Let the sides of triangle are a = 11 cm, b = 60 cm and c = 61cm.
Then, semi-perimeter of triangle,
\(s=\frac{a+b+c}{2}=\frac{11+60+61}{2}=\frac{132}{2}=66m\)
Now, area of triangle= \(=\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{66(66-11)(66-60)(66-61)}\)
\(=\sqrt{66\times55\times6\times5}=\sqrt{11\times6\times11\times5\times6\times5}\)
\(=11\times6\times5\times330cm^2\)
Here, we have to find the altitude of the smallest side, so we consider the base as smallest side.
∵ Area of a triangle = \(\frac{1}{2}\) x base x height
\(\therefore\ 330=\frac{1}{2}\times11\times b\Rightarrow b=\frac{330\times2}{11}=60cm\)
Hence, the altitude of the smallest side is 60 cm.

Let ABC be the field where wheat is grown. Also let ACD be the field which has been divided in two parts by joining C to the mid-point E of AD. For the area of triangle ABC, we have
a = 200 m, b = 240 m, c = 360 m
Therefore, s = \(\frac{200+240+360}{2} \mathrm{~m}=400 \mathrm{~m}\)
So, area for growing wheat
\(=\sqrt{400(400-200)(400-240)(400-360)} \mathrm{m}^{2}\)
\(=\sqrt{400 \times 200 \times 160 \times 40} \mathrm{~m}^{2}\)
\(=16000 \sqrt{2} \mathrm{~m}^{2}=1.6 \times \sqrt{2} \text { hectares }\)
= 2.26 hectares (nearly)
Let us now calculate the area of triangle ACD
Here, we have s \(=\frac{240+320+400}{2} \mathrm{~m}=480 \mathrm{~m}\)
So, area of \(\Delta \mathrm{ACD}=\sqrt{480(480-240)(480-320)(480-400)} \mathrm{m}^{2}\)
\(=\sqrt{480 \times 240 \times 160 \times 80} \mathrm{~m}^{2}=38400 \mathrm{~m}^{2}=3.84 \text { hectares }\)
We notice that the line segment joining the mid-point E of AD to C divides the triangle ACD in two parts equal in area. They have the bases AE and ED equal and, of course, they have the same height. 
Therefore, area for growing potatoes = area for growing onions
= (3.84 ÷ 2) hectares = 1.92 hectares.

For finding area of the park, we have
2s = 50 m + 80 m + 120 m = 250 m.
i.e., s = 125 m
Now, s – a = (125 – 120) m = 5 m,
s – b = (125 – 80) m = 45 m,
s – c = (125 – 50) m = 75 m.
Therefore, area of the park = \(\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{125 \times 5 \times 45 \times 75} \mathrm{~m}^{2}\)
\(=375 \sqrt{15} \mathrm{~m}^{2}\)
Also, perimeter of the park = AB + BC + CA = 250 m
Therefore, length of the wire needed for fencing = 250 m – 3 m (to be left for gate)
= 247 m
And so the cost of fencing = Rs.20 x 247 = RS. 4940

Suppose that the sides, in metres, are 3x, 5x and 7x (see Fig.).

Then, we know that 3x + 5x + 7x = 300 (perimeter of the triangle)
Therefore, 15x = 300, which gives x = 20.
So the sides of the triangle are 3 x 20 m, 5 x 20 m and 7 x 20 m
i.e., 60 m, 100 m and 140 m.
We have s \(=\frac{60+100+140}{2} \mathrm{~m}=150 \mathrm{~m}\)
and area will be \(\sqrt{150(150-60)(150-100)(150-140)} \mathrm{m}^{2}\)
\(=\sqrt{150 \times 90 \times 50 \times 10} \mathrm{~m}^{2}\)
\(=1500 \sqrt{3} \mathrm{~m}^{2}\)

21 cm²

12 cm²

Area of paper of shade I \(2\times \left( \frac { 1 }{ 2 } \times 16\times 16 \right) =256\)cm²
Similarly, Area of paper of shade II = 256 cm²
For area of paper of shade III
a = 8 cm, b = 6 cm, c = 6cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 8+6+6 }{ 2 } =10\) cm
\(\therefore \) Area of paper of shade III = \(\sqrt { s(s-a)(s-b)(s-c) } \)
 \(=\sqrt { 10(10-8)(10-6)(10-6) } \)
 \(=\sqrt { (10)(2)(4)(4) } =8\sqrt { 5 } \) 
= 17.89 cm²

For one triangular piece = 20 cm, b = 50 cm, c = 50 cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 20+50+50 }{ 2 } =\frac { 120 }{ 2 } =60\) cm
\(\therefore \)  Area of one triangle \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 60(60-20)(60-50)(60-50) } \)
\(=\sqrt { (60)(40)(10)(10) } =200\sqrt { 6 } \) cm²
\(\therefore \)  Area of 5 triangles of one colour \(=5(200\sqrt { 6 } )\) cm² = 1000\(\sqrt { 6 } \) cm²
Hence, 1000\(\sqrt { 6 } \) cm² cloth of each colour is required for the umbrella.

