Notes on Area of Triangle | Grade 10 > Compulsory Mathematics > Mensuration | KULLABS.COM

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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,

or, h2= c2 - (a - x)2 ............. (1)

Again, In the right-angled triangle ADC,

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.]

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

.

### Very Short Questions

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*}

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35cm

54cm

44cm

22cm

20 cm

24 cm

18 cm

23 cm

• ### Find the area of an equilateral triangle of perimeter 30 cm.

22(sqrt{3}) cm2

33(sqrt{3}) cm2

23(sqrt{3}) cm2

25(sqrt{3})cm2

5cm2

6cm2

4cm2

2cm2

27 cm2

25 cm2

26 cm2

28cm2

2.24 cm2

7.80 cm2

4.60 cm2

6.80 cm2

25 cm

24 cm

26 cm

22 cm

13 cm

9 cm

11 cm

14 cm

300 cm

350 cm

325 cm

360 cm

8.83 cm 2

9.92 cm2

9.10 cm2

9.12 cm2

270.63 cm2

270.66 cm2

224.55 cm2

260.53 cm2