Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 5 Geometry Ex 5.4 Text Book Back Questions and Answers, Notes.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 5 Geometry Ex 5.4

I. Construct the following quadrilaterals with the given measurements and also find their area.

Question 1.
ABCD, AB = 5 cm, BC = 4.5 cm, CD = 3.8 cm, DA = 4.4 cm and AC = 6.2 cm.
Answer:
Given ABCD, AB = 5 cm, BC = 4.5 cm, CD = 3.8 cm, DA = 4.4 cm and AC = 6.2 cm.
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.4 1
Steps:

  • Draw a line segment AB = 5 cm
  • With A and B as centers drawn arcs of radii 6.2 cm and 4.5cm respectively and let them cut at C.
  • Joined AC and BC.
  • With A and C as centrers drawn arcs of radii 4.4cm and 3.8 cm respectively and let them at D.
  • Joined AD and CD.
  • ABCD is the required quadrilateral.

Calculation of Area:
Area of the quadrilateral ABCD = \(\frac { 1 }{ 2 }\) × d × (h11 + h2) sq. units
= \(\frac { 1 }{ 2 }\) × 6.2 × (2.6 + 3.6)cm2 = 3.1 × 6.2 = 19.22 cm2

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.4

Question 2.
PLAY, PL = 7cm, LA = 6cm, AY = 6cm, PA = 8cm and LY = 7cm.
Answer:
Given PLAY, PL = 7cm, LA = 6cm, AY = 6cm, PA = 8cm and LY = 7cm
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.4 2
Steps:

  • Drawn a line segment PL = 7 cm
  • With P and L as centers, drawn arcs of radii 8 cm and 6 cm respectively, let them cut at A.
  • Joined PA and LA.
  • With L and A as centers, drawn arcs of radii 7 cm and 6 cm respectively and let them cut at Y.
  • Joined LY, PY and AY.
  • PLAY is the required quadrilateral.

Calculation of Area:
Area of the quadrilateral PLAY = \(\frac { 1 }{ 2 }\) × d × (h1 + h2) sq. units = \(\frac { 1 }{ 2 }\) × 8 × (5.1 + 1.4) cm2
\(\frac { 1 }{ 2 }\) × 8 × 6.5 cm2 = 26 cm2
Area of the quadrilateral = 26 cm2

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.4

Question 3.
PQRS, PQ = QR = 3.5 cm, RS = 5.2 cm, SP = 5.3 cm and ∠Q = 120°.
Answer:
Given PQ = QR = 3.5 cm, RS = 5.2 cm, SP = 5.3 cm and ∠Q = 120°
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.4 3
Steps:

  • Draw a line segment PQ = 3.5 cm
  • Made ∠Q = 120°. Drawn the ray QX.
  • With Q as centre drawn an arc of radius 3.5 cm. Let it cut the ray QX at R.
  • With R and P as centres drawn arcs of radii 5.2cm and 5.5 cm respectively and let them cut at S.
  • Joined PS and RS.
  • PQRS is the required quadrilateral.

Calculation of Area:
Area of the quadrilateral PQRS = \(\frac { 1 }{ 2 }\) × d × (h1 + h2) sq. units
= \(\frac { 1 }{ 2 }\) × 6 × (4.3 + 17)cm2
= 3 × 6cm2
= 18 cm2
Area of the quadrilateral PQRS = 18 cm2

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.4

Question 4.
MIND, MI = 3.6 cm, ND = 4 cm, MD = 4 cm, ∠M = 50° and ∠D = 100°.
Answer:
Given MI = 3.6 cm, ND = 4 cm, MD = 4 cm, ∠M = 50° and ∠D = 100°
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.4 4

Steps:

  • Drawn a line segment MI = 3.6 cm
  • At M on MI made an angle ∠IMX = 500
  • Drawn an arc with center M and radius 4 cm let it cut MX it D
  • At D on DM made an angle ∠MDY = 100°
  • With I as center drawn an arc of radius 4 cm, let it cut DY at N.
  • Joined DN and IN.
  • MIND is the required quadrilateral.

Calculation of Area:
Area of the quadrilateral MIND = \(\frac { 1 }{ 2 }\) × d × (h1 + h2) sq. units = \(\frac { 1 }{ 2 }\) × 3.2 × (2.7 + 33) cm2
= \(\frac { 1 }{ 2 }\) × 3.2 × 6 cm2 = 9.6 cm2
Area of the quadrilateral = 9.6 cm2

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.4

Question 5.
AGRI, AG = 4.5 cm, GR = 3.8 cm, ∠A = 60°, ∠G = 110° and ∠R = 90°.
Answer:
AG = 4.5 cm, GR = 3.8 cm, ∠A = 60°, ∠G = 110° and ∠R = 90°.
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.4 5
Steps:

  • Draw a line segment AG = 4.5 cm
  • At G on AG made ∠AGX = 110°
  • With G as centre drawn an arc of radius 3.8 cm let it cut GX at R.
  • At R on GR made ∠GRZ = 90°
  • At A on AG made ∠GAY = 90°
  • AY and RZ meet at I.
  • AGRI is the required quadrilateral.

