RD Sharma solution class 8 chapter 20 Mensuration I(Area of Trapezium and Polygon) Exercise 20.3

Exercise 20.3

Page-20.28

Question 1:

Find the area of the pentagon shown in fig. 20.48, if AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm, BF = 5 cm, CG = 7 cm and EH = 3 cm.

Answer 1:

$\text{The given figure is: }$


$$\begin{gathered} \text{Given:} \\ \text{AD = 10 cm, AG = 8 cm, AH = 6 cm, AF = 5 cm} \\ \text{BF = 5 cm, CG = 7 cm, EH = 3 cm} \\ \therefore \text{FG = AG - AF = 8 - 5 = 3 cm } \\ \text{And, GD = AD - AG = 10 - 8 = 2 cm} \\ \text{From given figure:} \\ \text{Area of Pantagon = (Area of triangle AFB) + (Area of trapezium FBCG) + (Area of triangle CGD) + (Area of triangle ADE)} \\ \text{= (}\frac{1}{2}\text{ × AF × BF) + [}\frac{1}{2}\times \text{(BF + CG) × (FG)] + (}\frac{1}{2}\times \text{GD × CG) + (}\frac{1}{2}\text{ × AD × EH)} \\ \text{= }(\frac{1}{2} \text{× 5 × 5) + [}\frac{1}{2}\times \text{(5 + 7) × (3)] + (}\frac{1}{2}\times 2 \text{× 7) + (}\frac{1}{2}\text{ × 10 × 3)} \\ \text{= (}\frac{25}{2}\text{) + [}\frac{36}{2}\text{] + (}\frac{14}{2}\text{) + (}\frac{30}{2}\text{)} \\ =12.5 + 18 + 7 + 15 \\ {\text{=52.5 cm}}^2 \end{gathered}$$

Question 2:

Find the area enclosed by each of the following figures [Fig. 20.49 (i)-(iii)] as the sum of the areas of a rectangle and a trapezium:

Answer 2:

$$\begin{gathered} \text{(i)} \\ \text{The given figure can be divided into a rectangle and a trapezium as shown below: } \end{gathered}$$

$$\begin{gathered} \text{From the above firgure:} \\ \text{Area of the complete figure = (Area of square ABCF)+(Area of trapezium CDEF)} \\ \text{=(AB×BC)+[}\frac{1}{2}\text{×(FC+ED)×(Distance between FC and ED)]} \\ \text{=(18×18)+[}\frac{1}{2}\text{×(18+7)×(8)]} \\ \text{=324+100} \\ {\text{=424 cm}}^2 \end{gathered}$$

$$\begin{gathered} \text{(ii)} \\ \text{The given figure can be divided in the following manner: } \end{gathered}$$

$$\begin{gathered} \text{From the above figure:} \\ \text{AB = AC-BC=28-20=8 cm} \\ \text{So that area of the complete figure = (area of rectangle BCDE)+(area of trapezium ABEF)} \\ \text{=(BC×CD)+[}\frac{1}{2}\text{×(BE+AF)×(AB)]} \\ \text{=(20×15)+[}\frac{1}{2}\text{×(15+6)×(8)]} \\ \text{=300+84} \\ {\text{=384 cm}}^2 \end{gathered}$$


$$\begin{gathered} \text{(iii)} \\ \text{The given figure can be divided in the following manner:} \end{gathered}$$

$$\begin{gathered} \text{From the above figure:} \\ \text{EF = AB = 6 cm} \\ \text{Now, using the Pythagoras theorem in the right angle triangle CDE:} \\ {\text{5}}^2{\text{ = 4}}^2{\text{+CE}}^2 \\ {\text{CE}}^2\text{ = 25-16=9} \\ \text{CE = }\sqrt{9 }\text{= 3 cm} \\ \text{And, GD=GH+HC+CD=4+6+4=14 cm} \\ \therefore \text{Area of the complete figure=(Area of rectangle ABCH)+(Area of trapezium GDEF)} \\ \text{=(AB×BC)+[}\frac{1}{2}\text{×(GD+EF)×(CE)]} \\ \text{=(6×4)+[}\frac{1}{2}\text{×(14+6)×(3)]} \\ \text{=24+30} \\ {\text{=54 cm}}^2 \end{gathered}$$

Question 3:

There is a pentagonal shaped park as shown in Fig. 20.50. 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 areas?

