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In the geometry of mathematics, we come across different types of shapes - some two dimensional and some three dimensional. Shapes that can be defined with the help of two parameters are called two-dimensional shapes and those which can be defined with three parameters form three-dimensional shapes. Two-dimensional shapes include square, rectangles, triangles, etc. A polygon is such a two-dimensional structure having straight sides. A triangle consists of three line segments placed at a particular angle for any pair of the three line segments. A triangle has three angles whose sum is constant which is 180°.

Let a triangle be △PQR.

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The sum of the angles of △PQR is 180°.

Hence, ∠P+∠Q+∠R=180°.

Based on side lengths of a triangle, triangles can be classified as-

### Equilateral Triangles

Triangles in which all the sides of the triangle are equal in length.

### Isosceles Triangles

Triangles in which any two sides of the triangle are equal resulting in equal base angles.

### Scalene Triangle

Triangles in which none of the sides has an equal length or all the sides of the triangle are different in lengths.

Based on the interior angles of a triangle, triangles can be classified as -

### Acute Angled Triangle

Triangles in which all the angles are less than 90°.

### Right Angled Triangle

Triangles in which any one of the angles is exactly 90°.

### Obtuse Angled Triangle

Triangle in which any one of the angles is more than 90°.

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If ∠A, ∠B, and ∠C are the interior angles of a triangle, ∠A+∠B+∠C=180°.

If b denotes the base of a triangle and h denotes the altitude of the triangle, the area of the triangle is given \[\frac{1}{2}\] х base х height square units.

If a,b and c are the sides of a triangle, the area of the triangle can be given by Heron’s formula = \[\sqrt{s(s-a)(s-b)(s-c)}\]

(where s stands for semi perimeter which is \[\frac{1}{2}\] (a+b+c)

If side lengths of a triangle are denoted by a, b and c, then the perimeter of the triangle is given by (a+b+c).

In a right-angled triangle, if p denotes perpendicular, b denotes base, and h denotes hypotenuse, h\[^{2}\] = b\[^{2}\] + p\[^{2}\]. This is known as Pythagoras’ Theorem.

1.Calculate the Area of a Right-angled Triangle in Which the Measurement of the Hypotenuse is 5cm and that of the Base is 3cm.

Ans: Given, base=3cm, hypotenuse=5cm.

We are given a right-angled triangle whose hypotenuse and base are known.

To find out the area of a triangle, we either need to know the values of base and height or the lengths of all the sides of the triangle. To find out the perpendicular of the triangle, we may use the Pythagoras theorem. The perpendicular of a right-angled triangle corresponds to the altitude of a triangle.

Let altitude/perpendicular be p cm.

For a right-angled triangle,

By Pythagoras Theorem,

perpendicular\[^{2}\] = hypotenuse\[^{2}\] - base\[^{2}\]

perpendicular\[^{2}\] = 5\[^{2}\] - 3\[^{2}\]

= 25 - 9

perpendicular\[^{2}\] = 16

perpendicular = 4

Area of the triangle = \[\frac{1}{2}\] х base х height = \[\frac{1}{2}\] х 3 х 4 = 6

Hence, the area of the triangle is 6 cm\[^{2}\] .

2. The Perpendicular of a Right-angled Triangle is 8 Units. The Area of the Triangle is Given to be 24 Square Units. Find the Base, Hypotenuse, and Perimeter of the Triangle.

Ans: Given, perpendicular=8 units, area=24 square units

We can find out the base of the triangle from the area formula and hypotenuse by Pythagoras Theorem consequently.

We know, Area = \[\frac{1}{2}\] х base х height

base = \[\frac{(2 \times Area)}{height}\]

So, = \[\frac{2 \times 24}{8}\] = 6

Now we know the base and perpendicular of the triangle.

By Pythagoras theorem,

hypotenuse\[^{2}\] = base\[^{2}\] + perpendicular\[^{2}\]

hypotenuse\[^{2}\] = 8\[^{2}\] + 6\[^{2}\]

= 64 + 36

hypotenuse\[^{2}\] = 100

hypotenuse = 10

Now we know all the three sides of the triangle.

perimeter=hypotenuse+base+perpendicular

=>perimeter=10+6+8 units

=>perimeter=24 units.

Hence, the base is 6 units, the hypotenuse is 10 units and the perimeter of the triangle is 24 units.

FAQ (Frequently Asked Questions)

1. Discuss the Various Properties of Triangles.

Ans: There are certain properties of the triangle which has been discussed below-

Triangles are polygons with the minimum number of sides.

Every triangle has three sides, three vertices, and three interior angles.

No matter how a triangle is constructed, the sum of the interior angles of the triangle will always be 180°.

The sum of lengths of any two sides of a triangle will always be greater than the length of its third side.

The difference of lengths between any two sides of a triangle will always be lesser than the length of its third side.

Any triangle can be divided into two right triangles in any given orientation.

The interior angles of an equilateral triangle are equal which is 60°.

2. What are Some of the Applications of Triangles?

Ans: Triangle is a very important shape and has numerous applications in Mathematics and real life.

Triangles are used in calculus. We use triangles as an elementary part of a different structure which helps in calculating the various properties of the structure. For instance, it is used to evaluate the area of a circle.

A polygon may be broken into simpler triangles and we can find the area of an n-sided polygon.

Triangle is an important shape in the construction of different real-life structures. It is one of the strongest shapes and has numerous applications in the field of construction.

Right-angled triangles combined with trigonometry is used to solve several real-life application based problems involving heights and distances and angle of elevation and angle of depression.