Perimeter of the rhombus land = 160 m
4a = 160 m
a = 40 m
One of the diagonal = 48 m
∴ Area of the land = 2 \(\times\) Area of the Δ ABC
s = \(\frac { 40+40+48 }{ 2 } =\frac { 128 }{ 2 } \)= 64 m
Area of Δ ABC =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 64(64-40)(64-40)(64-48) } \)
=\(\sqrt { 64\times 24\times 24\times 16 } =\sqrt { 589824 } \)
= 768 m²
∴ Area of the land = 2 \(\times\) 768 m² = 1536 m²

Area of the quadrilateral
= Area of Δ ABD + Area of Δ BCD
Δ ABD is right angled triangle
∴ Area = \(\frac{1}{2}\)b h
=
In Δ ABD, BD²= AD² + AB²
= 15²+ 20²= 225 + 400 = 625 m²
BD =\(\sqrt { 625 } \) = 25 m
∴ In ΔBCD, s = \(\frac { 25+25+17 }{ 2 } \)
= \(\frac { 68 }{ 2 } \) = 4
Area of Δ BCD = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 34(34-25)(34-26)(34-17) } \)
= \(\\ \sqrt { 34\times 9\times 8\times 17 } =\sqrt { 41616 } \)
∴ Area of the quadrilateral
= (150 + 204) m² = 354 m².

Area of the quadrilateral ABCD
= Area of the Δ ABD + Area of the Δ BCD
Sides of the triangle ABD are 13 cm, 14 cm, 15 cm.
s = \(\frac { 13+14+15 }{ 2 } \)cm
= \(\frac { 42 }{ 2 } \) = 21 cm
Area =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 21(21-13)(21-14)(21-15) } \)
=\(\sqrt { 21\times 8\times 7\times 6 } =\sqrt { 7056 } \)=84 cm²
sides of the triangle
BCD are 12 cm, 9 cm, 15 cm
∴ s = \(\frac { 12+9+15 }{ 2 } =\frac { 36 }{ 2 } \) = 18 cm
Area =\(\sqrt { 18(18-2)(18-9)(18-15) } \)
=\(\sqrt { 18\times 6\times 9\times 3 } =\sqrt { 2916 } \) = 54 cm²
∴ Area of the quadrilateral
= 84 cm² + 54 cm² =138 cm²

By the Pythagoras theorem
AB² = AD² + DB²
= 12²+16² = 144 + 256 = 400
AB = 20 cm
s = \(\frac { 34+20+42 }{ 2 } =\frac { 96 }{ 2 } \) = 48
∴ Area of the Δ ABC = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 48(48-34)(48-20)(48-42) } \)
= \(\sqrt { 48\times 14\times 28\times 6 } =\sqrt { 112896 } \)
= \(\\ \sqrt { 336\times 336 } \) = 336 sq.cm
Area of the triangle ABD

∴ Area of the unshaded region
= Area of Δ ABC - Area of Δ ABD
= 336 - 96 = 240 cm².

sides of a triangle = 48 cm, 20 cm, 52 cm,
s = \(\frac { 48+20+52 }{ 2 } =\frac { 120 }{ 2 } \) = 60 cm
(i) Area = \(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 60(60-48)(60-20)(60-52) } \)
=\(\sqrt { 60\times 12\times 40\times 8 } =\sqrt { 230400 } =\sqrt { 480\times 480 } \) = 480 sq.m
(ii) Area of the parallelogram = Area of the triangle (given)
bh = 480 cm²

h = 24 cm

Area of an isoeeles triangle
h = \(\sqrt { { 13 }^{ 2 }-{ 5 }^{ 2 } } =\sqrt { 169-25 } =\sqrt { 144 } \)=12 m
∴ Area of the triangular board
= \(\frac { 1 }{ 2 } \times bh=\frac { 1 }{ 2 } \) \(\times\) 10 \(\times\) 12 = 60 m²
cost of painting 1m²= Rs. 17.50
cost of painting 60m² = 60 \(\times\) 17.50 = Rs. 1050

Perimeter of an equilateral triangle = 180 cm
∴ one side (a) = \(\\ \frac { 180 }{ 3 } \) = 60 m.
Area of an equilateral triangle =\(\frac { \sqrt { 3 } }{ 4 } \) a² sq.units

= 900\(\sqrt { 3 } \) m²
= 900 \(\times\) 1.732 = 1558.8 m²

s = 600 m
side s are in the ratio 5 : 12 : 13
5x + 12x + 13x = 30x
s = 600 ⇒ \(\frac { 30x }{ 2 } \) = 600
30x = 1200
x = 40
∴ sides are 200 m, 480 m, 520 m.
∴ Area =\(\sqrt { s(s-a)(s-b)(s-c) } \)
=\(\sqrt { 600(600-200)(600-480)(600-520) } \)
=\(\sqrt { 600\times 400\times 120\times 80 } =\sqrt { 2304000000 } \)
=\(\sqrt { 48\times 48\times 1000\times 1000 } \) = 48 \(\times\) 1000 =48000 sq.m

side: 22 m, 120 m, 122 m
Using Heron's formula
s = \(\frac { 22+120+122 }{ 2 } =\frac { 264 }{ 2 } \) = 12 m
Area = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 132(132-22)(132-120)(132-122) } \)
= \(\sqrt { 132\times 110\times 12\times 10 } \)
= \(\sqrt { 1742400 } =\sqrt { 11\times 11\times 3\times 3\times 2\times 2\times 2\times 2\times 10\times 10 } \)
= 11 \(\times\) 3 \(\times\) 2 \(\times\) 10 = 1320 m²
cost of levelling 1 m² = Rs. 20
∴ cost oflevelling 1320 m²= 1320 \(\times\) 20 = Rs. 26400

(i) sides: 10 cm, 24 cm, 26 m
Using Heron's formula
Area of the triangle = \(\sqrt { s(s-a)(s-b)(s-c) } \) sq. units
s = \(\frac { a+b+c }{ 2 } =\left( \frac { 10+24+26 }{ 2 } \right) cm=\frac { 60 }{ 2 } \) = 30 cm
∴ Area =\(\sqrt { 30(30-10)(0-24)(0-26) } \)
= \(\sqrt { 30\times 20\times 6\times 4 } =\sqrt { 600\times 24 } =\sqrt { 14400 } \) = 120 cm²
(ii) Sides: 1.8 m, 8m, 8.2 m
s = \(\\ \frac { 1.8+8+8.2 }{ 2 } =\frac { 18 }{ 2 } \) = 9
∴ Area = \(\sqrt { s(s-a)(s-b)(s-c) } \)
= \(\sqrt { 9(9-1.8)(9-8)(9-8.2) } =\sqrt { 9\times 7.2\times 0.8 } \)
= \(\sqrt { 51.84 } \) = 7.2 m²