PLANES

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.

A plane surface forms the plane

figure of limited surfaces. We
know about the following plane
figures:

 Rectangle
 Square
 Triangle
 Circle

These are all closed plane figures

closed by line-segments. Now,
we have to know more about
such closed figures.

TRIANGLE
Definition of a Triangle
Triangle is a closed figure bounded by three straight lines called sides. It can also be
defined as polygon of three sides.

Area of triangle
The area of the triangle is given by the following
formulas:
Given the base and the altitude
 𝟏
𝐀 = 𝟐 𝐛𝐡
Given two sides and included angle

𝟏
𝐀= 𝐚𝐛𝐬𝐢𝐧(𝛉)
𝟐
Given three sides

𝑨 = √𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄)

Where, 𝑠 = called the semi-perimeter

Given one side and three angles (say angles A, B, and C, and side b are given)

𝒃𝟐 𝒔𝒊𝒏𝑨𝒔𝒊𝒏𝑪
𝑨=
𝟐𝒔𝒊𝒏𝑩
CENTERS OF TRIANGLE

Incenter
Incenter is the center of the inscribed circle (incircle) of the triangle, it is the point
of intersection of the angle bisectors of the triangle.

The radius of incircle is given by the formula

𝑨𝑻
𝒓=
𝒔
Where AT=Area of triangle and s=(a+b+c)/2

Circumcenter
Circumcenter is the point of intersection of perpendicular bisectors of the triangle.
It is also the center of the circumscribing circle (circumcircle).
 𝒂𝒃𝒄
𝑹=
𝟒𝑨𝑻

Where AT=Area of triangle

Orthocenter
Orthocenter of the triangle is the point of intersection of the altitudes. Like
circumcenter, it can be inside or outside the triangle as shown in the figure
below.

Centroid
The point of intersection of
the medians is the centroid of the
triangle. Centroid is the geometric
center of a plane figure.
Euler Line
The line that would pass through the orthocenter, circumcenter, and centroid of
the triangle is called the Euler line.

PROPERTIES OF A TRIANGLE
Side
Side of a triangle is a line segment that connects two vertices. Triangle has three
sides, it is denoted by a, b, and c in the figure below.

Vertex
Vertex is the point of intersection of two sides of triangle. The three vertices of
the triangle are denoted by A, B, and C in the figure below. Notice that the
opposite of vertex A is side a, opposite to vertex B is side B, and opposite to
vertex C is side c.

Included Angle or Vertex Angle

Included angle is the angle subtended by
two sides at the vertex of the triangle. It is
also called vertex angle. For convenience,
each included angle has the same notation
to that of the vertex, ie. angle A is the
included angle at vertex A, and so on. The
sum of the included angles of the triangle is
always equal to 180°.

𝑨 + 𝑩 + 𝑪 = 𝟏𝟖𝟎
 Altitude, h
Altitude is a line from vertex perpendicular to the opposite side. The altitudes of
the triangle will intersect at a common point called orthocenter.

If sides a, b, and c are known, solve one of the angles using Cosine Law then
solve the altitude of the triangle by functions of a right triangle. If the area of the
triangle At is known, the following formulas are useful in solving for the altitudes.

2𝐴
ℎ =
𝑎
Base
The base of the triangle is relative to which altitude is being considered. Figure
below shows the bases of the triangle and its corresponding altitude.

 If hA is taken as altitude then side a is the base

 If hB is taken as altitude then side b is the base
 If hC is taken as altitude then side c is the base

Median, m
Median of the triangle is a line from vertex to the midpoint of the opposite side. A
triangle has three medians, and these three will intersect at the centroid. The
figure below shows the median through A denoted by mA.
 Given three sides of the triangle, the median can be solved by two steps.
1. Solve for one included angle, say angle C, using Cosine Law. From the figure
above, solve for C in triangle ABC.
2. Using triangle ADC, determine the median through A by Cosine Law.
The formulas below, though not recommended, can be used to solve for the
length of the medians.
4𝑚 = 2𝑏 + 2𝑐 − 𝑎
4𝑚 = 2𝑎 + 2𝑐 − 𝑏
4𝑚 = 2𝑎 + 2𝑏 − 𝑐

Where mA, mB, and mC are medians through A, B, and C, respectively.

Angle Bisector
Angle bisector of a triangle is a line that divides one included angle into two equal
angles. It is drawn from vertex to the opposite side of the triangle. Since there are
three included angles of the triangle, there are also three angle bisectors, and
these three will intersect at the incenter. The figure shown below is the bisector
of angle A, its length from vertex A to side a is denoted as bA.
Perpendicular Bisector
Perpendicular bisector of the triangle is a perpendicular line that crosses through
midpoint of the side of the triangle. The three perpendicular bisectors are worth
noting for it intersects at the center of the circumscribing circle of the triangle.
The point of intersection is called the circumcenter. The figure below shows the
perpendicular bisector through side b.

