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.

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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.

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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

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

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

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

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

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

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

Let '2s' denotes the perimeter of the triangle the 2s = a +b + c

a + b = 2s - c
c + b = 2s - a
c + a = 2s - b

From equation (4)

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

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

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

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

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

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

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.]

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

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*}

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    44cm


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    20 cm


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    24 cm


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    22(sqrt{3}) cm2


    23(sqrt{3}) cm2


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    26 cm2


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    25 cm2


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    13 cm


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    14 cm 


    11 cm


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    325 cm


    350 cm


    300 cm


    360 cm


  • The two sides of a triangle are in the ratio 2:3 and its third side is 5cm. I the perimeter of the triangle is 15 cm. what is the  area of this triangle?

    9.92 cm2


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    9.12 cm2


  • Find the area of the triangular plane field having each side of 25m.

    270.66 cm2


    270.63 cm2


    224.55 cm2


    260.53 cm2


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