Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3

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

Question 1.
AIMS with \(\overline { AI } \) ∥ \(\overline { SM } \), AI = 6 cm, IM = 5 cm, AM = 9 cm and MS 6.5 cm.
Solution:
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 1
Given
AI = 6cm
IM = 5cm
AM = 9cm, and \(\overline { AI } \) ∥ \(\overline { SM } \)
MS = 6.5 cm
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 2
Construction:
Steps:

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

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

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Question 2.
BIKE with \(\overline { BI } \) ∥ \(\overline { EK } \), BI = 4 cm, IK = 3.5 cm, BK = 6 cm and BE = 3.5 cm
Solution:
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 3
Given:
In the trapezium BIKE,
BI = 4 cm
IK = 3.5 cm
BK = 6 cm
BE = 3.5 cm and \(\overline { BI } \) ∥ \(\overline { EK } \)
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 4
Construction:
Steps:

  1. Draw a line segment BI = 4 cm.
  2. With B and I as centres, draw arcs of radii 6 cm and 3.5 cm respectively and let them cut at K.
  3. Join BK and IK
  4. Draw KX parallel to BI
  5. With B as centre, draw an arc of radius 3.5 cm.cutting KX at E
  6. Join BE. BIKE is the required trapezium.

Calculation of area:
Area of the trapezium BIKE = \(\frac{1}{2}\) x h x (a + b) sq. units = \(\frac{1}{2}\) x 3.5 x (4 + 4.2)
= \(\frac{1}{2}\) x 3.5 x 8.2 = 14.35 sq.cm

Question 3.
CUTE with \(\overline { CD } \) ∥ \(\overline { ET } \), CU = 7 cm, ∠UCE = 80°, CE = 6 cm and TE = 5 cm.
Solution:
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 5
Given:
In the trapezium CUTE,
CU = 7 cm, ∠UCE = 80°,
CE = 6 cm, TE = 5 cm and \(\overline { CD } \) ∥ \(\overline { ET } \)
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 6
Construction:
Steps:

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

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

Question 4.
DUTY with \(\overline { DU } \) ∥ \(\overline { YT } \), DU = 8 cm, ∠DUT = 60°, UT = 6 cm and TY = 5 cm.
Solution:
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 7
Given:
In the trapezium DUTY
DU = 8 cm, ∠DUT = 60°,
UT = 6 cm, TY = 5 cm and \(\overline { DU } \) ∥ \(\overline { YT } \)
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 8
Construction:
Steps:

  1. Draw a line segment DU = 8 cm.
  2. Construct an angle ∠DUT = 60° at U
  3. With U as centre, draw an arc of radius 6 cm cutting UA at T.
  4. Draw TX parallel to DU
  5. With T as centre, draw an arc of radius 5 cm cutting TX at Y
  6. Join DE. DUTY is the required trapezium.

Calculation of area:
Area of the trapezium DUT Y = \(\frac{1}{2}\) x h x (a + b) sq. units= \(\frac{1}{2}\) x 5.2 x (8 + 5) sq. units = \(\frac{1}{2}\) x 5.2 x 13
= 33.8 sq.cm ,

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Question 5.
ARMY with \(\overline { AR } \) ∥ \(\overline { YM } \), AR = 7 cm, RM = 6.5 cm ∠RAY = 100° and ∠ARM = 60° 5
Solution:
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 9
Given:
In the trapezium ARMY
AR = 7 cm, RM = 6.5 cm,
∠RAY = 100° and ARM = 60°, \(\overline { AR } \) ∥ \(\overline { YM } \)
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 10
Construction:
Steps:

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

Calculation of area:
Area of the trapezium ARMY = \(\frac{1}{2}\) x h x (a + b) sq. units = \(\frac{1}{2}\) x 5.6 x (7 + 4.8) sq. units
= \(\frac{1}{2}\) x 5.6 x 11.8 = 33.04 sq.cm

Question 6.
BELT with \(\overline { BE } \) ∥ \(\overline { TL } \), BT = 7 cm ∠EBT = 85° and ∠BEL = 110°
Solution:
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 11
Given:
In the trapezium BELT
BE = 10 cm, BT = 7cm,
∠EBT = 85°, ∠BEL = 110° and \(\overline { BE } \) ∥ \(\overline { TL } \)
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 12
Construction:
Steps:

  1. Draw a line segment BE = 10 cm.
  2. Construct two angles ∠TBE = 85° and ∠BEL =110° respectively at the points B and E.
  3. With B as centre, draw an arc of radius 7 cm cutting BX at T.
  4. Draw TL ∥ BE
  5. BELT is the required trapezium

Question 7.
CITY with \(\overline { CI } \) ∥ \(\overline { YT } \) Cl = 7 cm, IT = 5.5 cm, TY = 4 cm and YC = 6 cm.
Solution:
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 13
Given:
In the trapezium CITY,
Cl = 7 cm
IT = 5.5 cm
TY = 4 cm
YC = 6 cm, and \(\overline { CI } \) ∥ \(\overline { YT } \)
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 14
Construction:
Steps:

  1. Draw a line segment Cl = 7 cm.
  2. Mark a point D on Cl such that CD = 4cm
  3. 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.
  4. With C as centre, draw an arc of radius 6 cm.
  5. Draw TY parallel to CL Let the line cut the previous arc at Y.
  6. Join CY. CITY is the required trapezium.

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

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Question 8.
DICE with \(\overline { DI } \) ∥ \(\overline { EC } \), DI = 6 cm, IC = ED = 5 cm and CE = 3 cm. Solution:
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 15
Given:
In the trapezium DICE,
DI = 6 cm
IC = ED = 5 cm
CE = 3 cm and \(\overline { DI } \) ∥ \(\overline { EC } \)
Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 3 Geometry Ex 3.3 16
Construction:
Steps:

  1. Draw a line segment DI = 6 cm.
  2. Mark a point M on DI such that DM = 3cm
  3. With D and I as centres, draw arcs of radii 5 cm each Let them cut at C. Join MC and IC.
  4. Draw CX parallel to DI
  5. With D as centre, draw an arc of radius 5 cm. Let it cut CX at E
  6. Join DE. DICE is the required trapezium.

Calculation of area:
Area of the trapezium DICE = \(\frac{1}{2}\) x h x (a + b) sq. units = \(\frac{1}{2}\) x 3.8 x (6 + 3) sq. units
= \(\frac{1}{2}\) x 3.8 x 9 = 17.1 sq. cm

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