\(1500\sqrt { 3 } \) cm²

a = 18 cm, b = 10 cm
Perimeter = 42 cm
\(\Rightarrow a+b+c=42\)
\(\\ \Rightarrow 18+10+c=42\)
\(\\ \Rightarrow 28+c=42\)
\(\\ \Rightarrow c=42-28\)
\(\Rightarrow \ c=14\quad \)cm
\(s=\frac { 42 }{ 2 } =21\)cm
\(\therefore \)   Area of the triangle \(=\sqrt { s(s-a)(s-b)(s-c) } \quad \quad \)
\(=\sqrt { 21(21-18)(21-10)(21-14) } \)
\(\\ =\sqrt { 21(3)(11)(7) }\)
\( \\ =\sqrt { (7)(3)(3)(11)(7) } \)
\(=(7)(3)\sqrt { 11 } =21\sqrt { 11 } \)cm²

(a) 114.89 cm²  
(b) \(40\sqrt { 3 } \)cm²  
(c) \(24\sqrt { 3 } \)cm²  
(d) 330 m²

For one black colour sheet  a= 4 cm, b = 6 cm
\(\therefore \) Area \(=\frac { a }{ 4 } \sqrt { 4{ b }^{ 2 }-{ a }^{ 2 } } \)     | Sheet is an isosceles triangle
 \(=\frac { 4 }{ 4 } \sqrt { 4{ (6) }^{ 2 }-4^{ 2 } } \)  \(=\sqrt { 128 } =8\sqrt { 2 } \) cm²
\(\therefore \) Area of 2 black colour sheets = 2\(\times 8\sqrt { 2 } \) = \( 16\sqrt { 2 } \) cm²
Similarly, area of 2 white colour sheets =  \( 16\sqrt { 2 } \) cm²
\(\therefore \) Total area of black and white colour sheets =  \( 16\sqrt { 2 } \) + \( 16\sqrt { 2 } \) =  \( 32\sqrt { 2 } \) cm²

Let ABC be the right angled triangle right angled at B. Let O be the centre of the circumcircle.
Then, O is the midpoint of the hypotenuse AC. I by geometry
OA = OB =OC
= Radius of the circumcircle = 3 cm
 
  \(\therefore \) Hypotenuse AC = Diameter of the circle
= 2 \(\times \) Radius of the circumcircle
= 2 \(\times \) 3 = 6 cm
Let BM be the perpendicular from B on AC.
\(\therefore \) BM =2 cm
\(\therefore \)  Area of the right angled triangle ABC
= \(\frac { 1 }{ 2 } \) \(\times \)Base \(\times \) Altitude
= \(\frac { 1 }{ 2 } \)\(\times \) AC \(\times \) BM = \(\frac { 1 }{ 2 } \)\(\times \) 6\(\times \) 2 = 6 cm².

Side (a) = \(\frac { 60 }{ 3 } =20\)m
\(\therefore \) Area =\(\quad =\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }=\frac { \sqrt { 3 } }{ 4 } { (20) }^{ 2 }=100\sqrt { 3 } \) m²

Area = \(\frac { AB\times BC }{ 2 } =\frac { 4\times 4 }{ 2 } \)= 8 cm²

Formula

Let the sides of triangle be a = 26 cm, b = 28 cm and c = 30 cm. Let s be the semi-perimeter of the triangle
Then, 
\(s=\frac{a+b+c}{2}=\frac{26+28+30}{2}=\frac{84}{2}=42cm\)
and area of triangle \(=\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{42\times(42-26)(42-28)(42-30)}\)
\(=\sqrt{42\times16\times14\times12}\)
\(=\sqrt{7\times6\times16\times7\times2\times6\times2}=336cm^2\)
Let h be the height of the parallelogram.
Then, area of parallelogram = base x height = 28 x b
∵ area of parallelogram =Area of triangle
ஃ  \(28\times b=336\Rightarrow b=\frac{336}{28}\Rightarrow b=12cm\)
Hence, the height of the parallelogram is 12 cm.​​​​​​​

Let given sides of the wall be a = 15m, b = 11m and c = 6 m.
ஃ  Semi-perimeter, \(s=\frac{a+b+c}{2}=\frac{15+11+6}{2}=\frac{32}{2}=16m\)
Now, area of the wall= \(=\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{16(16-15)(16-11)(16-6)}\)
\(=\sqrt{16\times1\times5\times10}=\sqrt{4\times4\times5\times5\times2}=5\times4\sqrt2\)
\(=20\sqrt{2}m^2\)
Hence, the area painted in colour is \(=20\sqrt{2}m^2\)