## RD Sharma Class 8 Solutions Chapter 20 Mensuration I (Area of a Trapezium and a Polygon) Ex 20.3

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3

**Other Exercises**

- RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.1
- RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.2
- RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3

**Question 1.**

**Find the area of the pentagon shown in the figure if AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm and BF = 5 cm, CG = 7 cm and EH = 3**

**Solution:**

In the figure, here are three triangles and one trapezium.

**Question 2.**

**Find the area enclosed by each of the following figures as the sum of the areas of a rectangle and a trapezium:**

**Solution:**

(i) In the figure ABCDEF,

Join CF, then, the figure consists one square and one trapezium ABCF is a square whose side = 18 cm

Area of the square = 18 x 18 cm² = 324 cm²

Area of trapezium FCDE = \(\frac { 1 }{ 2 }\) (CF + ED) x 8 cm²

= \(\frac { 1 }{ 2 }\) (18 + 7) x 8

= \(\frac { 1 }{ 2 }\) x 25 x 8 cm²

= 100 cm²

Total area of fig. ABCDEF = 324 + 100 = 424 cm²

(ii) In the figure ABCDEF,

Join BE.

The figure consists of one rectangle BCDE and one trapezium ABEF

Area of rectangle BCDE = BC x CD = 20 x 15 = 300 cm²

Area of trapezium ABEF,

= \(\frac { 1 }{ 2 }\) (BE + AF) x height

= \(\frac { 1 }{ 2 }\) (15 + 6) x 8 cm²

= \(\frac { 1 }{ 2 }\) x 21 x 8 cm²

= 84 cm²

Area of the figure ABCDEF = 300 + 84 = 384 cm²

(iii) In the figure ABCDEFGH,

HC= AB = EF = 6 cm

AH = BC = 4 cm

DE = GF = 5 cm

Join HC.

In right ∆CDE,

ED² = CD² + CE²

⇒ (5)² = (4)² + (CE)²

⇒ 25 = 16 + (CE)²

⇒ (CE)² = 25 – 16 = 9 = (3)²

CE = 3 cm

The figure consist a rectangle and a trapezium

Area of rectangle ABCH = AB x BC = 6 x 4 = 24 cm²

Area of trapezium GDEF,

= \(\frac { 1 }{ 2 }\) (GD + EF) x CE 1

= \(\frac { 1 }{ 2 }\) (GH + HC + CD + EF) x CE

= \(\frac { 1 }{ 2 }\) (4 + 6 + 4 + 6) x 3 cm²

= \(\frac { 1 }{ 2 }\) x 20 x 3 cm²

= 30 cm²

Total area of the figure ABCDEFGH = 24 + 30 = 54 cm²

**Question 3.**

**There is a pentagonal shaped park as shown in the figure. Jyoti and Kavita divided it in two different ways.**

**Find the area of this park using both ways. Can you suggest some another way of finding its area ?**

**Solution:**

In first case, the figure ABCDE is divided into 2 trapezium of equal area.

Now area of trapezium DFBC

Total area of the pentagon ABCDE = 2 x 168.75 = 337.5 m²

In second case, the figure ABCDE is divided into two parts, namely one square and other triangle.

Total area of pentagon ABCDE = 225 + 112.5 = 337.5 m²

**Question 4.**

**Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 cm, AO = 60 cm and AD = 90 cm.**

**Solution:**

In the figure ABCDEF,

AD = 90 cm

BL = 30 cm

AO = 60 cm

CN = 40 cm

AN = 50 cm

EO = 60 cm

AM = 20 cm

FM = 20 cm

AL = 10 cm

Area of ABCDEF = (150 + 800 + 900 + 200 + 1400 + 1600) cm² = 5050 cm²

**Question 5.**

**Find the area of the following regular hexagon:**

**Solution:**

In the regular hexagon MNOPQR There are two triangles and one rectangle.

Join MQ, MO and RP

NQ = 23 cm,

NA = BQ = \(\frac { 10 }{ 2 }\) = 5 cm

MR = OP = 13 cm

In right ∆BDQ,

PQ² = BQ² + BP²

⇒ (13)² = (5)² + BP²

⇒ 169 = 25 + BP²

⇒ BP² = 169 – 25 = 144 = (12)²

BP = 12 cm

PR = MO = 2 x 12 = 24 cm

Now area of rectangle RPOM = RP x PO = 24 x 13 = 312 cm²

Area of ∆PRQ = \(\frac { 1 }{ 2 }\) x PR x BQ

= \(\frac { 1 }{ 2 }\) x 24 x 5 = 60 cm²

Similarly area ∆MON = 60 cm²

Area of the hexagon MNOPQR = 312 + 60 + 60 = 432 cm²

Hope given RD Sharma Class 8 Solutions Chapter 20 Mensuration I Ex 20.3 are helpful to complete your math homework.

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