... given
seg BA ∥ seg CT ...... construction
∴ □ABCT is a parallelogram

∴ seg BA ≅ seg CT ...... opposite sides of a parallelogram
But seg BA ≅ seg CD ...... given ∴ seg CT ≅ seg CD

In ΔCTD, ∠CTD ≅ ∠CDT ...... (I) isosceles triangle theorem

Seg BA ∥ seg CT and line AD is a transversal.
∴ ∠BAT ≅ ∠CTD ...... (II) corresponding angles

From (I) and (II), ∠BAD ≅ ∠CDA, that is, ∠A ≅ ∠D.

Side BC ∥ side AD ∴ ∠ABC + ∠A = 180° and ∠DCB + ∠D = 180° ...... interior angles

∴ ∠ABC = 180° − ∠A = 180° − ∠D = ∠DCB

∴ ∠ABC ≅ ∠DCB

1In figure 5.38, points X, Y, Z are the midpoints of side AB, side BC and side AC of ΔABC respectively. AB = 5 cm, AC = 9 cm and BC = 11 cm. Find the length of XY, YZ, XZ.

Fig. 5.38

By the midpoint theorem, the segment joining the midpoints of two sides of a triangle is half the third side.

XY = ½ AC = ½ × 9 = 4.5 cm

YZ = ½ AB = ½ × 5 = 2.5 cm

XZ = ½ BC = ½ × 11 = 5.5 cm

∴ XY = 4.5 cm, YZ = 2.5 cm, XZ = 5.5 cm

2In figure 5.39, □PQRS and □MNRL are rectangles. If point M is the midpoint of side PR then prove that, (i) SL = LR, (ii) LN = ½ SQ.

Fig. 5.39

Given: □PQRS and □MNRL are rectangles; M is the midpoint of diagonal PR.

To prove: (i) SL = LR (ii) LN = ½ SQ

(i) Seg ML ⊥ side SR and seg PS ⊥ side SR ...... angles of rectangles
∴ seg ML ∥ seg PS ...... lines perpendicular to the same line

In ΔPSR, M is the midpoint of side PR and seg ML ∥ side PS.
∴ L is the midpoint of side SR ...... converse of midpoint theorem
∴ SL = LR

(ii) Similarly, seg MN ⊥ side QR and seg PQ ⊥ side QR ∴ seg MN ∥ seg PQ.
In ΔPQR, M is the midpoint of PR and MN ∥ PQ ∴ N is the midpoint of side QR.

In ΔRSQ, L and N are the midpoints of sides RS and RQ.
∴ LN = ½ SQ ...... midpoint theorem

3In figure 5.40, ΔABC is an equilateral triangle. Points F, D and E are midpoints of side AB, side BC, side AC respectively. Show that ΔFED is an equilateral triangle.

Fig. 5.40

Given: ΔABC is equilateral; F, D, E are midpoints of AB, BC, AC.

To prove: ΔFED is equilateral.

By the midpoint theorem,
FE = ½ BC, ED = ½ AB, FD = ½ AC

But AB = BC = AC ...... sides of an equilateral triangle
∴ ½ AB = ½ BC = ½ AC

∴ FE = ED = FD

∴ ΔFED is an equilateral triangle.

4In figure 5.41, seg PD is a median of ΔPQR. Point T is the midpoint of seg PD. Produced QT intersects PR at M. Show that PM/PR = 1/3. [Hint: draw DN ∥ QM.]

Fig. 5.41

Given: Seg PD is a median of ΔPQR; T is the midpoint of PD; ray QT meets PR at M.

To prove: PM/PR = 1/3

Construction: Draw seg DN parallel to seg QM, meeting side PR at N.

In ΔPDN, T is the midpoint of side PD and seg TM ∥ seg DN ...... construction
∴ M is the midpoint of side PN ...... converse of midpoint theorem
∴ PM = MN ...... (I)

In ΔRQM, D is the midpoint of side QR and seg DN ∥ seg QM
∴ N is the midpoint of side RM ...... converse of midpoint theorem
∴ MN = NR ...... (II)

From (I) and (II), PM = MN = NR

PR = PM + MN + NR = 3 PM

∴ PM/PR = 1/3

1Choose the correct alternative answer and fill in the blanks.

(i) If all pairs of adjacent sides of a quadrilateral are congruent then it is called — (D) rhombus
If all pairs of adjacent sides are congruent, all four sides are congruent.
(ii) If the diagonal of a square is 12√2 cm then the perimeter of square is — (C) 48 cm
Side = diagonal ÷ √2 = 12 cm; perimeter = 4 × 12 = 48 cm.
(iii) If opposite angles of a rhombus are (2x)° and (3x − 40)° then value of x is — (D) 40°
Opposite angles are congruent: 2x = 3x − 40 ∴ x = 40.

2Adjacent sides of a rectangle are 7 cm and 24 cm. Find the length of its diagonal.

Each angle of a rectangle is 90°, so the diagonal is the hypotenuse of a right angled triangle with sides 7 cm and 24 cm.

Diagonal² = 7² + 24² = 49 + 576 = 625 ...... Pythagoras theorem

∴ Diagonal = 25 cm

3If diagonal of a square is 13 cm then find its side.

