74.97 cm²

336 cm² , 22.4 cm

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

\(18\sqrt { 21 } \) cm²

For \(\Delta \)ABC
 a = 4 cm, b = 5 cm, c = 3 cm
\(\because \) a² + c² = b²
 
\(\therefore \) \(\Delta \)ABC is right angled with \(\angle B=90°\).
\(\therefore \)  Area of right triangle ABC = \(\frac { 1 }{ 2 } \times \)Base \(\times \) Height
\(=\frac { 1 }{ 2 } \times 3\times 4=6\) cm²
For \(\Delta \)ACD     
a = 4 cm, b = 5 cm, c = 5 cm
 \(\therefore s=\frac { a+b+c }{ 2 } =\frac { 4+5+5 }{ 2 } =\frac { 14 }{ 2 } =7\) cm
 \(\therefore \) Area of the ACD \(=\sqrt { s(s-a)(s-b)(s-c) } \)
 \(=\sqrt { 7(7-4)(7-5)(7-5) }\)
\( \\ =\sqrt { 7(3)(2)(22) }\)
\(=\ 2\sqrt { 21 } \) cm²
= 2\(\times \) 4.6 cm² (approx.) =9.2 cm² (approx.)
\(\therefore \) Area of the equilateral ABCD = Area of \(\Delta \) ABC+Area of \(\Delta \)ACD
= 6 cm² + 9.2 cm² = 15.2 cm² (approx.)

Let the given field be in the shape of a trapezium ABCD in which AB=25 m, CD= 10 m, BC = 13 m and AD = 14 m.
From D, draw DE 11 BC meeting AB at E. Also, draw DF \(\bot \) AB.
\(\therefore \) DE = BC =13 m
AE = AB - EB = AB - DC
= 25 - 10 = 15 m

For AED
a = 14 m, b = 13 m, c=  15 m
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 14+13+15 }{ 2 } =\frac { 42 }{ 2 } =21\) m
\(\therefore \)    Area of the \(\Delta \)AED  \(=\sqrt { s(s-a)(s-b)(s-c) } \)
 \(=\sqrt { 21(21-14)(21-13)(21-15) }\)
\( \\ =\sqrt { 21(7)(8)(6) } =\sqrt { \left( 7\times 3 \right) (7)\left( 4\times 2 \right) \left( 2\times 3 \right) } \)
\(=7\times 3\times 2\times 2=84\) m²
\(\Rightarrow \frac { 1 }{ 2 } \times \)AE\(\times \)DE = 84
\(\Rightarrow \ \frac { 1 }{ 2 } \times \)15\(\times \)DF = 84
\(\Rightarrow \) DF = \(\frac { 84\times 2 }{ 15 } \)
\(\Rightarrow \) DF=  \(\frac { 56 }{ 5 } \) m = 11.2 m
\(\Rightarrow\) Height of the trapezium is 11.2 m.
\(\therefore \) Area of parallelogram EBCD = Base \(\times \) Height
= EB\(\times \) DF = 10 \(\times \)\(\frac { 56 }{ 5 } \) = 112 m²
\(\therefore \) Area of the field = Area of AED + Area of parallelogram EBCD = 84 m² + 112 m² = 196 m².                 

48 cm²

Rs.10500

'a' = a, 'b' =a, 'c' =a
\(s=\frac { 'a'+'b'+'c' }{ 2 } =\frac { a+a+a }{ 2 } =\frac { 3a }{ 2 } \)
\(\therefore \) Area of the signal board
\(=\sqrt { s(s-'a')(s-'b')(s-'c') } \)
\(\\ =\sqrt { \frac { 3a }{ 2 } \left( \frac { 3a }{ 2 } -a \right) \left( \frac { 3a }{ 2 } -a \right) \left( \frac { 3a }{ 2 } -a \right) } \)
\(\\ =\sqrt { \frac { 3a }{ 2 } \left( \frac { a }{ 2 } \right) \left( \frac { a }{ 2 } \right) \left( \frac { a }{ 2 } \right) } =\sqrt { \frac { { 3a }^{ 4 } }{ 16 } } =\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }\)
Perimeter = 180 cm
\(\Rightarrow 'a'+'b'+'c'\ =\ 180\)
\(\\ \Rightarrow a+a+a=180\)
\(\\ \Rightarrow 3a=180\)
\(\\ \Rightarrow a=\frac { 180 }{ 3 } \)
\(\Rightarrow \)  a = 60 cm
\(\therefore \)  Area of the signal board
 \(=\frac { \sqrt { 3 } }{ 4 } \)  a² =\( \frac { \sqrt { 3 } }{ 4 } \) (60)² = \(900\sqrt { 3 } \) cm²
Alternatively,
\(s=\frac { 3a }{ 2 } =\frac { 3 }{ 2 } (60)=90\)cm
Area of the signal board
\(=\sqrt { s(s-'a')(s-'b')(s-'c') } \)
\(=\sqrt { 90(90-60)(90-60)(90-60) } \)
\(\\ =\sqrt { 90(30)(30)(30) } =900\sqrt { 3 } \) cm². 

Let the sides of the triangle be 13k, 14k, and 15k (in cm).Then,
Perimeter =  a + b + c = 13K + 14k + 15k = 42k cm    
According to the question, 42k = 84
\(\Rightarrow k=\frac { 84 }{ 42 } =2\)
\(\therefore \)  Sides are 26 cm, 28 cm, and 30 cm.
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 26+28+30 }{ 2 } \)
= 42 cm  
\(\therefore \)   Area \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 42(42-26)(42-28)(42-30) } \)
\(\\ =\sqrt { (42)(16)(14)(12) } \)
= 336 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

\(\therefore \)  Side = 8 cm
\(\therefore \) Perimeter = 6\(\times \)8 = 48 cm
Area of equilateral triangle OAB = \(\frac { \sqrt { 3 } }{ 4 } \)  (side)²
 
= \(\frac { \sqrt { 3 } }{ 4 } \)(8)² = 16 \(\sqrt { 3 } \) cm²
\(\therefore \) Area of the regular hexagon = 6 Area of equilateral triangle OAB
= 6\(\times \)16\(\sqrt { 3 } \) = 96\(\sqrt { 3 } \) 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.

Area \(=\frac { \sqrt { 3 } }{ 4 } { a }^{ 2 }=\frac { \sqrt { 3 } }{ 4 } { \left( 4 \right) }^{ 2 }=4\sqrt { 3 } \) cm²

Area = \(\frac { Base\times Perpendicular }{ 2 } =\frac { 6\times 8 }{ 2 } \) = 24 cm²

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