ML AGGARWAL CLASS 8 Chapter 18 Mensuration Exercise 18.2

  Exercise 18.2

Question 1

Sol :

Let ABCD is a rhombus 

$\begin{aligned} A B=B C &=C D=A D=13 \mathrm{~cm} . \\ A C &=10 \mathrm{Cm} . \end{aligned}$

'O' intersection point of diagnals 

$\overline{O A}=\overline{O C}=5 \mathrm{~cm} .$

In $\triangle^{L e}$ AOB

(i) $\overline{A B}^{2}=\overline{O A}^{2}+\overline{O B}^{2}$ (∵pythogorus theorem)

$13^{2}=5^{2}+\overline{O B}^{2}$

$169=25+\overline{O B}^{2}$

$\overline{O B}^{2}=169-25$

$\overline{O B}^{2}=144$

$\overline{O B}=\sqrt{144}$

$\overline{O B}=12 \mathrm{~cm}$

$\overline{B D}=2 \times \overline{O B}$

$=2 \times 12$

$\overline{B D}=24 \mathrm{Cm}$

(ii) Length of diagonal = 24cm

Area of rhombus =$\frac{1}{2}\times d_{1} \times d_{2}$

$=\frac{1}{2} \times 10 \times 24$

Area of rhombus $=120 \mathrm{~cm}^{2}$

Question 2

Sol :

Given ABCD is a trapezium 

Area of trapezium = $\frac{1}{2} \times$ (Sum of parallel sides) $\times$ (Distance between parllel sides)

$=\frac{1}{2} \times(1.5+8) \times 14$

Area of trapezium = 66.5$m^{2}$

Question 3

Sol :

Given 

Area of a trapezium = $360 \mathrm{~m}^{2}$

distance between two parallel side = 20m

Length of one parallel side = 25m

Let unknown parallel sides = x

Area of a trapezium $=\frac{1}{2}$ (sum of parallel sides) $\times$ (distance between parallel side)

$360=\frac{1}{2}(25+x) \times 20$

$(25+x)=\frac{360 \times 2}{20}$

25+x=36

x=36-25

x=11 m

∴ Another parallel side length =11 m

Question 4

Sol :

Given 

ABCD is a rhombus 

$\overline{B D}=13 C m$

$\overline{A B}=\overline{B C}=\overline{C D}=\overline{A D}=65 \mathrm{~cm}$

Altitude $\overline{A C}=5 \mathrm{~cm}$

(i) Area of rhombus =$\frac{1}{2} \times$ (product of diagonals)

$=\frac{1}{2} \times(13 \times 5)$

=$6.5 \times 5$

Area of rhombus = $32.5 \mathrm{~cm}^{2}$

(ii) Another diagonal $\overline{A C}=5 \mathrm{~cm} .$

Question 5

Sol :

(i) 

Area of trapezium ACDE

$=\frac{1}{2}(E D+A C) \times F G$

$=\frac{1}{2}(7+13) \times 6.5$

$=\frac{1}{2} \times 20 \times 6.5$

$=65 \mathrm{~m}^{2}$

(ii) Area of parallelogram ABCE = Base $\times$ distance between parallel sides 

$=7 \times 6.5$

$=45.5 \mathrm{~m}^{2}$

(iii) The area of triangle BCD= $\frac{1}{2} \times B C \times D H$

AC=AB+BC

13=7+BC

BC=13-7

BC=6 m

DH=GF=6.5 m

ஃ The area of triangle BCD= $\frac{1}{2} \times 6 \times 6.5$

=$3 \times 6.5$

=$19.5 \mathrm{~m}^{2}$

Question 6

Sol :

ABCD is a rhombus and EFG is triangle 

Given 

Area of rhombus = Area of a triangle 

$\frac{1}{2} \times d_{1} \times d_{2}$ = $\frac{1}{2} \times b \times h$

$\frac{1}{2} \times 22 \times d_{1}=\frac{1}{2} \times 24.8 \times 16.5$

$\begin{aligned} 22 \times d_{1} &=24.8 \times 16.5 \\ d_{1} &=\frac{24.8 \times 16.5}{22} \\ d_{1} &=18.6 \mathrm{~cm} . \end{aligned}$

Length of diagonal = 18.6 cm

Question 7

Sol :

Given

Perimeter of trapezium = 52 cm

Altitude = 8 cm

Length of parallel sides = perimeter - 2(parallel sides)

= 52 - 2$\times$ 10

=52- 20

Sum of parallel sides = 32 cm

Area of trapezium =$\frac{1}{2} \times$ (Sum of parallel side) $\times$ Altitude

$\begin{aligned}=& \frac{1}{2} \times 32 \times 8 \\=& 32 \times 4 \end{aligned}$

Area of trapezium = $128 \mathrm{~cm}^{2}$

Question 8

Sol :

