Targeting Mathematics Class 8 Solutions Pdf

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Mathematics Solutions Solutions for Class 8 Math Chapter 16 Area are provided here with simple step-by-step explanations. These solutions for Area are extremely popular among Class 8 students for Math Area Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Math Chapter 16 are provided here for you for free. You will also love the ad-free experience on Meritnation's Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Math are prepared by experts and are 100% accurate.

Page No 95:

Question 1:

If base of a parallelogram is 18 cm and its height is 11 cm, find its area.

Answer:

Area of a parallelogram =$\mathrm{Base}\times \mathrm{Height}=18\times 11=198{\mathrm{cm}}^2$

Page No 95:

Question 2:

If area of a parallelogram is 29.6 sq cm and its base is 8 cm, find its height.

Answer:

Area of a parallelogram = 29.6 sq cm
Area of a parallelogram =$\mathrm{Base}\times \mathrm{Height}$
$$\begin{gathered} \Rightarrow 29.6=8\times h \\ \Rightarrow h=\frac{29.6}{8}=3.7\mathrm{cm} \end{gathered}$$
Height = 3.7 cm

Page No 95:

Question 3:

Area of a parallelogram is 83.2 sq cm. If its height is 6.4 cm, find the length of its base.

Answer:

Area of a parallelogram =$\mathrm{Base}\times \mathrm{Height}$
$$\begin{gathered} \Rightarrow 83.2=b\times 6.4 \\ \Rightarrow b=13 \end{gathered}$$
Length of base = 13 cm

Page No 97:

Question 1:

Lengths of the diagonals of a rhombus are 15 cm and 24 cm, find its area.

Answer:

Area of rhombus =$\frac{1}{2}\times \left(\mathrm{Product}\mathrm{of}\mathrm{diagonals}\right)$
$\Rightarrow \mathrm{Area}\mathrm{of}\mathrm{rhombus}=\frac{1}{2}\times 15\times 24=180{\mathrm{cm}}^2$

Page No 97:

Question 2:

Lengths of the diagonals of a rhombus are 16.5 cm and 14.2 cm, find its area.

Answer:

Area of rhombus =

$\frac{1}{2}\times \left(\mathrm{Product}\mathrm{of}\mathrm{diagonals}\right)$ $\mathrm{Area}\mathrm{of}\mathrm{rhombus}=\frac{1}{2}\times 16.5\times 14.2=117.15{\mathrm{cm}}^2$

Page No 97:

Question 3:

If perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is the area of the quadrilateral ?

Answer:

Perimeter of the rhombus = 100 cm
$$\begin{gathered} \Rightarrow 4\times \mathrm{side}=100 \\ \Rightarrow \mathrm{side}=\frac{100}{4}=25\mathrm{cm} \end{gathered}$$
Thus, each side of the rhombus = 25 cm.
Diagonals of a rhombus bisect each other at 90$°$.
So, AO = OC =$\frac{48}{2}=24\mathrm{cm}$
In$△$AOB
We apply Pythagoras theorem,
$$\begin{gathered} {\mathrm{AO}}^2+{\mathrm{OB}}^2={\mathrm{AB}}^2 \\ \Rightarrow {24}^2+{\mathrm{OB}}^2={25}^2 \\ \Rightarrow {\mathrm{OB}}^2=625-576=49 \\ \Rightarrow \mathrm{OB}=7\mathrm{cm} \end{gathered}$$
So, DB =$2\times \mathrm{OB}=2\times 7=14\mathrm{cm}$
Area of rhombus =$\frac{1}{2}\times \left(\mathrm{Product}\mathrm{of}\mathrm{diagonals}\right)=\frac{1}{2}\times 14\times 48=336{\mathrm{cm}}^2$

Page No 97:

Question 4:

If length of a diagonal of a rhombus is 30 cm and its area is 240 sq cm, find its perimeter.

Answer:

Let the other diagonal be d cm
Area of a rhombus =$\frac{1}{2}\left(\mathrm{Product}\mathrm{of}\mathrm{diagonals}\right)$
$$\begin{gathered} \Rightarrow 240=\frac{1}{2}\times \left(30\times d\right) \\ \Rightarrow d=\frac{240\times 2}{30}=16 \end{gathered}$$
AC = 30 cm
DB = 16 cm
Diagonals of a rhombus bisect at right angles.
In$△$AOB,
$$\begin{gathered} {\mathrm{AO}}^2+{\mathrm{OB}}^2={\mathrm{AB}}^2 \\ \Rightarrow {15}^2+8^2={\mathrm{AB}}^2 \\ \Rightarrow {\mathrm{AB}}^2=225+64=289 \\ \Rightarrow \mathrm{AB}=\sqrt{289}=17\mathrm{cm} \end{gathered}$$
Thus, the side of the rhombus = 17 cm
Perimeter =$4\times 17=68$ cm

Page No 99:

Question 1:

In ☐ ABCD, l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm, find the area of ☐ ABCD.