SAMPLE PROBLEMS

1. Determine the area of the following triangles:

B B

Fig. 1 Fig. 2

a) For figure 1, B = 4 cm, Altitude = 9 cm

𝐴 = (4)(9) = 𝟏𝟖𝒄𝒎𝟐

b) For figure 2, B = 5 cm, Altitude = 6 cm

𝐴 = (5)(6) = 𝟏𝟓𝒄𝒎𝟐
2. Find the area of a right triangle whose sides measure x, 2x-1 and 2x+1

By Pythagorean Theorem:
𝑥 + (2𝑥 − 1) = (2𝑥 + 1)
𝑥=8
2𝑥 − 1 = 15
The area of triangle is:
1
𝐴 = (8)(15) = 𝟔𝟎 𝒔𝒒. 𝒖𝒏𝒊𝒕𝒔
2
3. Find the area of a triangle given three sides: a= 7m, b = 9m, and c =15m.

Using heron’s formula,

𝑠= = = 15.5
𝐴= 𝑠(𝑠 − 𝑎)(𝑠 − 𝑏)(𝑠 − 𝑐) = √15.5(15.5 − 7)(15.5 − 9)(15.5 − 15)

A=20.69m2

4. In triangle ABC, sin A: sin B : sin C = 1 : 2 : 3. If b = 4 cm. Then the perimeter

of the triangle is?

b= 4 cm

Using proportion: sinA:sinB:sinC = 1:2:3

Therefore, side a: side b : side c = 1 : 2 : 3 ; b = 4cm.

a = b/2 = 4/2 = 2 cm c = a x 3 = 2 x 3 = 6cm

P = a + b+ c = 2 + 4 + 6 = 12

P = 12 cm

5. Two sides of a triangle measure 67m and 50m. If its area is 1, 400 m2, find its
perimeter.

P = a + b +c

One side is missing, so we are looking for the missing side first.
 1
𝐴 = 𝑎𝑏𝑠𝑖𝑛𝜃
2
1
1400 = (67)(50)𝑠𝑖𝑛𝜃
2
𝜃 = 56.70°

Use cosine law to determine the missing side for the perimeter

𝑥 = 50 + 67 − 2 (50)(67)𝑐𝑜𝑠56.70°

𝑥 = 57.5385 𝑚

𝑃 = 50 + 67 + 57.5385

𝑷 = 𝟏𝟕𝟒. 𝟓𝟑𝟖𝟓 𝒎

6. In triangle ABC, BC = 216mm, AC = 301mm and angle A = 30˚. Find the area
of the triangle.

By the sine law,

𝑎 𝑏 𝑐
= =
𝑠𝑖𝑛𝐴 𝑠𝑖𝑛𝐵 𝑠𝑖𝑛𝐶
216 301 𝑐
= =
𝑠𝑖𝑛30 𝑠𝑖𝑛𝐵 𝑠𝑖𝑛𝐶
Solving for B, we obtain B = 44.1676˚

Solve for C = 180 ˚ - 30 ˚- 44.1676 ˚ = 105.8324 ˚

1
𝐴 = 𝑎𝑏𝑠𝑖𝑛𝐶
2
1
𝐴 = (216)(301)𝑠𝑖𝑛105.8324 ˚
2
𝐴 = 31, 274.7722 𝑚𝑚
 QUADRILATERALS

Quadrilateral is a polygon of four sides and four vertices. It is also

called tetragon and quadrangle. In the triangle, the sum of the interior angles is
180°; for quadrilaterals the sum of the interior angles is always equal to 360°

𝐴 + 𝐵 + 𝐶 + 𝐷 = 360

Common Quadrilaterals

Square

Area, A = a

Perimeter, P=4a

Diagonal, d = a√2

Rectangle
Area, A = ab

Perimeter, P=2(a+b)

Diagonal, d = √a + 𝑏

Rhombus

Area, A = a 𝑠𝑖𝑛𝜃 = 𝑎ℎ

Perimeter, P=4a

Shorter Diagonal, d = a√2 − 2𝑐𝑜𝑠𝜃

Longer Diagonal, d = a√2 + 2𝑐𝑜𝑠𝜃

Note: The diagonals of square and rhombus are perpendicular to each other.

Parallelogram
Area, A = ab𝑠𝑖𝑛𝜃 = 𝑏ℎ

Perimeter, P=2(a+b)

Shorter Diagonal, d = √a + 𝑏 − 2𝑎𝑏𝑐𝑜𝑠𝜃

Longer Diagonal, d = √a + 𝑏 + 2𝑎𝑏𝑐𝑜𝑠𝜃

Trapezoid
 𝟏
Area, A = (a + b)h
𝟐

Sample Problems

1. Square ABCD with a perimeter of 48 units. Find the number of units in BD.

Each side of the square must be 12 units.