Calculation of Area:
Area of the quadrilateral AGRI = \(\frac { 1 }{ 2 }\) × d × (h1 + h2) sq. units = × 6.8 × (2.9 + 2.4) cm2
= \(\frac { 1 }{ 2 }\) × 6.8 × 5.3 × cm2
Area of the quadrilateral = 18.02 cm2

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.4

II. Construct the following trapeziums with the given measures and also find their area.

Question 1.
AIMS with \(\overline{\mathrm{AI}}\) || \(\overline{\mathrm{SM}}\), AI = 6cm, IM = 5cm, AM = 9cm and MS = 6.5cm.
Answer:
Given AI = 6 cm, IM = 5 cm
AM = 9 cm, and \(\overline{\mathrm{AI}}\) || \(\overline{\mathrm{SM}}\)
MS = 6.5 cm

Rough Diagram
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.4 6

Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.4 7
Construction:
Steps:

  • Draw a line segment AI = 6cm.
  • With A and I as centres, draw arcs of radii 9 cm and 5 cm respectively and let them cut at M
  • Join AM and IM.
  • Draw MX parallel to AI
  • With M as centre, draw an arc of radius 6.5 cm cutting MX at S.
  • Join AS AIMS is the required trapezium.

Calculation of Area
Area of the trapezium AIMS = \(\frac { 1 }{ 2 }\) × h × (a + b) sq.units
= \(\frac { 1 }{ 2 }\) × 4.6 × (6 + 6.5)
= \(\frac { 1 }{ 2 }\) × 4.6 × 12.5
= 28.75 Sq.cm

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.4

Question 2.
CUTE with \(\overline{\mathrm{CD}}\) || \(\overline{\mathrm{ET}}\), CU = 7cm, ∠ UCE = 80° CE = 6cm and TE = 5cm.
Answer:
Given: In the trapezium CUTE,
CU = 7cm, ∠UCE = 80°,
CE = 6cm,TE = 5cm and \(\overline{\mathrm{CD}}\) || \(\overline{\mathrm{ET}}\)

Rough Diagram
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.4 8

Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.4 9
Construction:
Steps:

  • Draw a line segment CU = 7cm.
  • Construct an angle ∠UCE = 80° at C
  • With C as centre, draw an arc of radius 6 cm cutting CY at E
  • Draw EX parallel to CU
  • With E as centre, draw an arc 7 cm of radius 5 cm cutting EX at T
  • Join UT. CUTE is the required trapezium.

Calculation of area:
Area of the trapezium CUTE = \(\frac { 1 }{ 2 }\) × h × (a + b) sq. units
= \(\frac { 1 }{ 2 }\) × 5.9 × (7 + 5)sq. units
= 35.4 sq.cm

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.4

Question 3.
ARMY with \(\overline{\mathrm{AR}}\) || \(\overline{\mathrm{YM}}\), AR = 7cm, RM = 6.5 cm ∠RAY = 100° and ∠ARM = 60°
Answer:
Given: In the trapezium ARMY
AR = 7cm,RM = 6.5cm,
∠RAY = 100° and ARM 60°, \(\overline{\mathrm{AR}}\) || \(\overline{\mathrm{YM}}\)

Rough Diagram
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.4 10

Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.4 11

Construction:
Steps:

  • Draw a line segment AR = 7 cm.
  • Construct an angle ∠RAX = 100° at A
  • Construct an angle ∠ARN = 60° at R
  • With R as centre, draw an arc of radius 6.5 cm cutting RN at M
  • Draw MY parallel to AR
  • ARMY is the required trapezium.

Calcualtion of Area:
Area of the trapezium ARMY = \(\frac { 1 }{ 2 }\) × h × (a + b) sq. units
= \(\frac { 1 }{ 2 }\) × 5.6 × (7 + 4.8) sq. units
= \(\frac { 1 }{ 2 }\) × 5.6 × 1.18
= 33.04 sq.cm

Samacheer Kalvi 8th Maths Guide Chapter 5 Geometry Ex 5.4

Question 4.
CITY with \(\overline{\mathrm{CI}}\) || \(\overline{\mathrm{YT}}\), CI = 7cm, IT = 5 .5 cm, TY = 4cm and YC = 6cm.
Answer:
Given: In the trapezium CITY,
CI = 7cm,IT = 5.5cm, TY = 4cm
YC = 6cm, and \(\overline{\mathrm{CI}}\) || \(\overline{\mathrm{YT}}\)

Rough Diagram
Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.4 12

Samacheer Kalvi 8th Maths Guide Answers Chapter 5 Geometry Ex 5.4 13

Construction:
Steps:

  • Draw a line segment CI = 7 cm.
  • Mark a point D on CI such that CD = 4cm
  • With D and I as centres, draw arcs of radii 6 cm and 5.5 cm respectively.
    Let them cut at T. Join DT and IT.
  • With C as centre, draw an arc of radius 6 cm.
  • Draw TY parallel to Cl. Let the line cut the previous arc at Y.
  • Join CY. CITY is the required trapezium.

Construction of area:
Area of the trapezium CITY = \(\frac { 1 }{ 2 }\) × h × (a + b) sq. units
= \(\frac { 1 }{ 2 }\) × 5.5 × (7 + 4)sq.units
= \(\frac { 1 }{ 2 }\) × 5.5 × 11
= 30.25 sq.cm