## Area of Triangle

Subject: Compulsory Mathematics

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#### Overview

The area of a triangle is always half the product of the height and base.A triangle is a polygon with three edges and three vertices.A right triangle (or right-angled triangle, formerly called a rectangled triangle) has one of its interior angles measuring 90° (a right angle). The side opposite to the right angle is the hypotenuse, the longest side of the triangle.

##### Area of Triangle

Few classes back, we learned that $\text {area of a triangle} = \frac {1} {2} base \times height,$ when the measurements of base and height are given.

Similarly, the area of the triangle can be found when the length of sides is given.
Let the three sides of triangle ABC are a, b and c respectively. As AD⊥BC and AD is the height (h) of the triangle ABC.

Suppose the length of DC = x
Then the length of BD= a - x

Now, In the right-angled triangle ADB,

AB2 = BD2 + AD2
AB2 = BD2 + AD2
or, AD2= AB2- BD2
or, h2= c2 - (a - x)2 ............. (1)

Again, In the right-angled triangle ADC,

AC2 = AD2+ DC2
or, AD2= AC2- DC2
or h2= b2- x2..................... (2)

From equation (1) and (2) we get,

c2- (a-x)2= b2- x2
or, c2- (a - x)2= b2 - x2
or, c2 - ( a2 -2ax + x2 ) = b2- x2
or, c2 - a2 + 2ax -x2= b2 - x2
or, 2ax = b2- x2 - c2+ a2+ x2
or, 2ax = a2 + b2- c2

$or, x = \frac {a^2 + b^2 - c^2} {2a}$.................. (3)

Substituting the value of x from equation (3) in equation (2), we get

$h^2 = b^2 - \left (\frac {a^2 + b^2 - c^2} {2a} \right) ^ 2$

$or, h^2 = \left (b +\frac {a^2 + b^2 - c^2} {2a} \right) \left(b - \frac {a^2 + b^2 - c^2} {2a} \right)$

$or, h^2 = \left ( \frac {2ab + a^2 + b^2 - c^2} {2a}\right)\left ( \frac {2ab - a^2 - b^2 + c^2} {2a}\right)$

$or, h^2 = \frac {\{(a + b)^2 -c^2\}} {2a}$ $\frac {\{c^2 - (a -b)^2\}} {2a}$

$or, h^2 = \frac {(a + b + c) (a + b - c)} {2a}$$\frac {(c - a + b ) (c + a - b)} {2a}$

$or, h^2 = \frac {(a + b + c) (a + b - c)(c - a + b ) (c + a - b)} {4a^2}$

$or, h^2 = \frac {(a + b + c) (a + b + c - 2c)(c - a + b ) (c + a + b - 2b)} {4a^2}$ ............... (4)

Now, Substituting , a+b+c = 2s, we get,

$h^2 = \frac {2s ( 2s - 2c )( 2s - 2b )( 2s - 2a )} {4a^2}$

$or, h^2=\frac {2s.2( s - c ).2(s - b ).2(s - a )} {4a^2}$

$or, h^2=\frac {16s( s - a ) (s - b ) (s - c )} {4a^2}$

$or, h = \sqrt {\frac {4s( s - a ) (s - b ) (s - c )} {a^2} }$

$or, h = \frac{2}{a}\sqrt {s( s - a ) (s - b ) (s - c )}$

Now, $\text{Area of}\; \Delta \text {ABC} = \frac {1}{2} base \times height$

$= \frac {1} {2} a \times \frac {2} {a} \sqrt {s( s - a ) (s - b ) (s - c )}$

$\boxed {\therefore \text {Area of}\; \Delta \text {ABC} =\sqrt {s( s - a ) (s - b ) (s - c )}}$

$\text { Semi-perimeter of triangle } = \frac {a + b + c} {2}$

$\text { Area of equilateral triangle } = \frac {\sqrt {3} } {4} a^2$

Area of isosceles triangle = $\frac{b} {4}$ $\sqrt{4a^2-b^2}$[ where a is the base and b is the lenght of equal sides.]

##### Things to remember

a. Area of triangle ABC =$\sqrt{s(s-a)(s-b)(s-c)}$.

b.This formula is known as hero's formula hero was a greek mathematician.

c. $\text {Area of a triangle} = \frac {1} {2} base \times height,$

• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Questions and Answers

Solution:

Here,
a = 21 cm
b = 13 cm
c = 20 cm
Then,

$s= \frac{a+b+c}{2}\\ \:\:\: = \frac{21+13+20}{2}cm\\ \:\:\: = \frac{54}{2}cm \\ \:\:\: = 27 \: cm$

\begin{align*} Area \: of \: triangle &= \sqrt{s(s-a)(s-b)(s-c)} \\ &= \sqrt{27(27-21)(27-13)(27-20)} cm^2\\ &= \sqrt{27 \times 6 \times 14 \times 7}cm^2 \\ &= \sqrt{15876}cm^2 \\ &= 126 \: cm^2 \end{align*}

Solution:

Let length of equilateral triangle = a
According to the question,

\begin{align*} a+a+a &= 18 cm \\ 3a &= 18cm \\ a&=\frac{18}{3}\\ \therefore a &= 6 \: cm \\ \: By \: formula, \\ Area \: of \: equilateral \Delta &= \frac{\sqrt{3}}{4}a^2 \\ &= \frac{\sqrt{3}}{4}\times 6^2 \; cm^2 \\ &= \frac{\sqrt{3}}{4} \times 36cm^2 \\ &= 9\sqrt{3}cm^2 \: _{Ans} \end{align*}

Solution:

Suppose side of an equilateral triangle = a

\begin{align*} By \: formula, \: \\ Area \: of \: equilateral \Delta &= \frac{\sqrt{3}}{4}a^2\\ or, 36\sqrt{3}cm^2 &= \frac{\sqrt{3}}{4}a^2 \\ or, \frac{36\sqrt{3}\times 4}{\sqrt{3}} &= a^2\\ or, a^2 &= 144 cm^2 \\ or, a &= \sqrt{144}cm \\ \therefore a &= 12 \: cm \: _{Ans} \end{align*}

Solution:

Ratio of sides of the given triangle is 5 : 4 : 3.
Let its side be 5x, 4x and 3x.