12² + 12² = (BD)²; 144 + 144 = (BD)²;
BD = 𝟏𝟐√𝟐 units

2. Parallelogram ABCD, sides as marked.

Find AD.

6x - 10 = 3x + 5 (opposite sides)
3x - 10 = 5; 3x = 15; x = 5
AD = 4x - 5 = 4(5) - 5 = 15

3. Rectangle MATH, diagonals = 36 in.

Find x and y.

Each diagonal segment = 18.

2x + 4y = 18 and 4x - y = 18
Solve simultaneously.
x = 5 (as shown at the right)
2x + 4y = 18
2(5) + 4y = 18
4y = 8
y=2
4. Parallelogram DEFG, ∠s as marked.
Find m∠F.

∠D supplementary to ∠E
(2x+12) + (5x) = 180
7x + 12 = 180; 7x = 168; x = 24
m∠F = 2(24) + 12 = 60º

5. The distance from a point inside a square to its vertices are 3, 4 and 5m
consecutively. Find the length of the side of the square.

d = 3 + 5 = 8m
3 4
𝑑 = 𝑎 √2
5
8 = 𝑎 √2

𝒂 = 𝟓. 𝟔𝟓𝒎

6. How much would be the perimeter of a square be increased if its area is

doubled?

Get the side of the doubled area to know how much the perimeter would
be increased.

𝐴𝑠𝑞𝑢𝑎𝑟𝑒 = 𝑎

2= 𝑎

𝑎 = √2 = 1.414

𝒂 = 𝟏. 𝟒𝟏𝟒

7. The sides of a parallelogram measure 68 cm and 83 cm and one of the diagonals

measures 42cm. Solve for the largest interior angle of the parallelogram.

Check whether the diagonal is short or long:

𝑑 < √𝑎 + 𝑏 𝑑 > √𝑎 + 𝑏
 42 < 68 + 83 = 107.2986 𝑐𝑚 ; 𝑆ℎ𝑜𝑟𝑡𝑒𝑟 𝑑𝑖𝑎𝑔𝑜𝑛𝑎𝑙

By cosine law, solve for the smallest interior angle A (A = C) .

(𝐷𝐵) = (𝐵𝐴) + (𝐴𝐷) − 2(𝐵𝐴)(𝐴𝐷) cos 𝐴

42 = 68 + 83 − 2(68)(83) cos 𝐴

A = 30.2699˚

To solve for the largest interior angle, since it is a supplement. Subtract the
smallest angle to 180˚.

B = 180˚ - 30.2699˚

B = 149.7031˚

8. The sides of a parallelogram measure 25 cm and 40 cm. One of its interior

angles measures 78˚. If the consecutive midpoints of the sides are connected, find
the area of the figure thus formed.

The area of the shaded figure is

𝐴 = 𝐴 − 𝐴∆

𝐴 = (40)(25)𝑠𝑖𝑛78° − 𝑠𝑖𝑛78° (2) −

𝑠𝑖𝑛78° (2)

𝑨𝒔𝒉𝒂𝒅𝒆𝒅 = 𝟒𝟖𝟗. 𝟎𝟕𝟑𝟖 𝒄𝒎𝟐

9. Each side of a rhombus measures 105m. If the distance between its parallel
side is 20m, find its area.

The area of the rhombus given its side and altitude is

A = bh

b = 105m, h = 20m
 A = 105(20)

A = 2, 100 m2

10. The diagonals of rhombus measure 50mm and 96mm. Find its area.

To find the area of rhombus given diagonals:

𝐴= 𝑑 𝑑

1
𝐴= (50)(96)
2
A = 2, 400 mm2.

11. The bases of the trapezoid are 18m and 26m. If the bases are 20m apart, find
the area of the trapezoid.

𝐴 = (𝑏 + 𝑏 )ℎ

𝐴 = (18 + 26)20

𝐴 = 440 𝑚

12. The base width of a trapezoidal channel is 3m and the sides are sloping at 2
vertical to 1 horizontal. Water is flowing at depth of 1.20m, find the area of flow.

By ratio and proportion:

.
=

𝑥 = 0.60𝑚

The top width T is

T = 2x + 3

T = 2(0.60) + 3

T = 4.2m

The area of the trapezoid given the bases and the altitude is

𝐴 = (𝑏 + 𝑏 )ℎ
 1
𝐴 = (4.20 + 3)1.20
2
𝐴 = 4.32 𝑚