1. The document presents a new formula for calculating the area of an isosceles triangle derived by the author through applying Pythagorean theorem.
2. The formula is: Area of isosceles triangle = b × (4a^2 - b^2)/4, where a is the length of two equal sides and b is the length of the unequal side (base).
3. The author provides examples and step-by-step working to demonstrate the application of the new formula, as well as verification by comparing the results to existing Heron's formula.
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International Journal of Engineering Inventions (IJEI) provides a multidisciplinary passage for researchers, managers, professionals, practitioners and students around the globe to publish high quality, peer-reviewed articles on all theoretical and empirical aspects of Engineering and Science.
The peer-reviewed International Journal of Engineering Inventions (IJEI) is started with a mission to encourage contribution to research in Science and Technology. Encourage and motivate researchers in challenging areas of Sciences and Technology.
International Journal of Computational Engineering Research (IJCER) is dedicated to protecting personal information and will make every reasonable effort to handle collected information appropriately. All information collected, as well as related requests, will be handled as carefully and efficiently as possible in accordance with IJCER standards for integrity and objectivity.
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journal publishing, how to publish research paper, Call For research paper, international journal, publishing a paper, IJERD, journal of science and technology, how to get a research paper published, publishing a paper, publishing of journal, publishing of research paper, reserach and review articles, IJERD Journal, How to publish your research paper, publish research paper, open access engineering journal, Engineering journal, Mathemetics journal, Physics journal, Chemistry journal, Computer Engineering, Computer Science journal, how to submit your paper, peer reviw journal, indexed journal, reserach and review articles, engineering journal, www.ijerd.com, research journals,
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Deadline : 31st May 2013
Notification : 15th June 2013
Revision : 20th June 2013
Publication : 01st July 2013
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C0211014019
1. Research Inventy: International Journal Of Engineering And Science
Vol.2, Issue 11 (April 2013), Pp 14-19
Issn(e): 2278-4721, Issn(p):2319-6483, Www.Researchinventy.Com
14
Invention Of The Plane Geometrical Formulae - Part II
Mr. Satish M. Kaple
Asst. Teacher Mahatma Phule High School, KherdaJalgaon (Jamod) - 443402 Dist- Buldana,
Maharashtra (India)
Abstract: In this paper, I have invented the formulae for finding the area of an Isosceles triangle. My finding is
based on pythagoras theorem.
I. INTRODUCTION
A mathematician called Heron invented the formula for finding the area of a triangle, when all the three
sides are known. Similarly, when the base and the height are given, then we can find out the area of a triangle.
When one angle of a triangle is a right angle, then we can also find out the area of a right angled triangle. Hence
forth, We can find out the area of an equilateral triangle by using the formula of an equilateral triangle. These
some formulae for finding the areas of a triangles are not exist only but including in educational curriculum also.
But, In educational curriculum. I don’t appeared the formula for finding the area of an isosceles triangle with
doing teaching – learning process . Hence, I have invented the new formula for finding the area of an isosceles
triangle by using Pythagoras theorem.
I used pythagoras theorem with geometrical figures and algebric equations for the invention of the new
formula of the area of an isosceles triangle. I Proved it by using geometrical formulae & figures, 20 examples
and 20 verifications (proofs).
Here myself is giving you the summary of the research of the plane geometrical formulae- Part II
II. METHOD
First taking an isosceles triangle ABC A
B C
Fig. No. -1
Now taking a, a & b for the lengths of three sides of ABC
A
a a
B b C
Fig. No. – 2
2. Invention Of The Plane Geometrical…
15
Draw perpendicular AD on BC.
Fig. No. - 3
ABC is an isosceles triangle and it is an acute angle also.
In ABC,
Let us represent the lengths of the sides of a triangle with the letters a,a,b. Side AB and side AC are congruent
side. Third side BC is the base. AD is perpendicular to BC.
Hence, BC is the base and AD is the height.
Here, taking AB=AC = a
Base , BC = b Height, AD = h
In ABC, two congruent right angled triangle are also formed by the length of perpendicular AD drawn on the
side BC from the vertex A. By the length of perpendicular AD drawn on the side BC, Side BC is divided into
two equal parts of segment. Therefore, these two equal segments are seg DB and seg DC. Similarly, two a right
angled triangles are also formed, namely, ADB and ADC which are congruent.
Thus,
DB = DC = 1/2 × BC
DB = DC = 1/2 × b = b/2
ADB and ADC are two congruent right angled triangle.
Taking first right angled ADC,
In ADC, Seg AD and Seg DC are both sides forming
the right angle. Seg AC is the hypotenuse.
