<P>

Given :

∠C = 90°

AB = 9 m

BC = 12 m

CD = 5 m

AD = 8 m

To FIND :

Area of the park .

Solution :

In Δ BCD :

Base = 5 m

Height = 12 m

Hypotenuse = ?

By Pythagorus THEOREM :

\longmapsto\tt{{(H)}^{2}={(B)}^{2}+{(P)}^{2}}⟼(H)

2

=(B)

2

+(P)

2

\longmapsto\tt{{(H)}^{2}={(5)}^{2}+{(12)}^{2}}⟼(H)

2

=(5)

2

+(12)

2

\longmapsto\tt{{(H)}^{2}=25+144}⟼(H)

2

=25+144

\longmapsto\tt{{(H)}^{2}=169}⟼(H)

2

=169

\longmapsto\tt{H=\sqrt{169}}⟼H=

169

\longmapsto\tt\bf{H=13\:m}⟼H=13m

Now ,

a = 5 m

b = 12 m

c = 13 m

\longmapsto\tt{s=\dfrac{a+b+c}{2}}⟼s=

2

a+b+c

\longmapsto\tt{s=\dfrac{5+12+13}{2}}⟼s=

2

5+12+13

\longmapsto\tt{s=\cancel\dfrac{30}{2}}⟼s=

2

30

\longmapsto\tt\bf{s=15\:m}⟼s=15m

\longmapsto\tt{Area=\sqrt{s(s-a)(s-b)(s-c)}}⟼Area=

s(s−a)(s−b)(s−c)

\longmapsto\tt{\sqrt{15(15-5)\:(15-12)\:(15-13)}}⟼

15(15−5)(15−12)(15−13)

\longmapsto\tt{\sqrt{15\:(10)\:(2)\:(2)}}⟼

15(10)(2)(2)

\longmapsto\tt{5\times{3}\times{2}}⟼5×3×2

\longmapsto\tt\bf{30\:{m}^{2}}⟼30m

2

Area of Δ BCD is 30 m² .

Similarly ,

In Δ ABD :

a = 9 m

b = 8 m

c = 13 m

\longmapsto\tt{s=\dfrac{a+b+c}{2}}⟼s=

2

a+b+c

\longmapsto\tt{s=\dfrac{9+8+13}{2}}⟼s=

2

9+8+13

\longmapsto\tt{s=\cancel\dfrac{30}{2}}⟼s=

2

30

\longmapsto\tt\bf{s=15\:m}⟼s=15m

\longmapsto\tt{Area=\sqrt{s(s-a)(s-b)(s-c)}}⟼Area=

s(s−a)(s−b)(s−c)

\longmapsto\tt{\sqrt{15(15-8)\:(15-9)\:(15-13)}}⟼

15(15−8)(15−9)(15−13)

\longmapsto\tt{\sqrt{15\:(7)\:(6)\:(2)}}⟼

15(7)(6)(2)

\longmapsto\tt{3\times{2}\sqrt{5\times{7}}}⟼3×2

5×7

\longmapsto\tt\bf{6\sqrt{35}\:{m}^{2}}⟼6

35

m

2

Area of Δ ABD is 30 m² .

Total Area of Park :

\longmapsto\tt{Area\:of\:\triangle\:BCD+Area\:of\:\triangle\:ABD}⟼Areaof△BCD+Areaof△ABD

\longmapsto\tt{30+6\sqrt{35}}⟼30+6

35

\longmapsto\tt\bf{65.4\:{m}^{2}}⟼65.4m

2

So , The Area Occupied by the park is 65.4 m² ..