Answer:

Draw perpendicular from c to line AB. Name the point E.
CE = AD = 8 cm
EB =

$\mathrm{AB}-\mathrm{AE}=\mathrm{AB}-\mathrm{CD}=13-9=4\mathrm{cm}$

Area of rectangle AECD =

$l\times b=9\times 8=72{\mathrm{cm}}^2$

Area of Triangle BEC =

$\frac{1}{2}\times b\times h=\frac{1}{2}\times 8\times 4=16{\mathrm{cm}}^2$

Area of ☐ ABCD = Area of AECD + Area of triangle BEC = 72 + 16 = 88 cm²

Page No 99:

Question 2:

Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm respectively and its height is 4.2 cm, find its area.

Answer:

Area of trapezium =$\frac{1}{2}\left(\mathrm{sum}\mathrm{of}\mathrm{parallel}\mathrm{sides}\right)\times \mathrm{height}$
$$\begin{gathered} =\frac{1}{2}\times \left(8.5+11.5\right)\times 4.2 \\ =\frac{1}{2}\times 20\times 4.2 \\ =42{\mathrm{cm}}^2 \end{gathered}$$

Page No 99:

Question 3:

☐ PQRS is an isosceles trapezium l(PQ) = 7 cm. seg PM ⊥ seg SR, l(SM) = 3 cm, Distance between two parallel sides is 4 cm, find the area of ☐ PQRS

Answer:

Draw a perpendicular from Q to line MR. Where it meets the line MR name it point N.
MN = PQ = 7 cm
In$△$PMS,
$$\begin{gathered} {\mathrm{PM}}^2+{\mathrm{SM}}^2={\mathrm{PS}}^2 \\ \Rightarrow 4^2+3^2={\mathrm{PS}}^2 \\ \Rightarrow {\mathrm{PS}}^2=16+9=25 \\ \Rightarrow \mathrm{PS}=5\mathrm{cm} \end{gathered}$$
PQRS is an isosceles trapezium so, PS = QR = 5 cm
PM = QN = 4 cm
So, NR = SM = 3 cm
SR = SM + MN + NR = 3 + 7 + 3 = 13 cm
Area of trapezium PQRS =$\frac{1}{2}\times \left(\mathrm{sum}\mathrm{of}\mathrm{parallel}\mathrm{sides}\right)\times \mathrm{height}$
$$\begin{gathered} =\frac{1}{2}\times \left(7+13\right)\times 4 \\ =40{\mathrm{cm}}^2 \end{gathered}$$

Page No 101:

Question 1:

Sides of a triangle are cm 45 cm, 39 cm and 42 cm, find its area.

Answer:

a = 45 cm
b = 39 cm
c = 42 cm
$s=\frac{a+b+c}{2}=\frac{45+39+42}{2}=63$
$$\begin{gathered} \mathrm{Area}=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{63\left(63-45\right)\left(63-39\right)\left(63-42\right)} \\ =\sqrt{63\times 18\times 24\times 21} \\ =756{\mathrm{cm}}^2 \end{gathered}$$

Page No 101:

Question 2:

Look at the measures shown in the adjacent figure and find the area of ☐ PQRS.

Answer:

Join PR.
In triangle PSR,
Applying Pthygoras theorem,
$$\begin{gathered} {\mathrm{PS}}^2+{\mathrm{SR}}^2={\mathrm{PR}}^2 \\ \Rightarrow {36}^2+{15}^2={\mathrm{PR}}^2 \\ \Rightarrow {\mathrm{PR}}^2=1296+225=1521 \\ \Rightarrow \mathrm{PR}=\sqrt{1521}=39\mathrm{m} \end{gathered}$$
In triangle PQR,
$$\begin{gathered} s=\frac{56+25+39}{2}=60 \\ \mathrm{Area}=\sqrt{60\left(60-56\right)\left(60-25\right)\left(60-39\right)} \\ =\sqrt{60\times 4\times 35\times 21} \\ =\sqrt{176400} \\ =420{\mathrm{m}}^2 \end{gathered}$$
Area of triangle PSR =$\frac{1}{2}\times 15\times 36=270{\mathrm{m}}^2$
Area of PQRS = Area of triangle PSR + Area of triangle PQR
= 270 + 420
= 690 m²

Page No 101:

Question 3:

Some measures are given in the adjacent figure, find the area of ☐ABCD.

Answer:

Area of

$△$

BDC =

$\frac{1}{2}\times 13\times 60=390{\mathrm{m}}^2$

Area of

$△$

BAD =

$\frac{1}{2}\times AB\times AD=\frac{1}{2}\times 40\times 9=180{\mathrm{m}}^2$

Area of ☐ABCD = Ar of

$△$

BDC + Ar of

$△$

BAD
= 390 + 180
= 570 m²

Page No 102:

Question 1:

Find the areas of given plots. (All measures are in metres.)
(1)

(2)

Answer:

(1)
Ar of

$△$

PQA =

$\frac{1}{2}\times \mathrm{QA}\times \mathrm{PA}=\frac{1}{2}\times 30\times 50=750$${\mathrm{m}}^2$