A
Here, AC =a
Height , AD = h h a
DC = b/2 and m ADC = 900
D b/2 C
Fig. No - 4
According to Pythagoras Theorem,
(hypotenuse) 2
= ( one side forming the right angle) 2
+ ( second side forming the right angle) 2
In short,
( Hypotenuse ) 2
= ( one side) 2
+ ( second side) 2
AC2
= AD2
+ DC2
AD2
+ DC2
= AC2
h2
+ ( b/2 ) 2
= a2
h2
= a2
– (b/2) 2
h2
= a2
– b2
4
h2
= a2
× 4 – b2
4 4
h2
= 4a2
– b2
4 4
h2
= 4a2
– b2
3. Invention Of The Plane Geometrical…
16
4
Taking the square root on both side,
h2
= 4a2
– b2
4
h2
= 1 × (4 a2
– b2
)
4
h2
= 1 × 4a2
- b2
4
The square root of h2
is h and the square root of ¼ is ½
.·. h = ½ × 4a2
– b2
.·. Height, h = ½ 4a2
– b2
.·. AD =h = ½ 4a2
– b2
Thus,
Area of ABC = ½ × Base × Height
= ½ × BC × AD
=½ × b × h
But Height, h = ½ 4a2
– b2
.·. Area of ABC = ½ × b × ½ 4a2
– b 2
.·. Area of ABC = b × 1 4a2
– b2
2 2
= b × 1 × 4a2
– b2
2 × 2
= b 4a2
– b2
4
.·. Area of an isosceles ABC = b 4a2
– b2
4
For example- Now consider the following examples:-
Ex. (1) If the sides of an isosceles triangle are 10 cm, 10 cm and 16 cm.
Find it’s area D
DEF is an isosceles triangle.
In DEF given alongside, 10cm 10 cm
l ( DE) = 10 cm.
l l ( DF) = 10 cm. l ( EF) = 16 cm
E 16 cm F
Let,
a = 10 cm
4. Invention Of The Plane Geometrical…
17
Base, b = 16 cm.
By using The New Formula of an isosceles triangle,
.·. Area of an isosceles DEF = A (DEF)
= b 4a2
- b2
4
= 16 × 4(10)2
– (16)2
4
= 4 × 4 × 100 – 256
= 4 × 400 – 256
= 4 × 144
The square root of 144 is 12
= 4 × 12 = 48 sq.cm.
.·. Area of an isosceles DEF = 48 sq.cm.
Verification :-
Here,
l (DE) = a = 10 cm.
l ( EF) = b = 16 cm.
l ( DF) = c = 10 cm.
By using the formula of Heron’s
Perimeter of DEF = a + b + c
= 10 + 16 + 10 = 36 cm
Semiperimeter of DEF,
S = a + b + c
2
S = 36
2
S = 18 cm.
.·.Area of an isosceles DEF = s (s– a) (s– b) (s– c)
= 18 × (18 – 10) × (18 –16) × (18–10)
= 18 × 8 × 2 × 8
= (18 × 2) × (8 × 8)
= 36 × 64
= 36 × 64
The square root of 36 is 6 and the square root of 64 is 8
= 6 × 8 = 48 sq.cm
.·. Area of DEF = 48 sq.cm
Ex. (2) In GHI, l (GH) = 5 cm, l (HI) = 6 cm and l (GI) = 5 cm.
Find the area of GHI.
GHI is an isosceles triangle.
In GHI given alongside,
5. Invention Of The Plane Geometrical…
18
l ( GH ) = 5 cm. Fig No- 6
l ( HI ) = 6 cm.
l ( GI ) = 5 cm
Let,
a = 5 cm
Base, b = 6 cm.
By using The New Formula of area of an isosceles triangle,
.·. Area of an isosceles GHI = b 4a2
– b2
4
= 6 × 4 × (5)2
– (6)2
4
The simplest form of 6 is 3
4 2
= 3 × ( 4 × 25) – 36
2
= 3 × 100 – 36
2
= 3 × 64
2
The square root of 64 is 8
= 3 × 8 = 3 × 8 = 24
2 2 2
= 12 sq.cm.
.·. Area of an isosceles GHI = 12 sq.cm.
Verification :-
Here,
l (GH) = a = 5 cm.
l (HI) = b = 6 cm.
l (GI) = c = 5 cm.
By using the formula of Heron’s
Perimeter of GHI = a + b + c
= 5 + 6 + 5
= 16 cm
Semiperimeter of GHI,
S = a + b + c
2
S = 16
2
S = 8 cm.
.·.Area of an isosceles GHI =
6. Invention Of The Plane Geometrical…
19
s (s– a) (s– b) (s– c)
= 8 × (8 – 5) × (8 –6) × (8–5)
= 8 × 3 × 2 × 3
= (8 × 2) × (3 × 3)
= 16 × 9
= 144
The square root of 144 is 12
= 12 sq.cm
.·. Area of an isosceles GHI = 12 sq.cm.
Explanation:-
We observe the above solved examples and their verifications, it is seen that the values of solved examples by
using the new formula of an isosceles triangle and the values of their verifications are equal.
Hence, The new formula of the area of an isosceles triangle is proved.
III. CONCLUSIONS
Area of an isosceles triangle = b × 4a2
– b2
4
From the above new formula , we can find out the area of an isosceles triangle. This new formula is useful in
educational curriculum, building and bridge construction and department of land records. This new formula is
also useful to find the area of an isosceles triangular plots of lands, fields, farms, forests, etc. by drawing their
maps.
REFERENCES
1 Geometry concepts and Pythagoras theorem.