8th Maths : Chapter 2 : Measurements : Perimeter and Area of Combined shapes: Numerical Example solved problems

__Example 2.5__

Find the
perimeter and area of the given Fig.2.23. (π = 22/7)

**Solution:**

Radius of
a circular quadrant, *r* = 3.5 *cm* and side of a square, *a* = 3.5 *cm*.

The given
figure is formed by the joining of 4 quadrants of a circle with each side of a square.
The boundary of the given figure consists of 4 arcs and 4 radii.

(i) Perimeter
of the given combined shape

= 4 ×
length of the arcs of the quadrant of a circle + 4 ×
radius

= 22 + 14
= 36 *cm* (approximately)

(ii) Area
of the given combined shape

= area of
the square + 4 × area of the quadrants of the circle

*A *= 12.25+38.5 = 50.75* cm*^{2}* *(approximately)

** **

__Example 2.6__

Nishanth
has a key-chain which is in the form of an equilateral triangle and a semicircle
attached to a square of side 5 *cm* as
shown in the Fig. 2.24. Find its area. ( *π*
=
3.14, √3 = 1.732)

*Solution:*

Side of the
square = 5 *cm*

Diameter
of the semi-circle = 5 *cm* ⇒ Radius = 2.5 *cm*

Side of the
equilateral triangle = 5 *cm*

∴
Area of the key-chain = area of the semi circle + area of the square + area of the
equilateral triangle

= 9.81+
25 +
10.83

= 45.64 *cm*^{2} (approximately)

** **

__Example 2.7____ __

A 3-fold
invitation card is given with measures as in the Fig. 2.25. Find its area.

*Solution:*

Figures I and II are trapeziums separately as well as combinedly.

The parallel
sides of the combined trapezium (I and II) are 5 *cm* and 16 *cm* and its height,
*h* = 8 + 8 =
16*cm* , length of the rectangle (III) =
16 *cm* and its breadth = 8 *cm*

Area of the
combined invitation card = area of the combined trapeziu*m + *area of the rectangle

= 1/2 ×*h*×(*a* +*b*)+ *l* ×*b*

= 1/2×16×(5 +16)+16×8

= 168 +
128 =
296 *cm*^{2}

**Aliter:**

Area of the
invitation card = area of the outer rectangle – area of the right angled triangle

*= *[*l *×*b*]* *–
[ 1/2* *×*b*×*h *]

= [ 24×16
] –
[ 1/2 ×11×16
]

= 384 − 88 = 296 *cm*^{2}

** **

__Example 2.8__

Seenu wants
to buy a floor mat for his kitchen at home as given in Fig. 2.27. If the cost of
the mat is ₹ 20 per square foot, what will be the cost of the entire mat?

*Solution:*

The mat given
in the figure can be split into two rectangles as follows:

Area of the
entire mat = area of the I rectangle + area of the II rectangle

*= l*_{1}* *×*b*_{1}* *+* l*_{2}* *×*b*_{2}

= 5 ×2
+
9 ×2
=
10 +18
=28
*sq.feet*

Cost per
sq. foot = ₹ 20

The total cost of the entire mat = 28 × ₹ 20 = ₹ 560

**Try these**

In the above example split the given mat into two trapeziums and
verify your answer.

**Solution:**

Area of the mat = Area of I trapezium + Area of II trapezium

= [ 1/2 × *h*_{1} × (*a*_{1} + *b*_{1}
)] + [ 1/2 × *h*_{2} × (*a*_{2} + *b*_{2}
)] sq. units

= [ 1/2 × 2 × (7 + 5)] + 1/2
× 2 × (9 + 7) sq. feet

= 12 + 16 = 28 sq.feet

∴ Cost per sq.feet = ₹ 20

Cost for 28 sq. feet = ₹20
× 28 = ₹560

∴ Total cost for the entire mat = ₹560

Both the answers are the same.

** **

__Example 2.9__

Find the
area of the shaded region in the square of side 10 *cm* as given in the Fig. 2.29. (π = 22/7)

*Solution:*

Mark the
unshaded parts of the given figure as I, II, III and IV

Area of the I and III parts = Area of the square – Area of 2 semicircles

= 10×10
–
22/7 ×5×5
= 100 – 78.57 = 21.43 *cm*^{2}.

Similarly,
the area of the II and IV parts = 21.43 *cm*^{2}

Area of the
unshaded parts (I, II, III and IV)

= 21.43 ×2
= 42.86 *cm*^{2} (approximately)

Area of the
shaded part = area of the square – area of the unshaded parts

= 100 – 42.86
= 57.14 *cm*^{2} (approximately)

**Do you Know:**

1. The area of the unshaded regions in each of the squares of side
*a* units are the same in all the cases
given below.

2. If the biggest circle is cut from a square of side ‘*a*’ *units*,
then the remaining area in the square is approximately 3/14 *a*^{2} *sq.units*. ( π = 22/7)

3. The area of the biggest circle cut out from the square of ‘a’
units *= *11/14 *a*^{2} *sq. units* (approximately)

4. In the given figure if *π =* 22/7 , the area of the
unshaded part of a square of side *a* units
is approximately 3/7 *a*^{2} *sq.units* and that of the shaded part is approximately
4/7 *a*^{2} *sq.units*.

** **

__Example 2.10__

Find the
area of an irregular polygon field whose measures are as given in the Fig. 2.30.

*Solution:*

The given
field has four triangles (I, III, IV and V) and a trapezium (II).

∴ The total area of the field = 15 + 65
+ 16 + 65 + 65 = 226 *m*^{2}

^{}

Tags : Questions with Answers, Solution | Measurements | Chapter 2 | 8th Maths , 8th Maths : Chapter 2 : Measurements

Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail

8th Maths : Chapter 2 : Measurements : Perimeter and Area of Combined shapes - Example solved problems | Questions with Answers, Solution | Measurements | Chapter 2 | 8th Maths

**Related Topics **

Privacy Policy, Terms and Conditions, DMCA Policy and Compliant

Copyright © 2018-2023 BrainKart.com; All Rights Reserved. Developed by Therithal info, Chennai.