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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:

Given

AI = 6cm

IM = 5cm

AM = 9cm, and \(\overline { AI } \) ∥ \(\overline { SM } \)

MS = 6.5 cm

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 = \(\frac{1}{2}\) 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

Question 2.

BIKE with \(\overline { BI } \) ∥ \(\overline { EK } \), BI = 4 cm, IK = 3.5 cm, BK = 6 cm and BE = 3.5 cm

Solution:

Given:

In the trapezium BIKE,

BI = 4 cm

IK = 3.5 cm

BK = 6 cm

BE = 3.5 cm and \(\overline { BI } \) ∥ \(\overline { EK } \)

Construction:

Steps:

- Draw a line segment BI = 4 cm.
- With B and I as centres, draw arcs of radii 6 cm and 3.5 cm respectively and let them cut at K.
- Join BK and IK
- Draw KX parallel to BI
- With B as centre, draw an arc of radius 3.5 cm.cutting KX at E
- 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:

Given:

In the trapezium CUTE,

CU = 7 cm, ∠UCE = 80°,

CE = 6 cm, TE = 5 cm and \(\overline { CD } \) ∥ \(\overline { ET } \)

Construction:

Steps:

- Draw a line segment CU = 7 cm.
- 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 of radius 5 cm cutting EX at T
- 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:

Given:

In the trapezium DUTY

DU = 8 cm, ∠DUT = 60°,

UT = 6 cm, TY = 5 cm and \(\overline { DU } \) ∥ \(\overline { YT } \)

Construction:

Steps:

- Draw a line segment DU = 8 cm.
- Construct an angle ∠DUT = 60° at U
- With U as centre, draw an arc of radius 6 cm cutting UA at T.
- Draw TX parallel to DU
- With T as centre, draw an arc of radius 5 cm cutting TX at Y
- 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 ,

Question 5.

ARMY with \(\overline { AR } \) ∥ \(\overline { YM } \), AR = 7 cm, RM = 6.5 cm ∠RAY = 100° and ∠ARM = 60° 5

Solution:

Given:

In the trapezium ARMY

AR = 7 cm, RM = 6.5 cm,

∠RAY = 100° and ARM = 60°, \(\overline { AR } \) ∥ \(\overline { YM } \)

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.

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:

Given:

In the trapezium BELT

BE = 10 cm, BT = 7cm,

∠EBT = 85°, ∠BEL = 110° and \(\overline { BE } \) ∥ \(\overline { TL } \)

Construction:

Steps:

- Draw a line segment BE = 10 cm.
- Construct two angles ∠TBE = 85° and ∠BEL =110° respectively at the points B and E.
- With B as centre, draw an arc of radius 7 cm cutting BX at T.
- Draw TL ∥ BE
- 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:

Given:

In the trapezium CITY,

Cl = 7 cm

IT = 5.5 cm

TY = 4 cm

YC = 6 cm, and \(\overline { CI } \) ∥ \(\overline { YT } \)

Construction:

Steps:

- Draw a line segment Cl = 7 cm.
- Mark a point D on Cl 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.

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

Question 8.

DICE with \(\overline { DI } \) ∥ \(\overline { EC } \), DI = 6 cm, IC = ED = 5 cm and CE = 3 cm. Solution:

Given:

In the trapezium DICE,

DI = 6 cm

IC = ED = 5 cm

CE = 3 cm and \(\overline { DI } \) ∥ \(\overline { EC } \)

Construction:

Steps:

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