Since AB = 9 m and BC = 40 m, Ð B = 90°, we have

\( \mathrm{AC} =\sqrt{9^{2}+40^{2}} \mathrm{~m} \)
\(=\sqrt{81+1600} \mathrm{~m} \)
\(=\sqrt{1681} \mathrm{~m}=41 \mathrm{~m} \)
Therefore, the first group has to clean the area of triangle ABC, which is right angled.
\(Area of \Delta \mathrm{ABC}=\frac{1}{2} \times base \times height \)
\(=\frac{1}{2} \times 40 \times 9 \mathrm{~m}^{2}=180 \mathrm{~m}^{2} \)
The second group has to clean the area of triangle ACD, which is scalene having sides 41 m, 15 m and 28 m.
Here, \(s=\frac{41+15+28}{2} \mathrm{~m}=42 \mathrm{~m}\)
Therefore, area of \(\Delta \mathrm{ACD}=\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{42(42-41)(42-15)(42-28)} \mathrm{m}^{2} \)
\( =\sqrt{42 \times 1 \times 27 \times 14} \mathrm{~m}^{2}=126 \mathrm{~m}^{2} \)
So first group cleaned 180 m² which is (180 – 126) m², i.e., 54 m² more than the area cleaned by the second group.
Total area cleaned by all the students = (180 + 126) m² = 306 m².

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.

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 \(\sqrt{150(150-60)(150-100)(150-140)} \mathrm{m}^{2}\)
\(=\sqrt{150 \times 90 \times 50 \times 10} \mathrm{~m}^{2}\)
\(=1500 \sqrt{3} \mathrm{~m}^{2}\)

33750 cm²

30 cm^2.

48 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

In right triangle PSQ,  PQ\(\frac { 1 }{ 2 } \)² = PS² + QS²     |By Pythagoras Theorem
= (12)² + (16)²
= 144 + 256 =400
\(\Rightarrow \)  PQ = \(\sqrt { 400 } \) = 20 cm
Now, for  \(\Delta \)PQR
a = 20cm, b = 48cm, c = 52cm
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 20+48+52 }{ 2 } =60\) cm
\(\therefore \)  Area of  \(\Delta \)PQR \(=\sqrt { s(s-a)(s-b)(s-c) } \)
\(=\sqrt { 60(60-20)(60-48)(60-52) }\)
\( \\ =\sqrt { (60)(40)(12)(8) } \)
 \(=\sqrt { \left( 6\times 10 \right) \left( 4\times 10 \right) \left( 6\times 2 \right) \left( 8 \right) } \)
\(=6\times 10\times 8=480\) cm²
Area of \(\Delta \) PSQ = \(\frac { 1 }{ 2 } \)\(\times \)Base\(\times \)Altitude
=\(\frac { 1 }{ 2 } \)\(\times \)16\(\times \)12=96 cm²
\(\therefore \) Area of the shaded portion =Area of \(\Delta \)PQR - Area of \(\Delta \)PSQ
= 480 - 96 = 384 cm²

Let the sides forming the right angle be a cm and b cm. Then,
      
a + b + 10 = 24
\(\Rightarrow \)  a + b = 14               ....(1)
Also,  a² + b² = (10)²     |By Pythagoras Theorem
\(\Rightarrow \)   a² + b² = 100      ...(2)
We know that   (a + b)² = a² + b² + 2ab
\(\Rightarrow \)  (14)² = 100+2ab
\(\Rightarrow \) 2ab = 96
\(\Rightarrow \)  ab=48                   ...(3)
Also,    (a - b)² = a² + b²- 2ab
= 100 - 2\(\times \)48
= 100 - 96 = 4       | if a>b  
\(\Rightarrow \)  a-  b = 2
Solving (1) and (4), we get   a=8cm, b=6cm
\(\therefore \)   Area = \(\frac { 1 }{ 2 } \)ab = \(\frac { 1 }{ 2 } \).8.6 = 24 cm²

Area = 6\(\left\{ \frac { \sqrt { 3 } }{ 4 } { (8) }^{ 2 } \right\} \) = \(96\sqrt { 3 } \) cm²

Area = \(\frac { (45.8+81.2)\times 22 }{ 2 } \)
= 127\(\times \)11 = 1397 cm²

Formula

See Hero's formula.

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

\(\frac { \sqrt { 3 } }{ 4 } { (6) }^{ 2 }=9\sqrt { 3 }\) cm².

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

Formula

Formula