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.

96 cm²

Here we have perimeter of the triangle = 32 cm, a = 8 cm and b = 11 cm.
Third side c = 32 cm – (8 + 11) cm = 13 cm
So, 2s = 32, i.e., s = 16 cm,
s – a = (16 – 8) cm = 8 cm,
s – b = (16 – 11) cm = 5 cm,
s – c = (16 – 13) cm = 3 cm.
Therefore, area of the triangle = \(\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{16 \times 8 \times 5 \times 3} \mathrm{~cm}^{2}=8 \sqrt{30} \mathrm{~cm}^{2}\)

21 cm²

12 cm²

For one tile a = 9 cm, b = 28 cm, c = 35 cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 9+28+35 }{ 2 } =36\) cm
\(\therefore \) Area of one tile  \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 36(36-9)(36-28)(36-35) } \)
\(\\ =\sqrt { 36(27)(8)(1) } =\sqrt { 36\left( 9\times 3 \right) \left( 4\times 2 \right) } \)
 \(=6\times 3\times 2\sqrt { 6 } =36\sqrt { 6 } \) cm²
\(\therefore \)  Area of 16 tiles \(=36\sqrt { 6 } \times 16=576\sqrt { 6 } \)cm²
 \(\therefore \) Cost of polishing the tiles at the rate of 50 p per cm².
\(=576\sqrt { 6 } \times 50\) p = Rs. \(\frac { 576\sqrt { 6 } \times 50 }{ 100 } \) 
= Rs. \(288\sqrt { 6 } \) = Rs. 705.60

a = 12 cm, b = 12 cm   Perimeter = 30 cm

\(\Rightarrow \)  a +  b + c = 30
\(\Rightarrow \) 12 + 12 + c = 30
 \(\Rightarrow \) 24 + c = 30
 \(\Rightarrow \)  c = 30 - 24
 \(\Rightarrow \) c = 6 cm
\(​​s=\frac { 30 }{ 2 } \) cm = 15 cm
\(\therefore \)  Area of the triangle \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 15(15-12)(15-12)(15-6) } \)
\(=\sqrt { 15(3)(3)(9) } =9\sqrt { 15 } \) 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

Let a = 51 m, b = 37 m, and c = 20 m
Then, s=  \(\frac { a+b+c }{ 2 } \) = \(\frac { 51+37+20 }{ 2 } =\frac { 108 }{ 2 } \) = 54 m
\(\therefore \)  Area of the triangular field = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 54(54-51)(54-37)(54-20) } \) 
=\(\sqrt { 54\times 3\times 17\times 34 } =\sqrt { 2\times 3\times 3\times 3\times 3\times 17\times 2\times 17 } \)
= \(2\times 3\times 3\times 17=306\)m²
Space occupied by one rose bed = 6 m²
Number of rose beds that can be prepared in the field = \(\frac { Area\ of\ the\ field }{ Space\ occupied\ by\ one\ rose\ bed } \)
\(=\frac { 306 }{ 6 } =51\)

For \(\Delta \)ABD
a = 13 cm, b = 14 cm, c = 15 cm

\(\therefore s=\frac { a+b+c }{ 2 } s=\frac { 13+14+15 }{ 2 } \) = 21 cm
\(\therefore \)   Area = \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 21(21-13)(21-14)(21-15) } \)
\(\sqrt { 21\times 8\times 7\times 6 } =84\) cm²
For\(\Delta \) BCD
1 = 9 cm, b = 12 cm, c = 15 cm
\( \therefore s=\frac { a+b+c }{ 2 } s=\frac { a+12+15 }{ 2 } \)=18 cm
\(\therefore \)Area \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt {18(18-9)(18-12)(18-15) } \)
\(\sqrt { 18\times 9\times 6\times 3 } =54\) cm²
Now, area of quadrilateral ABCD = Area of \(\Delta \)ABD + Area of \(\Delta \)BCD
= 84 cm² + 54 cm² = 138 cm²

For \(\Delta \) ABC
a = 34 cm, b = 42 cm, c = 20 cm

\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 34+42+20 }{ 2 } =48\) cm
\(\therefore \)  Area of  \(\Delta \) ABC = \(\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 48(48-34)(48-42)(48-20) } \)
\(=\sqrt { 48(14)(6)(28) } =\quad 336\) cm²
\(\therefore \) Area of parallelogram ABCD = 2 area of triangle ABC
= 2\(\times \) 336 cm² = 672 cm²

Let the third side be x cm. Then, 12 + 12 + x = 30
24 + x = 30
x = 6 cm
So, a = 12 cm, b = 12 cms, c = 6 cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 12+12+6 }{ 2 } \)
= 15 cm
\(\therefore \)   Area \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 15(15-12)(15-12)(15-6) } \)
 \(=9\sqrt { 15 } \) cm²

b + b + 6 = 24
 \(\Rightarrow \)   b = 9 cm
\(\therefore \) Area \(=\frac { 9 }{ 4 } \sqrt { 4{ b }^{ 2 }-{ a }^{ 2 } } =\frac { 6 }{ 4 } \sqrt { 4{ (9) }^{ 2 }-{ (6) }^{ 2 } } \)
 \(=\frac { 3 }{ 2 } \sqrt { 288 } =\frac { 3 }{ 2 } .12\sqrt { 2 } \)
\( 18\sqrt { 2 } \)  cm²

See Hero's formula.

Area = \(\frac { 1 }{ 2 } \times 10\times 8=40\)cm²

\(\frac { \sqrt { 3 } { a }^{ 2 } }{ 4 } =16\sqrt { 3 } \quad \Rightarrow \)  a=8
Perimeter 3a = 3 \(\times \) 8 = 24 m

Formula