Exercise 16.2
Question 1
(i) Area of parallelogram =$b\times h$
Base (b)= 8cm
Height (h)= 4.5cm
Area of parallelogram = $8 \times 4.5$
= $36 \mathrm{~cm}^{2}$
(ii)Base $(b)=2 \mathrm{~cm}$
Height $(h)=4.4 \mathrm{~cm}$
Area of parallelogram =$b\times h$
$=2 \times 4.4$
Area of parallelogram $=8.8 \mathrm{~cm}^{2}$
(iii) $\begin{aligned} \text { Base }(b) &=2.5 \mathrm{~cm} \\ \text { Height }(h) &=3.5 \mathrm{~cm} \end{aligned}$
$\begin{aligned} \text { Area } &=b \times h \\ &=2.5 \times 35 \end{aligned}$
Area of parallelogram = $8.75 \mathrm{Cm}^{2}$
Question 2
(i)Area of triangle = $\frac{1}{2} \times$ base $\times$ height
Base (b) = 6.4cm
Height (h)=6cm
Area of triangle = $\frac{1}{2} \times 6.4 \times 6$
$=6.4 \times 3$
Area of triangle = 19.2 $\mathrm{cm}^{2}$
(ii)Base (b) = 5cm
Height (h)= 6cm
Area of triangle = $\frac{1}{2} b h$
=$\frac{1}{2} \times 5 \times 6$
=$5 \times 3$
Area of triangle = $15 \mathrm{~cm}^{2}$
(iii)Base $(b)=4.5 \mathrm{~cm}$
Height $(h)=6 \mathrm{~m}$
$\begin{aligned} \text { Areaof triangle } &=\frac{1}{2} b h \\ &=\frac{1}{2} \times 4.5 \times 6 \\ &=4.5 \times 3 \end{aligned}$
Area of triangle = $13.5 \mathrm{~cm}^{2}$
Question 3
(i) $41 \mathrm{cm}^{2}$
(ii) 12.3cm
(iii)10.3cm
(iv)5.8cm
Question 4
(i)193.72 $\mathrm{cm}^{2}$
(ii) 11.6cm
(iii) 15.5cm
(iv)80cm
Question 5
(Diagram to be added)
Area of parallelogram = $base\times height$
Base =6cm, height =3cm
Area of parallelogram = $6 \times 3=18 \mathrm{~cm}^{2}$---①
Consider
Base = 4cm , height = h
Area of parallelogram = $4 \times h$ ----②
①=②
$4 \times h=18$
$h=\frac{18}{4}$
$n=4.5 \mathrm{~cm}$
Question 6
Consider
Base (b) = 9cm , Height (h)= 6cm
Area of triangle = $\frac{1}{2} b h$
$=\frac{1}{2} \times 9 \times 6$
$=4 \times 3$
Area of triangle = 27$Cm^{2}$----①
Consider
Base (b)= 7.5m, height (h)=?
