Types of Triangles: Important Terms, Scalene, Acute, and Obtuse Triangle

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There are three types of triangles: Scalene Triangle, Isosceles Triangle, and Equilateral Triangle. Triangles form the foundation stone of geometry. Mainly, their utilisation is maximum in the field of architecture, construction, interior designing. Although, the concept of triangles holds crucial importance for those in the field of engineering. Triangles have several key advantages for architects, such as their commonality, structural soundness, and ease of use every day. A triangle’s strength comes from its shape, which spreads forces evenly. 

What is a Triangle?

A triangle is defined as a specific type of two dimensional polygon having only three sides, three vertices and three angles. Based on the lengths and measures of the sides and angles of a triangle, there are six basic types of triangles.

Triangles can have the same sides or they can have different sides. They can have the same angles or they can have different angles. Angles can be greater than, less than, or equal to 90 degrees.

A triangle having the same angle and different side length is known as a similar triangle. Therefore, the bigger triangle and the smaller triangle always have the same side ratio. The simplest polygon is the triangle. Polygons are closed, flat, two-dimensional shapes that have many corners. Three corners are necessary to make a polygon since one or two corners cannot form a closed shape.

Angle sum is the property of triangles that the sum of all internal angles equals 180°.

Important Terms Related to Triangles

Median: A median is the line segment connecting the vertex of the triangle to the midpoint of the opposite side of the triangle.

Angle Bisector: An angle bisector is also defined as the angle in an equilateral triangle or the non-congruent angle in an isosceles triangle.

Centroid: An intersection of a triangle’s medians (the lines connecting its vertex to its opposite vertex) is a triangle’s centroid. It divides each of the medians in the ratio 2:1, which means it is located at 13% of the distance between each side.

Altitude: A triangle’s altitude is formed by a line segment that crosses through its vertex and is perpendicular to the line containing its base. The extended base of the altitude is found where it crosses the altitude.

Orthocentre: A triangle's orthocentre is the intersection of the perpendiculars drawn from its vertices to its opposite sides.

Types of Triangles

Refer to the table below for the types of triangles and their properties.

Types Name of Triangle Properties of Triangles Diagrams
On the basis of sides Scalene Triangle
  • All sides are unequal 
  • No two sides are the same length
  • Each interior angle is different

Not all sides are non congruent 

  • The smallest side is one which is opposite to the smallest angle 
  • The longest side is the one which is opposite to the longest angle
  • No line of symmetry
  • Cannot be bisected into two equal halves
  • Each angle on a scalene triangle is half that of its opposite side when inscribed in a circle
 

Scalene Triangle

On the basis of sides Isosceles Triangle
  • Only two sides are equal
  • Only one side is unequal
  • The angles opposite to the equal sides of an isosceles triangle are equal
  • Two sides are congruent
  • One side in non congruent
  • A congruent angle is an angle opposite to the equal sides
  • Third unequal angle is an acute or obtuse
  • Circumcentre lies inside its triangle if all three angles are acute
  • Chords are the sides of the circumcentre

Isosceles Triangle

On the basis of sides Equilateral Triangle
  • All sides are equal
  • Each interior angle is of 60 degrees
  • Median, altitude, angle bisector, perpendicular bisector lie at one line
  • The angle bisector form the symmetry line of an equilateral triangle
  • Each altitude is the median

Equilateral Triangle

On the basis of interior angles Acute Triangle
  • All the angles are less than 90 degrees
  • All angles are acute
  • Circumcentre lies insides
  • Angle bisector intersects incentre
Acute Triangle
On the basis of interior angles Obtuse Triangle
  • One angle of all the angles is greater than 90 degrees
  • Two acute angles 
  • One obtuse angles
  • Median intersect at the centroid
  • Circumcentre lies outside

Obtuse Triangle

On the basis of interior angles Right Triangle
  • One of the angles is equal to 90 °
  • Other two angles are acute 
  • Hypotenuse is the side opposite to the right angle
  • Perpendicular is the side next to 90 ° angle
  • Base is the side on which triangle rests

Right Triangle

Things to Remember

  • An apex angle is a triangle whose two base angles are not congruent.
  • Area, A = ½ × base × height
  • By dividing the base in half and the apex angle in half, an isosceles triangle’s height is divided in half as well.
  • From the median, we can get two equal sides. 
  • Circumcentre lies outside the triangle, if the angle is 90°. 
  • An isosceles triangle is divided into two equal triangles by the median.
  • Median is the line which triangles into two. 
  • Orthocentre, circumcentre, incenter and centroid coincide at the same point. 
  • A circumcircle’s radius is always half its hypotenuse, and its centre is the hypotenuse’s midpoint.
  • The right angle of the hypotenuse produces three similar triangles.

Sample Questions

Ques. With the help of the following figure prove that BF/FE = BE/EC. Given that DE // AC and DF // AE. (3 Marks)

Ans. From the above triangle ABC, we can see that. 

Given, 

DE // AC

DF // AE

Prove: BBF/FE = BE/EC

Bu using Basic Proportionality Theorem,

BD/DA = BE/EC………. Equation (i)

Again applying Basic Proportionality Theorem,

BD/DA = BF/FE………. Equation (ii)

From equation (i) and equation (ii),

BE/EC = BF/FE

Hence proved.