Given 

area of trapezium = $540 \mathrm{cm}^{2}$

Altitude =18 cm

Ratio of length of parallel sides = 7:5

Let length of parallel sides = 7x, 5 x

∴ Area of trapezium = $\frac{1}{2} \times$ (Sum of parallel side) $\times$ Altitude

$540=\frac{1}{2} \times(7 x+5 x) \times 18$

$540=\frac{1}{2}(12 x) \times 18$

$540= 6 \times 18 \times x$

$x=\frac{540}{6 \times 18}$

x=5 cm

Length of parallel sides = $7 x=7 \times 5=35 \mathrm{~cm}$

$5 x=5 \times 5=25 \mathrm{~cm} .$

Question 9

Sol :

(i) 

Area enclosed by shape = Area of square AHGE + Area of triangle BCH + Area of square DCGE

$=5 \times 5+\frac{1}{2} \times 2 \times 9+4 \times 3$

=25+9+12

$=46 \mathrm{~cm}^{2}$

(ii) 

Area enclosed by shape = Area of square ABCD - [Area of triangle EFDH+ Area of triangle HIJ]

=$9 \times 9-\left[\frac{1}{2} \times 5 \times 7+\frac{1}{2} \times 5 \times 7\right]$

$=81-(5 \times 7) \Rightarrow 81-35=46 \mathrm{~cm}^{2}$

Question 10

Sol :

(i) 

In $\Delta^{l e} A B D$

$A B^{2}+A D^{2}=D B^{2}$ (∵Pythagoras theorem)

$40^{2}+A D^{2}=41^{2}$

$A D^{2}=41^{2}-40^{2}$

=1681-1600

$A D^{2}=81$

$A D=\sqrt{81}$

AD=9 cm

(ii) Area of trapezium = $\frac{1}{2} \times$ (Sum of parallel side) $\times$ Altitude

$=\frac{1}{2}(15+40) \times 9$

=$\frac{1}{2} \times 55 \times 9$

Area of trapezium = $247.5 \mathrm{~cm}^{2}$

(iii) Area of triangle BCD= Area of trapezium ABCD -[area of triangle ADB]

$=247.5-\left[\frac{1}{2} \times A B \times A D\right]$

$=247.5-\left[\frac{1}{2} \times 40 \times 9\right]$

=247.5-[180]

Area of triangle BCD = 67.5$\mathrm{~cm}^{2}$

 

Question 11

Sol :

 Area of section ①

Area of trapezium = $\frac{1}{2} \times$ (Sum of parallel side) $\times$ Altitude

$=\frac{1}{2} \times(28+20) \times 4$

$=96 \mathrm{~cm}^{2}$

$\therefore$ Area section (1) $=96 \mathrm{~cm}^{2}$

Ares of section ② 

Area of trapezium = $\frac{1}{2} \times$ (Sum of parallel side) $\times$ Altitude

$=\frac{1}{2} \times(24+32) \times 4$

Ares of section ②  $=112 \mathrm{~cm}^{2}$

Section ③ Dimension are same as section ①

ஃ Area of section ③$=96 \mathrm{~cm}^{2}$

Section ④ Dimension are same as section ②

∴ Area of section ④ = 112$\mathrm{~cm}^{2}$

Question 12

Sol :

From $\Delta^{\operatorname{le}} A B D$

$B D^{2}=A B^{2}+A D^{2}$

$B D^{2}=6^{2}+88^{2}$

$B D^{2}=36+64$

$B D^{2}=100$

BD=10 cm

From $\Delta l e$ BDC

$B C^{2}=B D^{N}+D C^{2}$

$26^{2}=10^{2}+D C^{2}$

$676=100+D C^{2}$

$D C^{2}=676-100$

$D C^{2}=576$

$D C=\sqrt{576}$

$D C=24 \mathrm{Cm}$

Area of quadrilateral ABCD = Area of $\Delta^{l e} BAD$ + Area of $\Delta l e BDC$

$=\frac{1}{2}(A B \times A D)+\frac{1}{2}(B D \times D C)$

$=\frac{1}{2}(6 \times 8)+\frac{1}{2}(10 \times 24)$

$=\frac{1}{2}(48)+\frac{1}{2}(240)$

=24+120

Area of quadrilateral ABCD = $144 \mathrm{~cm}^{2}$

Question 13

Sol :

Given ABCDEFGH a regular octagon 

Area of octagon ABCDEFGH = Area of square ABCH + Area of square HCDG + Area of square GDEF

= $2 \times$ Area of square ABCH + Area of square HCDG 

$=2 \times\left(\frac{1}{2} \times(8+15) \times 6\right]+(8 \times 15)$

$=\left(2 \times \frac{1}{2} \times 23 \times 6\right)+(8 \times 15)$

$=\quad 23 \times 6+8 \times 15$

=138+120

$=258 \mathrm{~m}^{2} \mathrm{2}$

Question 14

Sol :

