48 m²

74.97 cm²

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

\(1500\sqrt { 3 } \) 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 = \(\sqrt{s(s-a)(s-b)(s-c)}\)
\(=\sqrt{16 \times 8 \times 5 \times 3} \mathrm{~cm}^{2}=8 \sqrt{30} \mathrm{~cm}^{2}\)

\(18\sqrt { 2 } \)cm²

30 cm^2.

480 cm²

Rs. 6705

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 \(\Delta \) 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

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

For triangular ground   a = 5m, b = 7m, c = 8m
\(\therefore s=\frac { a+b+c }{ 2 } =\frac { 5+7+8 }{ 2 } \) m = 10 m
\(\therefore \) Area \(=\sqrt { s(s-a)(s-b)(s-c) } \)
 \(=\sqrt { 10(10-5)(10-7)(10-8) }\)
\( \\ =\sqrt { 300 } =10\sqrt { 3 } \)
=10\(\times \)1.73 =17.3 m²
\(\therefore \) Cost of levelling  = 17.3 \(\times \) 10 = Rs.173

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

Area of \(\Delta \)ABC = \(=\frac { AB\times BC }{ 2 } =\frac { 24\times 10 }{ 2 } \) = 120 m².

Formula

Perimeter = 4\(\times \)5=20 cm.

Longest side = \(\frac { 5 }{ 3+4+5 } \times 144\)=60 m.

s=\(\frac { 20+15+9 }{ 2 } \)=22 cm

Formula

 \(\because \)   7²+24²=25²
\(\therefore \)   Triangle is right angled with hypotenuse 25 cm.
\(\therefore \)    Area =\(\frac { 7\times 24 }{ 2 } \)= 84 cm²

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

Formula