Let ABCD be the field.
Perimeter = 400 m
So, each side = 400 m ÷ 4 = 100 m.
i.e. AB = AD = 100 m.
Let diagonal BD = 160 m.
Then semi-perimeter s of D ABD is given by
\(s=\frac{100+100+160}{2} \mathrm{~m}=180 \mathrm{~m}\)
Therefore, area of \(\Delta \mathrm{ABD}=\sqrt{180(180-100)(180-100)(180-160)}\)
\(=\sqrt{180 \times 80 \times 80 \times 20} \mathrm{~m}^{2}=4800 \mathrm{~m}^{2}\)
Therefore, each of them will get an area of 4800 m².

11.2 cm

33750 cm²

\(18\sqrt { 2 } \)cm²

12 cm²

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

\(288\sqrt { 21 } \) cm²

\(100\sqrt { 3 } \)  cm²

'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. 

For triangle   a = 15 cm, b = 14 cm, c = 13 cm
\( \therefore \ =\frac { a+b+c }{ 2 } s=\frac { 15+14+13 }{ 2 } =21\)c m
\(\therefore \)  Area = \(\sqrt { s(s-a)(s-b)(s-c) } \)
 \(=\sqrt { 21(21-15)(21-14)(12-13) } \)
\(=\sqrt { 21(6)(7)(8) } =84\) cm²
Let the height of the parallelogram be h cm.
Then, area of the parallelogram = Base \(\times \) Height = 15 \(\times \) h = 15 cm²
According to the question, Area of the parallelogram = Area of the triangle
\(\Rightarrow \)  15h = 84
\(\Rightarrow \)  \(\frac { 84 }{ 15 } \) = 5.6 cm
Hence, the height of the parallelogram is 5.6 cm.

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 = \(\frac { Area\ of\ the\ field }{ Space\ occupied\ by\ one\ rose\ bed } \)
\(=\frac { 306 }{ 6 } =51\)

Let the side of the equilateral triangle be a cm.
Then it's area = \(\frac { \sqrt { 3 } }{ 4 } \) a² cm²
 \(\frac { \sqrt { 3 } }{ 4 } \) a² = \(81\sqrt { 3 } \)   \(\Rightarrow \) a²= 81\(\times \)4
\(\Rightarrow \)  a = \(\sqrt { 81\times 4 } \)
\(\Rightarrow \)  a = 9\(\times \)2 = 18 cm
\(\therefore \) Perimeter of the equilateral triangle = 31 = 3 \(\times \)18 = 54 cm

Formula

Formula

8\(\times \)4.5 = 9 \(\times \) altitude
\(\Rightarrow \)  Altitude = 4 cm

Formula

Perimeter = 17+33+20 = 70 cm

Formula