Jaspreet's diagram 

Area of ABCD = Area square  ABCF+ Areaof square FCDE

$=2 \times($ Areaof $\square A B C F)$ (∵ Both are symmetric)

$=2 \times\left(\frac{1}{2} \times(A B+C F) \times A F\right)$

=$2 \times \frac{1}{2} \times(18+32) \times \frac{18}{2}$

= $50 \times 9$

Area of ABCDE = $450 \mathrm{~cm}^{2}$

Rahul's diagram 

-----------------------

Area of pentagon ABCDE = area of triangle DEC + area of square ECBA 

$=\frac{1}{2}(E C \times D F)+B C \times A B$

$=\frac{1}{2} \times 18 \times 14+18 \times 18$

=126+324

$=450 \mathrm{~cm}^{2}$

We can find area of pentagon in this way 

Mahesh 's diagram 

Question 15

Sol :

Area of shaded pentagon ABECD = Area of square ABCD - [Area of triangle BEC]

$=18 \times 10-\left[\frac{1}{2} \times 8 \times E B\right] \rightarrow(1)$

From triangle BEC

$B C^{2}=E C^{2}+E B^{2}$

$10^{2}=8^{2}+E B^{2}$

$E B^{2}=100-64$

$E B^{2}=36$

$E B=\sqrt{36}$

$E B=6 \mathrm{Cm}$

Sub EB value is eq.. ①

∴ Area of shaded pentagon ABECD = $180-\left[\frac{1}{2} \times 8 \times 6\right]$

=$180-[4 \times 6]$

=180-24

∴ Area of shade pentagon ABECD = $156 \mathrm{~cm}^{2}$

Question 16

Sol :

Given 

ABCDE is a polygon 

$\begin{aligned} A D &=8 \mathrm{~cm} \\ A H &=6 \mathrm{~m} \\ A G &=4 \mathrm{~cm} \\ A F &=3 \mathrm{Cm} . \\ B F &=2 \mathrm{~cm}, \mathrm{CH}=3 \mathrm{~cm}, \quad E G=2.5 \mathrm{Cm} . \end{aligned}$

Area of polygon ABCDE = area of triangle ABF + Area of square BCHF + Area of triangle CHD+ Area of triangle AGE

$=\frac{1}{2}(A F \times B F)+\frac{1}{2}(B F+C H) \times F H+\frac{1}{2}(D H \times C H)+$\frac{1}{2}(A D \times E G)$

$\begin{aligned} A D &=A H+H D \\ 8 &=6+H D \\ & H D=8-6 \\ H D &=2 C m \end{aligned}$

AH=AF+FH

6=3+FH

FH=6-3

FH=3 cm

∴ Area of polygon ABCDE = $\frac{1}{2}(3 \times 2)+\frac{1}{2}(5) \times 3+\frac{1}{2} \times 2 \times 3+\frac{1}{2}(8 \times 2.1)$

=3+7.5+3+10

Area of Polygon ABCDE $=23.5 \mathrm{cm}^{2}$

Question 17

Sol :

Given PQRSTU is a polygon 

PS=11 cm

PY=9 cm

PX=8 cm

PW=5 cm

PV=3 cm

QV=5 cm

UW=4 cm

RX=6 cm

TY=2 cm

$\begin{aligned} VX&=PX-PY \\ &=8-3 \\ VX &=5 \mathrm{~cm} \end{aligned}$

WY=PY-PW=9-5=4 cm

$\begin{aligned} XS &=P S-PX \\ &=11-8 \\ XS &=3 cm \end{aligned}$

$\begin{aligned} YS &=PS-PY \\ &=11-9 \\ YS &=2 a m \end{aligned}$

=Area of polygon PQRSTU = Area of triangle PQN + Area of square  QRXV + Area of triangle XRS + Area of triangle PUW + Area of square UMNT + Area of triangle YST

=$\left(\frac{1}{2} \times PV \times QV\right)+$ $\frac{1}{2}(Q V+R X) \times(V X)$ +$\frac{1}{2}(R X)(XS)$ + $\frac{1}{2} \times$ PW $\times$ VW +$\frac{1}{2}(V W+XT) \times(YW)$ +$\frac{1}{2}(YS) \times(Y T)$

= $\frac{1}{2} \times 3 \times 5+\frac{1}{2}(5+6) \times 5+\frac{1}{2}(6 \times 3)+\frac{1}{2}(5 \times 4)+\frac{1}{2}(4 \times 2 ) \times 4$ + $\frac{1}{2}(2 \times 2)$

= $\frac{1}{2}(15+55+18+20+24+4)$

= $\frac{1}{2}(136)$

= $68 \mathrm{~cm}^{2}$