$semi\:perimeter = \frac{5x+4x+3x}{2} = 6x$

\begin{align*} Area \: of \: \Delta &= 54 cm^2 \\ or, \sqrt{s(s-a)(s-b)(s-c)} &= 54 \\ or, \sqrt{6x(6x-5x)(6x-4x)(6x-3x)} &= 54 \\ or, \sqrt{6x \times x \times 2x \times 3x } &= 54 \\or, \sqrt{36x^4} &= 54 \\or, 6x^2 &= 54 \\ x^2 &= 9 \\ \therefore x &= 3 \: cm \end{align*}

Now, its sides are 15 cm, 12 cm and 9 cm.

$perimeter = sum \: of \: its \: sides = 15 + 12 + 9 = 36 \: cm$

Solution:

Area of given isoceles triangle is 12 cm2
Length of its base = 8 cm
Let, the other two equal sides be x.

Now,

$semi \: perimeter \: of \: triangle = \frac{x+x+8}{2} = \frac{2x+8}{2} = \frac{2(x+4)}{2} = x+4$

\begin{align*} Area \: of \: triangle &= \sqrt{s(s-a)(s-b)(s-c)} \\ 12 &= \sqrt{(x+4)(x+4-x)^2 (x+4-8) }\\ or, 12^2 &= (x+4)\times(4)^2 \times (x-4) \\ or, 144 &= 16(x^2 - 16)\\ or, x^2 -16 &= 9 \\ or, x^2 &= 9+16\\ or, x &= \sqrt{25}\\ \therefore x &=5 \\ \therefore \text{Length of other}& \: equal \: sides = 5 \: cm \: _{Ans} \end{align*}

Solution:

One of the side of triangle = 126 cm
Difference between hypotenuse and other side = 42 cm
$i.e. h -b =42 \\ \: \: \: \: \: \: \: \: \: or, h = 42+b$

Now,

\begin{align*} h^2 &= p^2+b^2 \\ h^2 &= (126)^2 + b^2 \\ or, (42+b)^2 &= 15876 + b^2 \\ or, 1764 + 84b + b^2 &= 15876 + b^2 \\ or, 84b &= 15,876 -1764 \\ or, b &= \frac{14112}{84}\\ \therefore b &= 168 cm \: _{Ans} \\ \therefore h&= 168 + 42 \\ &= 210 cm \: _{Ans} \\ \: \\ Area \: of \: \Delta &= \frac{1}{2} \times base \times height \\ &= \frac{1}{2} \times 168 \times 126 \\ &= 10,584 \:cm^2 \: _{Ans} \end{align*}

Solution:

Area of given isoceles triangle = 12 cm2
and given equal sides are 5 cm.

Let, base of the triangle be x cm.

Now,

$Semi- perimeter = \frac{5+5+x}{2} = \frac{10+x}{2} cm$

\begin{align*} Area \: of \: triangle &= \sqrt{s(s-a)(s-b)(s-c)} \\ 12 &= \sqrt{ \left( \frac{10+X}{2}\right) \left(\frac{10+x}{2} - 5 \right) \left( \frac{10+x}{2}-x\right) \left( \frac{10+x}{2}-5 \right)} \\12 &= \sqrt{ \left( \frac{10+X}{2}\right) \left(\frac{10+x -10}{2} \right) \left( \frac{10+x-2x}{2}\right) \left( \frac{10+x-10}{2} \right)} \\ or, 12 &= \sqrt{\left(\frac{10+x}{2}\right)\left( \frac{10-x}{2} \right) \frac{x}{2} \frac{x}{2}} \\ or, 12 &= \sqrt{\frac{10^2 - x^2}{4} \times \frac{x^2}{4}} \\ Put & \: x^2=m \\ \therefore 12 &= \frac{(100-m)m}{16} \\ squaring \: on \:b&oth \: side, we \: get \\ 144 &= \frac{100m - m^2}{16} \\ or, 2304 &= 100m-m^2 \\ or, m^2 - 10&0m + 2304 = 0 \end{align*}

Which is quadratic equation in m.

\begin{align*} \therefore m &= \frac{-(-100) \pm \sqrt{(-100)^2 -4 .1.2304}}{2.1} \\ &= \frac{100 \pm \sqrt{10000-9216}}{2}\\ &= \frac{100 \pm \sqrt{784} }{2}\\&=\frac{100 \pm 28}{2}\\ Taking \: (+ ve) \: s&ign \: and \: reserving \: m, we \:get \\ x^2&=\frac{128}{2}\\or, x &= \sqrt{64}\\ \therefore x &= 8 \\ \: \\ Taking \: (-ve) si&gn \: and \: reserving\: m, we\: get\\ x^2&=\frac{72}{2}\\or, x &= \sqrt{36}\\ \therefore x &= 6 \end{align*}

##### Quiz

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