Ques. A vertical pole of length 6 metre casts a shadow 4 metre long on the ground and at the same time a tower casts a shadow 28 metre long. Find the height of the tower. (4 Marks)

Ans. Given,

Length of the vertical pole, L1 = 6 m

Shadow of the pole, s = 4 m

Let the height of the tower be ‘x' metres.

Length of the shadow of the tower, L2= 28 m

In ΔABC and ΔDFE,

∠C = ∠E (angle of elevation)

∠B = ∠F = 90°

Using AA similarity criterion,

ΔABC ~ ΔDFE

The corresponding sides of two similar triangles are proportional.

AB/DF = BC/EF

6/x = 4/28

x = (6 ×28)/4

x = 6 × 7

x = 42

Hence, the height of the tower = 42 m.

Ques. Find PR, if ΔABC ~ ΔQRP, area (ΔABC) / area (ΔPQR) = 9/4 , AB = 18 cm and BC = 15 cm. (4 Marks)

Ans. Given that ΔABC ~ ΔQRP.

Area (ΔABC) / Area (ΔQRP) =9/4

AB = 18 cm and BC = 15 cm

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Area (ΔABC) / Area (ΔQRP) = BC2/RP2

9/4 = (15)2/RP2

RP2 = (4/9) × 225

PR2 = 100

So,

PR = 10 cm

Also, read: Area of a Triangle

Ques. Find the value of x, if DE || BC in ΔABC. (3 Marks)

Ans. In ΔABC, 

Given: DE || BC

Find: x

Using Thales’ Theorem, we get,

AD/BD = AE/EC

x/x+1 = x+3/x+5

x(x + 5) = (x + 3)(x + 1)

x2 + 5x = x2 + 3x + x + 3

x2 + 5x – x2 – 3x – x = 3

x = 3 

So, the value of x is 3 cm. 

Ques. Hypotenuse of a right triangle is 25 cm and out of the remaining two sides, one is longer than the other by 5 cm. Find the lengths of the other two sides. (4 Marks)

Ans. Let the base, AB = x cm

Altitude, BC = (x + 5) cm

In right Δ,

Using Pythagoras’ theorem

AB2 + BC2 = AC2

(x)2 + (x + 5)2 = 252

x2 + x2 + 10x + 25 – 625 = 0

2x2 + 10x – 600 = 0

x2 + 5x – 300 = 0 (Dividing both sides by 2)

x2 + 20x – 15x – 300 = 0

x(x + 20) – 15(x + 20) = 0

(x – 15)(x + 20) = 0

x – 15 = 0 or x + 20 = 0

x = 15 or x = -20

The base cannot be negative.

So,

x = 15 cm

Length of the other side = 15 + 5 = 20 cm

Two sides are = 15 cm and 20 cm.

Ques. Prove that ΔADE ~ ΔGCF, if AB ⊥ BC, FG ⊥ BC and DE ⊥ AC. (3 Marks)

Ans. In right ΔABC,

∠A + ∠C = 90° (i)

In right ΔAED,

∠A + ∠2 = 90°

From (i) and (ii), ∠C = ∠2

Similarly, ∠A = ∠1

In ΔADE & ΔGCF

∠A = 1 (Proved)

∠C = 2 (Proved)

∠AED = ∠GFC

∴ ΔADE – ΔGCF (Hence Proved)

Find more sample questions here: Triangles Important Questions

CBSE X Related Questions

1.
A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

      2.
      An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
      An umbrella has 8 ribs which are equally spaced

          3.

          The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

          Length (in mm)

          Number of leaves

          118 - 126

          3

          127 - 135 

          5

          136 - 144

          9

          145 - 153

          12

          154 - 162

          5

          163 - 171

          4

          172 - 180

          2

          Find the median length of the leaves. 
          (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

              4.
              The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

                  5.
                  Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
                  (i) 2, 4, 8, 16, . . . .
                  (ii) \(2, \frac{5}{2},3,\frac{7}{2}\), . . . .
                  (iii) – 1.2, – 3.2, – 5.2, – 7.2, . . . .
                  (iv) – 10, – 6, – 2, 2, . . .
                  (v) 3, \(3 + \sqrt{2} , 3 + 3\sqrt{2} , 3 + 3 \sqrt{2}\) . . . .
                  (vi) 0.2, 0.22, 0.222, 0.2222, . . . .
                  (vii) 0, – 4, – 8, –12, . . . .
                  (viii) \(\frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}, \frac{-1}{2}\), . . . .
                  (ix) 1, 3, 9, 27, . . . .
                  (x) a, 2a, 3a, 4a, . . . .
                  (xi) a, \(a^2, a^3, a^4,\)  . . . .
                  (xii) \(\sqrt{2}, \sqrt{8} , \sqrt{18} , \sqrt {32}\) . . . .
                  (xiii) \(\sqrt {3}, \sqrt {6}, \sqrt {9} , \sqrt {12}\) . . . . .
                  (xiv) \(1^2 , 3^2 , 5^2 , 7^2\), . . . .
                  (xv) \(1^2 , 5^2, 7^2, 7^3\), . . . .

                      6.

                      Form the pair of linear equations for the following problems and find their solution by substitution method.

                      (i) The difference between two numbers is 26 and one number is three times the other. Find them.

                      (ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

                      (iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

                      (iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km.

                      (v) A fraction becomes\(\frac{ 9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.

                      (vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

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