Mathematics Solutions Solutions for Class 8 Maths Chapter 16 - Area

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Mathematics Solutions Solutions for Class 8 Maths Chapter 16 Area are provided here with simple step-by-step explanations. These solutions for Area are extremely popular among class 8 students for Maths 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 Maths 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 8 Maths 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 = $Base \times Height = 18 \times 11 = 198 {cm}^{2}$

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 = $Base \times Height$
$\Rightarrow 29.6 = 8 \times h \Rightarrow h = \frac{29.6}{8} = 3.7 cm$
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 = $Base \times Height$
$\Rightarrow 83.2 = b \times 6.4 \Rightarrow b = 13$
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 = $\frac{1}{2} \times (Product of diagonals)$
$\Rightarrow Area of rhombus = \frac{1}{2} \times 15 \times 24 = 180 {cm}^{2}$

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 = $\frac{1}{2} \times (Product of diagonals)$

$Area of rhombus = \frac{1}{2} \times 16.5 \times 14.2 = 117.15 {cm}^{2}$

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

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

Page No 99:

Question 1:

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

Answer:

Draw perpendicular from c to line AB. Name the point E.
CE = AD = 8 cm
EB = $AB - AE = AB - CD = 13 - 9 = 4 cm$
Area of rectangle AECD = $l \times b = 9 \times 8 = 72 {cm}^{2}$
Area of Triangle BEC = $\frac{1}{2} \times b \times h = \frac{1}{2} \times 8 \times 4 = 16 {cm}^{2}$
Area of â˜ 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 = $\frac{1}{2} (sum of parallel sides) \times height$
$= \frac{1}{2} \times (8.5 + 11.5) \times 4.2 = \frac{1}{2} \times 20 \times 4.2 = 42 {cm}^{2}$

Page No 99:

Question 3:

â˜ 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 â˜ 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,
${PM}^{2} + {SM}^{2} = {PS}^{2} \Rightarrow 4^{2} + 3^{2} = {PS}^{2} \Rightarrow {PS}^{2} = 16 + 9 = 25 \Rightarrow PS = 5 cm$
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 = $\frac{1}{2} \times (sum of parallel sides) \times height$
$= \frac{1}{2} \times (7 + 13) \times 4 = 40 {cm}^{2}$

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

Page No 101:

Question 2:

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

Answer:

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

Answer:

Area of $△$ BDC = $\frac{1}{2} \times 13 \times 60 = 390 m^{2}$
Area of $△$ BAD = $\frac{1}{2} \times A B \times A D = \frac{1}{2} \times 40 \times 9 = 180 m^{2}$
Area of â˜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 = $\frac{1}{2} \times QA \times PA = \frac{1}{2} \times 30 \times 50 = 750$ $m^{2}$
Ar of $△$ PBT = $\frac{1}{2} \times PB \times BT$ = $\frac{1}{2} \times 60 \times 30 = 900 m^{2}$
Ar of $△$ SBT = $\frac{1}{2} \times SB \times BT = \frac{1}{2} \times 90 \times 30 = 1350 m^{2}$
Ar of $△$ RCS = $\frac{1}{2} \times RC \times CS = \frac{1}{2} \times 25 \times 60 = 750 m^{2}$
Ar of QRCA = $\frac{1}{2} (QA + RC) \times AC = \frac{1}{2} \times (50 + 25) \times 60 = 2250$ 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
$Ar = π r^{2} = π \times {(28)}^{2} = \frac{22}{7} \times 28 \times 28 = 2464 sq cm$

(2) 10.5 cm
$Ar = π r^{2} = π \times {(10.5)}^{2} = \frac{22}{7} \times 10.5 \times 10.5 = 346.5 sq cm$

(3) 17.5 cm
$Ar = π r^{2} = π \times {(17.5)}^{2} = \frac{22}{7} \times 17.5 \times 17.5 = 962.5 sq cm$

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
$Ar = π r^{2} \Rightarrow 176 = \frac{22}{7} \times r^{2} \Rightarrow r^{2} = \frac{176 \times 7}{22} = 56 \Rightarrow r = \sqrt{56} d = 2 r = 2 \sqrt{56}$

(2) 394.24 sq cm
$Ar = π r^{2} \Rightarrow 394.24 = \frac{22}{7} \times r^{2} \Rightarrow r^{2} = \frac{394.24 \times 7}{22} = 125.44 \Rightarrow r = 11.2 d = 2 r = 2 \times 11.2 = 22.4 cm$

(3) 12474 sq cm
$Ar = π r^{2} \Rightarrow 12474 = \frac{22}{7} \times r^{2} \Rightarrow r^{2} = \frac{12474 \times 7}{22} = 3969 \Rightarrow r = 63 d = 2 r = 2 \times 63 = 126 cm$

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 =
$Ar of garden with road - Ar of garden = π R^{2} - π r^{2} = π (R^{2} - r^{2}) = \frac{22}{7} ({(24.5)}^{2} - 21^{2}) = \frac{22}{7} \times (600.25 - 441) = \frac{22}{7} \times 159.25 = 500.5 m^{2}$

Page No 104:

Question 4:

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

Answer:

Circumference = 88 cm
$\Rightarrow 2 π r = 88 \Rightarrow r = \frac{88 \times 7}{2 \times 22} = 14 cm$
$Area of the circle = π r^{2} = \frac{22}{7} \times {(14)}^{2} = 616 {cm}^{2}$

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