Area of triangle = $=\frac{1}{2} \times 7.5 \times h$
①=②
$\begin{aligned} \frac{1}{2} \times 7.5 \times h &=27 \\ \frac{h}{2} &=\frac{27}{7.5} \end{aligned}$
h= 7 .2m
Height corresponding to the base 7.5m = 7.2m
Question 7
(Diagram to be added)
Base = 8cm
Hypotenuse = 17cm
Height = h
$A C^{2}=A B^{2}+B C^{2}$ (Pythagoras Theorem)
$\begin{aligned} 17^{2}=& h^{2}+8^{2} \\ h^{2}=& 289-64 \end{aligned}$
$h^{2}=225$
$h=\sqrt{225}$
$h=15 \mathrm{~cm}$
Height of right angled triangle = 15cm
Area of triangle $=\frac{1}{2} b h$
$=\frac{1}{2} \times 8 \times 15$
$=4 \times 15$
Area of triangle = $60 \mathrm{~cm}^{2}$
Question 8
(Diagram to be added)
Given
(i) $\triangle A B C$ is Right angled triangle
$\begin{aligned} A C^{2} &=-A B^{2}+B C^{2} \\ A C^{2} &=6^{2}+8^{2} \\ A C &=\sqrt{36+64} \\ &=\sqrt{100} \\ A C &=10 \mathrm{Cm} \end{aligned}$
(ii) Area of triangle = $\frac{1}{2} b h$
$=\frac{1}{2} \times 8 \times 6$
$=8 \times 3$
Area of triangle = $24 \mathrm{Cm}^{2}$ ----①
Area of triangle = $\frac{1}{2} \times A C \times B N$
=$\frac{1}{2} \times 10 \times B N$----②
①=②
$\begin{aligned} \frac{1}{2} \times 10 \times B N &=24 \\ B N &=\frac{48}{10} \\ B N &=4.8\mathrm{cm}\end{aligned}$
Question 9
(Diagram to be added)
Given AB = 10cm
EF= 16cm
M is mid point of DC
$D M=M C=\frac{10}{2}=5 \mathrm{Cm}$
Area of $\triangle A E B$ = Area of parallelogram ABCD
$\frac{1}{2} \times A B \times E F=A B \times G F$
$G F=8 \mathrm{~cm}$
∴ EF= GF + EG
$16=8+E G$
$E G=8 \mathrm{cm}$
Given Area of Triangle AEB= Area of Parallelogram ABCD
∴ Area of trapezium AMCB is common in both Triangle AED and Parallelogram ABCD
∴ Area of $\triangle A D M=$ Area of $\triangle M E C$
$=\frac{1}{2} \times M C \times E G$
$=\frac{1}{2} \times 5 \times 8$
$=5 \times 4$
∴Area of △ADM $=20 \mathrm{c} \mathrm{m}^{2}$
Question 10
(Diagram to be added)
ABCD is a rectangle of size = $18 \operatorname{cm} \times 10 \mathrm{~cm}$
$\angle E=90^{\circ}$
In $\triangle E C B$
$B C^{2}=E C^{2}+E B^{2}$
$10^{2}=8^{2}+E B^{2}$
$E B^{2}=100-64$
$E B^{2}=36$
$E B=\sqrt{36}$
$E B=6 \mathrm{Cm}$
Area of shaded region= Area of $\square A B C D$ - Area of $\triangle E C B$
=$l \times b-\left(\frac{1}{2} b h\right)$
=$18 \times 10-\left(\frac{1}{2} \times 6 \times 8\right)$
=$180-24$
Area of shaded region = $156 \mathrm{cm}^{2}$
Question 11
(i) (Diagram to be added)
Area of shaded region = Area of $\square A B C D$ - ( Area of $\Delta^{l \varphi} E C B$+ Area of DEF )
$18 \times 10-\left(\frac{1}{2} \times 10 \times 8+\frac{1}{2} \times 6 \times 10\right)$
$(\because D C=D E+E C)$ $18=10+E C$ =$E C=8 \mathrm{~cm}$)
$=180-(40+30)$
$=180-70$
Area of shaded region = $110 \mathrm{Cm}^{2}$
(ii)(Diagram to be added)
Area of shaded region = Area of $\square A B C D-$ [Area of $\Delta A E \mathrm{~F}+$ Area of $\triangle E C B$ Area of $\triangle$ DCF ]
=$A B \times B C-\left[\frac{1}{2} \times A E \times A F+\frac{1}{2} \times E B \times B C+\frac{1}{2} \times D F \times D C\right]$
=$20 \times 20-\left[\frac{1}{2} \times 10 \times 10+\frac{1}{2} \times 10 \times 20+\frac{1}{2} \times 10 \times 20\right]$
$=400-[10 \times 5+10 \times 10+10 \times 10]$
$=400-[50+100+100]$
=400-[250]
Area of shaded region = $150 \mathrm{~cm}^{2}$
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