Triangle
Triangle is a polygon with three sides. Some Properties of a triangle are :
1. The sum of all angles of a triangle is always 180◦.
2. The sum of any two sides is always greater than the third side.
3. If an angle of a triangle has a obtuse angle then rest of the two angles has to be
acute.
4. The angle opposite the longest side in the triangle is the greatest angle of the triangle.
Classification of triangles
Depending on the types of the angles and sides a triangle has, triangle can be classifieds as follows:1. Acute angled triangle : when all the angles of a triangle are less than 90 degrees.
2. Obtuse angled triangle: When anyone angle of the triangle is greater than 90 degrees.
3. Right Angled Triangle : When one of the angle of the triangle is equal to 90 degrees.
4. Scalene Triangle: When all the sides of the triangle are different.
5. isosceles Triangle : When any two sides of the triangle are equal to each other.
6. Equilateral Triangle : When all the sides of the triangle are equal to each other.
Area of Triangle
1. Area of Triangle is = (1/2) x B x H
Where ,
B is the length of the base and
H is the height of the altitude drawn to the base from the opposite vertex .
2. Area of triangle using Hero's Formula ( if all the sides of the triangle are given)
Area of triangle = √s (s-a)(s-b)(s-c)
Where,
a, b, c, are the length of the sides of the triangle
and s = ( a + b + c)/2
3. Perimeter of the triangle with sides a, b and c
Perimeter = (a + b + c)
Important Formula relating to Triangles
1. Area of equilateral triangle
Area = ( √3)/4)a2
Where,
a is the side of an equilateral triangle
2. Area of Isosceles Triangle
Area = (b/4){√4a2 - b2}
Where,
a is the length of two equal sides and b is the length of the third side.
3. For a right angled triangle with sides a, b and c and right angled between the sides a and c, the relation among these sides is defined by Pythagoras theorem.
b2 =
a2 + c2
Few Solved examples
Problem 1 : The base of a triangular field is 240 m and the altitude drawn on it from the opposite vertex is 50 m. Calculate the area of the field.
Solution:
Given that,
the base = 240m
the altitude = 50m
therefore using the area formula for height and base and putting the values, we above
area = (1/2)bh = (1/2)x240x50 = 120x50= 6000 square meter
Problem 2 :
There is an equilateral triangle with side equal to 75m. Calculate the area of the field and the amount to water the field if the rate of watering the field is 5per square meter.
Solution:
Using the formula of calculating the area of an equilateral field given above,
i.e. area = ( √3/4)a2
We have,
area = ( 1.732/4)(75)(75) { taking √3 = 1.732)
area = 2435.62 square meter
Now the rate of watering the field is 5 per square meter, then the amount required to water the field is
2435.62 x 5 = 12178.12
Problem 3:
The heights of two similar right-angled triangle ΔLMN and ΔOPQ are 48cm and 36cm . If OP = 12 cm , then LM is
(a) 20 cm (b) 12 cm
(c) 10√6/3 (d) 16 cm
Answer: (d) 16 cm
Solution:
If two triangle are similar then their respective sides are in the same ratio.
Then, LM/OP = 48/36
=> LM/12 = 48/36
=> LM = (4/3)x12
=> LM = 16 cm
To learn about circle : Mensuration Circle
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