Ar of

$△$

PBT =

$\frac{1}{2}\times \mathrm{PB}\times \mathrm{BT}$

 =

$\frac{1}{2}\times 60\times 30=900{\mathrm{m}}^2$

Ar of

$△$

SBT =

$\frac{1}{2}\times \mathrm{SB}\times \mathrm{BT}=\frac{1}{2}\times 90\times 30=1350{\mathrm{m}}^2$

Ar of

$△$

RCS =

$\frac{1}{2}\times \mathrm{RC}\times \mathrm{CS}=\frac{1}{2}\times 25\times 60=750{\mathrm{m}}^2$

Ar of QRCA =

$\frac{1}{2}\left(\mathrm{QA}+\mathrm{RC}\right)\times \mathrm{AC}=\frac{1}{2}\times \left(50+25\right)\times 60=2250$

 m²
Ar os PQRST = Ar of

$△$

PQA + Ar of

$△$

PBT + Ar of

$△$

SBT + Ar of

$△$

RCS + Ar of QRCA
= 750 + 900 + 1350 + 750 + 2250
= 6000 m²

(2) Disclaimer: The information given is insufficient to solve the question.

Page No 104:

Question 1:

Radii of the circles are given below, find their areas.
(1) 28 cm
(2) 10.5 cm
(3) 17.5 cm

Answer:

(1) 28 cm

$$\begin{gathered} \mathrm{Ar}=\mathrm{\pi }{r}^2=\mathrm{\pi }\times {\left(28\right)}^2 \\ =\frac{22}{7}\times 28\times 28 \\ =2464\mathrm{sq}\mathrm{cm} \end{gathered}$$

(2) 10.5 cm

$$\begin{gathered} \mathrm{Ar}=\mathrm{\pi }{r}^2=\mathrm{\pi }\times {\left(10.5\right)}^2 \\ =\frac{22}{7}\times 10.5\times 10.5 \\ =346.5\mathrm{sq}\mathrm{cm} \end{gathered}$$

(3) 17.5 cm

$$\begin{gathered} \mathrm{Ar}=\mathrm{\pi }{r}^2=\mathrm{\pi }\times {\left(17.5\right)}^2 \\ =\frac{22}{7}\times 17.5\times 17.5 \\ =962.5\mathrm{sq}\mathrm{cm} \end{gathered}$$

Page No 104:

Question 2:

Areas of some circles are given below find their diameters.
(1) 176 sq cm
(2) 394.24 sq cm
(3) 12474 sq cm

Answer:

(1) 176 sq cm

$$\begin{gathered} \mathrm{Ar}=\mathrm{\pi }{r}^2 \\ \Rightarrow 176=\frac{22}{7}\times r^2 \\ \Rightarrow r^2=\frac{176\times 7}{22}=56 \\ \Rightarrow r=\sqrt{56} \\ d=2r=2\sqrt{56} \end{gathered}$$

(2) 394.24 sq cm

$$\begin{gathered} \mathrm{Ar}=\mathrm{\pi }{r}^2 \\ \Rightarrow 394.24=\frac{22}{7}\times r^2 \\ \Rightarrow r^2=\frac{394.24\times 7}{22}=125.44 \\ \Rightarrow r=11.2 \\ d=2r=2\times 11.2=22.4\mathrm{cm} \end{gathered}$$

(3) 12474 sq cm

$$\begin{gathered} \mathrm{Ar}=\mathrm{\pi }{r}^2 \\ \Rightarrow 12474=\frac{22}{7}\times r^2 \\ \Rightarrow r^2=\frac{12474\times 7}{22}=3969 \\ \Rightarrow r=63 \\ d=2r=2\times 63=126\mathrm{cm} \end{gathered}$$

Page No 104:

Question 3:

Diameter of the circular garden is 42 m. There is a 3.5 m wide road around the garden. Find the area of the road.

Answer:

Diameter of the garden(d) = 42 m
Radius, r= 21 m
Diameter of the garden including the road = 42 + 3.5 + 3.5 = 49 m
Radius of the garden with the road = 24.5 m
Area of road =
$$\begin{gathered} \mathrm{Ar}\mathrm{of}\mathrm{garden}\mathrm{with}\mathrm{road}-\mathrm{Ar}\mathrm{of}\mathrm{garden} \\ =\mathrm{\pi }{R}^2-\mathrm{\pi }{r}^2 \\ =\mathrm{\pi }\left(R^2-r^2\right) \\ =\frac{22}{7}\left({\left(24.5\right)}^2-{21}^2\right) \\ =\frac{22}{7}\times \left(600.25-441\right) \\ =\frac{22}{7}\times 159.25 \\ =500.5{\mathrm{m}}^2 \end{gathered}$$

Page No 104:

Question 4:

Find the area of the circle if its circumfence is 88 cm.

Answer:

Circumference = 88 cm
$$\begin{gathered} \Rightarrow 2\mathrm{\pi }r=88 \\ \Rightarrow r=\frac{88\times 7}{2\times 22}=14\mathrm{cm} \end{gathered}$$
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{circle}=\mathrm{\pi }{r}^2=\frac{22}{7}\times {\left(14\right)}^2 \\ =616{\mathrm{cm}}^2 \end{gathered}$$

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