Answer 3:

$\text{A pentagonal park is given below: }$




$$\begin{gathered} \text{Jyoti and Kavita divided it in two different ways. } \\ \text{(i) Jyoti divided is into two trapeziums as shown below: } \end{gathered}$$

$$\begin{gathered} \text{It is clear that the park is divided in two equal trapeziums whose parallel sides are 30 m and 15 m. } \\ \text{And, the distance between the two parallel lines: }\frac{15}{2}\text{=7.5 m} \\ \text{∴ Area of the park=2×(Area of a trapazium)} \\ =2\times [\frac{1}{2}\times (30+15)\times (7.5\left)\right] \\ {\text{=337.5 m}}^2 \end{gathered}$$

$$\begin{gathered} \text{(ii)} \\ \text{Kavita divided the park into a rectangle and a triangle, as shown in the figure. } \end{gathered}$$

$$\begin{gathered} \text{Here, the height of the triangle = 30-15=15 m} \\ \text{∴ Area of the park=(Area of square with sides 15 cm)+(Area of triangle with base 15 m and altitude 15 m)} \\ \text{=(15×15)+(}\frac{1}{2}\text{×15×15)} \\ =225+112.5 \\ {\text{=337.5 m}}^2 \end{gathered}$$

Page-20.29

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.

Answer 4:

$\text{The given polygon is: }$



$$\begin{gathered} \text{Given:} \\ \text{AL=10 cm, AM=20 cm, AN=50 cm} \\ \text{AO=60 cm, AD=90 cm} \\ \text{Hence, we have the following:} \\ \text{MO=AO-AM=60-20=40 cm} \\ \text{OD=AD-AO=90-60=30 cm} \\ \text{ND=AD-AN=90-50=40 cm} \\ \text{LN=AN-AL=50-10=40 cm} \\ \text{From given figure:} \\ \text{Area of Polygon=(Area of triangle AMF)+(Area of trapezium MOEF)+(Area of triangle EOD)+(Area of triangle DNC)+}(Area of trapezium \text{NLBC})+(Area of triangle \text{ALB}) \\ \text{=(}\frac{1}{2}\text{×AM×MF)+[}\frac{1}{2}\times \text{(MF+OE)×(OM)]+(}\frac{1}{2}\text{×OD×OE)+(}\frac{1}{2}\text{×DN×NC)}+[\frac{1}{2}\times \text{(LB+NC)×(NL)]+(}\frac{1}{2}\times \text{AL×LB)} \\ \text{=}(\frac{1}{2}\text{×20×20)+[}\frac{1}{2}\times \text{(20+60)×(40)]+(}\frac{1}{2}\times 30\text{×60)+(}\frac{1}{2}\text{×40×40)}+[\frac{1}{2}\times \text{(30+40)×(40)]+(}\frac{1}{2}\times 10\times \mathrm{30)} \\ \text{=200+1600+900+800+1400+150} \\ {\text{=5050 cm}}^2 \end{gathered}$$

Question 5:

Find the area of the following regular hexagon.

Answer 5:

$\text{The given figure is: }$

Join QN.



$$\begin{gathered} \text{It is given that the hexagon is regular. So, all its sides must be equal to 13 cm.} \\ \text{Also, AN = BQ} \\ \text{QB+BA+AN = QN} \\ \text{AN+13+AN = 23} \\ \text{2AN = 23-13 = 10} \\ \text{AN = }\frac{10}{2}\text{ = 5 cm} \\ \text{Hence, AN = BQ = 5 cm} \\ \text{Now, in the right angle triangle MAN:} \\ {\text{MN}}^2{\text{=AN}}^2{\text{+AM}}^2 \\ {\text{13}}^2{\text{=5}}^2{\text{+AM}}^2 \\ {\text{AM}}^2\text{=169-25=144} \\ \text{AM=}\sqrt{144}\text{=12cm.} \\ \text{∴ OM = RP = 2×AM = 2×12 = 24 cm} \\ \text{Hence, area of the regular hexagon = (area of triangle MON)+(area of rectangle MOPR)+(area of triangle RPQ)} \\ \text{=(}\frac{1}{2}\text{×OM×AN)+(RP×PO)+(}\frac{1}{2}\text{×RP×BQ)} \\ \text{=(}\frac{1}{2}\text{×24×5)+(24×13)+(}\frac{1}{2}\text{×24×5)} \\ \text{=60+312+60} \\ {\text{=432 cm}}^2 \end{gathered}$$