Question

A park, in the shape of a quadrilateral ABCD, has angle C = 90° l, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

[NCERT MATHS CLASS 9TH CH 12 - HERON's FORMULA EX 12.2 Q2]

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Answer 1

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 :

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

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

$\longmapsto\tt{{(H)}^{2}=25+144}$

$\longmapsto\tt{{(H)}^{2}=169}$

$\longmapsto\tt{H=\sqrt{169}}$

$\longmapsto\tt\bf{H=13\:m}$

Now ,

• a = 5 m
• b = 12 m
• c = 13 m

$\longmapsto\tt{s=\dfrac{a+b+c}{2}}$

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

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

$\longmapsto\tt\bf{s=15\:m}$

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

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

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

$\longmapsto\tt{5\times{3}\times{2}}$

$\longmapsto\tt\bf{30\:{m}^{2}}$

Area of Δ BCD is 30 m² .

Similarly ,

In Δ ABD :

• a = 9 m
• b = 8 m
• c = 13 m

$\longmapsto\tt{s=\dfrac{a+b+c}{2}}$

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

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

$\longmapsto\tt\bf{s=15\:m}$

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

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

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

$\longmapsto\tt{3\times{2}\sqrt{5\times{7}}}$

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

Area of Δ ABD is 30 m² .

Total Area of Park :

$\longmapsto\tt{Area\:of\:\triangle\:BCD+Area\:of\:\triangle\:ABD}$

$\longmapsto\tt{30+6\sqrt{35}}$

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

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

Answer 2

Answer:

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