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*}