Let the side be x cm. Then x² + x² = 13² ...... Pythagoras theorem

2x² = 169 ∴ x² = 169/2

∴ x = 13/√2 = (13√2)/2 ≈ 9.19

∴ Side = 13/√2 cm = (13√2)/2 cm

4Ratio of two adjacent sides of a parallelogram is 3 : 4, and its perimeter is 112 cm. Find the length of its each side.

Let the adjacent sides be 3k cm and 4k cm.

Perimeter = 2(3k + 4k) = 14k = 112 ∴ k = 8

3k = 24 and 4k = 32

∴ The sides are 24 cm, 32 cm, 24 cm and 32 cm

5Diagonals PR and QS of a rhombus PQRS are 20 cm and 48 cm respectively. Find the length of side PQ.

Diagonals of a rhombus are perpendicular bisectors of each other. Let them meet at O.

PO = ½ × 20 = 10 cm, QO = ½ × 48 = 24 cm, ∠POQ = 90°

PQ² = PO² + QO² = 10² + 24² = 100 + 576 = 676 ...... Pythagoras theorem

∴ PQ = 26 cm

6Diagonals of a rectangle PQRS are intersecting in point M. If ∠QMR = 50° then find the measure of ∠MPS.

∠PMS = ∠QMR = 50° ...... vertically opposite angles

Diagonals of a rectangle are congruent and bisect each other.
∴ PM = ½ PR = ½ QS = SM

In ΔPMS, PM = SM ∴ ∠MPS = ∠MSP ...... isosceles triangle theorem

∠MPS + ∠MSP + ∠PMS = 180°
2∠MPS + 50° = 180° ∴ ∠MPS = 65°

∴ ∠MPS = 65°

7In the adjacent Figure 5.42, if seg AB ∥ seg PQ, seg AB ≅ seg PQ, seg AC ∥ seg PR, seg AC ≅ seg PR then prove that seg BC ∥ seg QR and seg BC ≅ seg QR.

Fig. 5.42

Given: seg AB ∥ seg PQ, seg AB ≅ seg PQ, seg AC ∥ seg PR, seg AC ≅ seg PR

To prove: seg BC ∥ seg QR and seg BC ≅ seg QR

In □ABQP, seg AB ∥ seg PQ and seg AB ≅ seg PQ ...... given
∴ □ABQP is a parallelogram ...... test for parallelogram
∴ seg AP ∥ seg BQ and seg AP ≅ seg BQ ...... (I)

In □ACRP, seg AC ∥ seg PR and seg AC ≅ seg PR ...... given
∴ □ACRP is a parallelogram
∴ seg AP ∥ seg CR and seg AP ≅ seg CR ...... (II)

From (I) and (II), seg BQ ∥ seg CR and seg BQ ≅ seg CR

∴ □BQRC is a parallelogram ...... test for parallelogram

∴ seg BC ∥ seg QR and seg BC ≅ seg QR ...... opposite sides of a parallelogram

8*In the Figure 5.43, □ABCD is a trapezium. AB ∥ DC. Points P and Q are midpoints of seg AD and seg BC respectively. Then prove that, PQ ∥ AB and PQ = ½(AB + DC).

Fig. 5.43 with construction

Given: □ABCD is a trapezium, AB ∥ DC; P and Q are midpoints of AD and BC.

To prove: PQ ∥ AB and PQ = ½(AB + DC)

Construction: Draw ray AQ to meet ray DC at point E.

In ΔABQ and ΔECQ,
∠ABQ ≅ ∠ECQ ...... alternate angles, AB ∥ DE, transversal BC
seg BQ ≅ seg CQ ...... Q is the midpoint of BC
∠AQB ≅ ∠EQC ...... vertically opposite angles

∴ ΔABQ ≅ ΔECQ ...... A-S-A test
∴ AQ = QE and AB = CE ...... c.s.c.t.

In ΔADE, P is the midpoint of AD and Q is the midpoint of AE.
∴ PQ ∥ DE and PQ = ½ DE ...... midpoint theorem

PQ ∥ DE, that is, PQ ∥ DC; and DC ∥ AB ∴ PQ ∥ AB

PQ = ½ DE = ½ (DC + CE) = ½ (DC + AB) ...... CE = AB

∴ PQ = ½(AB + DC)

9In the adjacent figure 5.44, □ABCD is a trapezium. AB ∥ DC. Points M and N are midpoints of diagonal AC and DB respectively then prove that MN ∥ AB.

Fig. 5.44 with construction

Given: □ABCD is a trapezium, AB ∥ DC; M and N are midpoints of diagonals AC and DB.

To prove: MN ∥ AB

Construction: Draw ray DM to meet side AB at point E.

In ΔDMC and ΔEMA,
∠DCM ≅ ∠EAM ...... alternate angles, DC ∥ AB, transversal AC
seg CM ≅ seg AM ...... M is the midpoint of AC
∠DMC ≅ ∠EMA ...... vertically opposite angles

∴ ΔDMC ≅ ΔEMA ...... A-S-A test
∴ DM = ME ...... c.s.c.t.

In ΔDEB, M is the midpoint of DE and N is the midpoint of DB.
∴ MN ∥ EB ...... midpoint theorem

E lies on AB, so MN ∥ AB.

5. Quadrilaterals

Practice Set 5.1

Practice Set 5.2

Practice Set 5.3

Practice Set 5.4

Practice Set 5.5

Problem Set 5