MODULE 1: MENSURATION OF PLANE FIGURES

EXPECTED LEARNING OUTCOMES:

At the end of this lesson, the students should be able:

1. Describe a common mathematical terms, concepts and principles of Solid

Mensuration.
2. Compute areas of different plane figures.

PLANE FIGURES

- are flat two-dimensional (2D) shape. A plane figure can be made of straight lines,
curved lines, or both straight and curved lines.

INTERIOR AND EXTERIOR ANGLES OF A POLYGON

For a regular polygon (all sides and angles are equal) of insights, the interior
𝟑𝟔𝟎˚
angle is given by 𝟏𝟖𝟎˚ − where n is to the number of sides. The exterior angle of
𝒏
𝟑𝟔𝟎˚
a regular polygon is given by .
𝒏

EXAMPLE:
1. Calculate the interior and exterior angle of a square.
SOLUTION:
𝟑𝟔𝟎˚ 𝟑𝟔𝟎˚
Interior angle = 𝟏𝟖𝟎˚ − = 𝟏𝟖𝟎˚ −
𝒏 𝟒

Interior angle = 𝟗𝟎˚

𝟑𝟔𝟎˚ 𝟑𝟔𝟎˚
Exterior angle = =
𝒏 𝟓

Exterior angle = 𝟕𝟐˚

2. Calculate the interior and exterior angle of a regular pentagon.

SOLUTION:
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 Since pentagon is a 5-sided polygon, n=5
𝟑𝟔𝟎˚ 𝟑𝟔𝟎˚
Interior angle = 𝟏𝟖𝟎˚ − = 𝟏𝟖𝟎˚ −
𝒏 𝟓

Interior angle = 𝟏𝟎𝟖˚

𝟑𝟔𝟎˚ 𝟑𝟔𝟎˚
Exterior angle = =
𝒏 𝟒

Exterior angle = 𝟗𝟎˚

COMPUTATION OF AREAS AND PERIMETERS OF COMMON SHAPES

The picture shows the formula for both perimeter and area for common shapes:

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COMPOSITE AREAS
Often a closed figure can be identified as compromising two or more different
common figures. Such figures are called composite figures. The area of a composite
figure is the sum of the areas of the individual common figures.
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑒 𝑓𝑖𝑔𝑢𝑟𝑒 = 𝑠𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎𝑠 𝑜𝑓 𝑎𝑙𝑙 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙 𝑐𝑜𝑚𝑚𝑜𝑛 𝑓𝑖𝑔𝑢𝑟𝑒𝑠

𝐴 = 𝐴 + 𝐴 + 𝐴 +. . . + 𝐴

EXAMPLE 1:

Calculate the area and the perimeter of the

composite figure shown.

SOLUTION:

The composite figure is composed of rectangle and a triangle. To compute for the
total area, summate the area of the rectangle and the square.

For the Area:

𝐴 =𝐴 +𝐴

𝐴 = 𝐿 𝑥 𝑊 = 5𝑚 𝑥 12𝑚 = 60 𝑚

1 1
𝐴 = 𝑥 𝑏 𝑥 ℎ = 𝑥 3𝑚 𝑥 10𝑚 = 15 𝑚
2 2
𝐴 = 60 𝑚 + 15 𝑚 = 𝟕𝟓 𝒎𝟐

For the Perimeter:

Perimeter = 3m + 10m +5m +5m +12m

Perimeter = 35m

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EXAMPLE NO. 02:

Calculate the area and the perimeter of the backyard given

the following dimensions:

SOLUTION:

For the Area:

Since the figure is the trapezoid, use the formula for trapezoid.

1 1
𝐴 = 𝑥 (𝑎 + 𝑏) 𝑥 ℎ = 𝑥 (2.1 + 8.7) 𝑥 4.5
2 2

𝑨𝒃𝒂𝒄𝒌𝒚𝒂𝒓𝒅 = 𝟐𝟒. 𝟑𝟎 𝒎𝟐

For the Perimeter:

From Pythagorean Theorem:

x 4.5m X2 = (4.5)2 + (3)2

3m
X= 5.41 m
Perimeter = 2.1m +2x + 8.7m = 2.1+(2)(5.41) + 8.7
Perimeter = 21.62 meters

CIRCLES

1. AREA OF A CIRCLE:

𝐴 = 𝑑 𝑂𝑅 𝜋𝑟 where: d=diameter of the circle ; r= radius

2. CIRCUMFERENCE OF A CIRCLE:
𝐶 = 2𝞹𝒓 𝑶𝑹 𝞹𝒅

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EXAMPLE:
Compute for the area and circumference of the circle.
SOLUTION:
From the figure, given r=12 cm
For the Area:
𝐴 = 𝜋(12𝑐𝑚)
𝑨𝑪𝑰𝑹𝑪𝑳𝑬 = 𝟒𝟓𝟐. 𝟑𝟗 𝒄𝒎𝟐
For the Circumference:
𝐶 = 2𝞹𝒓 = 2𝞹(𝟏𝟐𝒄𝒎)
𝑪𝑪𝑰𝑹𝑪𝑳𝑬 = 𝟕𝟓. 𝟒𝟎 𝒄𝒎
3. ARC LENGTH
o An arc is part of the circumference of a circle.
o The arc length of a circle is calculated by finding the circumference of a
circle and multiplying by the fraction of the angle that it forms at the centre
of the circle.

𝜃
𝐴𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ = 𝑥 2𝜋𝑟
360

Where:
r= radius of the circle
θ= the angle the ends of the arc make with the centre of the circle

EXAMPLE:
Compute for the arc lengths for each of the following:
a. b.

SOLUTION:
a. θ= 360˚-90˚=270˚

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 𝜃 270
𝐴𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ = 𝑥 2𝜋𝑟 = 𝑥 2 𝑥 𝜋 𝑥 2.4𝑐𝑚
360 360
𝑨𝒓𝒄 𝒍𝒆𝒏𝒈𝒕𝒉 = 𝟏𝟏. 𝟑𝟏 𝒄𝒎
b. θ= 360˚- 45˚=315˚
𝜃 315
𝐴𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ = 𝑥 2𝜋𝑟 = 𝑥 2 𝑥 𝜋 𝑥 7.2𝑐𝑚
360 360

𝑨𝒓𝒄 𝒍𝒆𝒏𝒈𝒕𝒉 = 𝟑𝟗. 𝟓𝟖 𝒄𝒎

4. SECTOR
o A sector is a part of the area of a circle, as shown by the shaded area in the
diagram. The angle θ, at the centre of the circle, subtends the arc at AB.
The area of a sector is calculated using:

𝜃
𝐴 = 𝑥 𝜋𝑟
360

5. SEGMENT
o A segment is formed by joining two points on the circumference with a
straight line, known as a chord. This forms two areas, with the smaller area
being the minor segment and the larger area the major
segment. This is shown by the blue (minor segment) and
pink (major segment) sections in the diagram. The area of
the minor segment can be calculated by:
𝐴 =𝐴 − 𝐴
𝜃 1
𝐴 = 𝑥 𝜋𝑟 − 𝑥 𝑟 𝑠𝑖𝑛𝜃
360 2
𝜃 1
𝐴 =𝑟 ( 𝑥 𝜋 − 𝑥 𝑠𝑖𝑛𝜃)
360 2

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EXAMPLE:

From the circle shown, which has a radius of 8 cm, Compute

the following:

a. the area of the minor sector

b. the area of the minor segment of the circle.

SOLUTION:
a. Use formula for area of sector:
𝜃 95
𝐴 = 𝑥 𝜋𝑟 = 𝑥 𝜋 (8)
360 360
𝑨𝑺𝑬𝑪𝑻𝑶𝑹 = 𝟓𝟑. 𝟎𝟔 𝒄𝒎𝟐

b. Use fo.rmula for area of segment:

𝜃 1
𝐴 =𝑟 ( 𝑥 𝜋 − 𝑥 𝑠𝑖𝑛𝜃)
360 2
95 1
𝐴 = (8) ( 𝑥 𝜋 − 𝑥 𝑠𝑖𝑛(95))
360 2
𝑨𝑴𝑰𝑵𝑶𝑹 𝑺𝑬𝑮𝑴𝑬𝑵𝑻 = 𝟐𝟏. 𝟏𝟖 𝒄𝒎𝟐