Question

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Answer 1

Step-by-step explanation:

given :

• A park, in the shape of a quadrilateral ABCD

• c = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

to find :

• How much area does it occupy = ?

• How much area does it occupy = ?

solution :

• please check the full attached file

Answer 2

Given :

• A park in the shape of a quadrilateral ABCD, has ∠C = 90° ,AB = 9 cm, BC = 12 cm, CD = 5 cm and AD = 8 cm.

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To Find :

• How much area does this park occupy.

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Solution :

${\blue{✇}}$Here :

After constructing the quadrilateral the figure will be divided into 2 parts a right - angled triangle and another triangle .

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${\blue{✇}}$Formula Used :

$\large{\color{cyan}{\bigstar}} \: \: {\underline{\boxed{\color{red}{\sf{ Area{\small_{(Triangle) }} = \dfrac{1}{2} \times Base \times Height }}}}}$

$\large{\color{cyan}{\bigstar}} \: \: {\underline{\boxed{\color{red}{\sf{ Area{\small_{(Triangle) }} = \sqrt{s (s - a) (s - b) (s - c) }}}}}}$

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${\blue{✇}}$Calculating the Area of ΔBCD :

~ Here :

• ➙ Base = 12 m
• ➙ Height = 5 m

~ Calculating the Area :

${\longmapsto{\qquad{\rm{ Area{\small_{(Triangle) }} = \dfrac{1}{2} \times Base \times Height }}}} \\ \\ \ {\longmapsto{\qquad{\rm{ Area{\small_{(Triangle) }} = \dfrac{1}{2} \times 12 \times 5 }}}} \\ \\ \ {\longmapsto{\qquad{\rm{ Area{\small_{(Triangle) }} = \dfrac{1}{\cancel2} \times \cancel{60} }}}} \\ \\ \ {\longmapsto{\qquad{\rm{ Area{\small_{(Triangle) }} = 1 \times 30 }}}} \\ \\ \ {\qquad{\sf{ Area \: of \: BCD \: = {\green{\sf{ 30 \: m² }}}}}}$

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${\blue{✇}}$Calculating the Area of Δ ABD :

~ Here :

• ➙ Side 1 = 8 cm
• ➙ Side 2 = 9 cm
• ➙ Side 3 = BD

~ Calculating the 3rd side using Pythagoras theorem :

${\longrightarrow{\qquad{\sf{ H² = B² + H² }}}} \\ \\ \ {\longrightarrow{\qquad{\sf{ BD² = (12)² + (5)² }}}} \\ \\ \ {\longrightarrow{\qquad{\sf{ BD² = 144 + 25 }}}} \\ \\ \ {\longrightarrow{\qquad{\sf{BD² = 169 }}}} \\ \\ \ {\longmapsto{\qquad{\sf{ BD = \sqrt{169}}}}} \\ \\ \ {\qquad{\sf{ 3rd \: Side \: = {\red{\sf{ 13 \: m}}}}}}$

~ Now, The Area of triangle ABD :

Semi - Perimeter :

${\longrightarrow{\qquad{\sf{S = \dfrac{a + b + c}{2}}}}} \\ \\ \ {\longrightarrow{\qquad{\sf{ S = \dfrac{8 + 9 + 13}{2}}}}} \\ \\ \ {\longrightarrow{\qquad{\sf{ S = \cancel\dfrac{30}{2}}}}} \\ \\ \ {\qquad{\sf{ Semi - \: Perimeter \: = {\pink{\sf{ 15 \: m}}}}}}$

Area :

${\longmapsto{\qquad{\sf{ Area{\small_{(Triangle) }} = \sqrt{s (s - a) (s - b) (s - c) }}}}} \\ \\ \ {\longmapsto{\qquad{\sf{ Area{\small_{(Triangle) }} = \sqrt{15 (15 - 8) (15 - 9) (15 - 13) }}}}} \\ \\ \ {\longmapsto{\qquad{\sf{ Area{\small_{(Triangle) }} = \sqrt{15 \times 7 \times 6 \times 2 }}}}} \\ \\ \ {\longmapsto{\qquad{\sf{Area{\small_{(Triangle) }} = 6 \times \sqrt{35} }}}} {\bigg\lgroup {\pink{\sf{ Taking \: \sqrt{35} = 5.91 }}} \bigg\rgroup } \\ \\ \ {\longmapsto{\qquad{\sf{ Area{\small_{(Triangle) }} = 6 \times 5.91 }}}} \\ \\ \ {\qquad{\sf{ Area \: = {\purple{\sf{ 35.46 \: m² }}}}}}$

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~ Calculating Area of Quadrilateral :

${\longrightarrow{\qquad{\sf{ Area{\small_{(Quadrilateral) }} = Triangle_1 + Triangle_2 }}}} \\ \\ \ {\longrightarrow{\qquad{\sf{ Area{\small_{(Quadrilateral )}} = 30 \: m² + 35.46 \: m² }}}} \\ \\ \ {\qquad{\pink{:\longmapsto}}{\underline{\overline{\boxed{\color{orange}{\sf{ 65.5 \: m²(Approx.) }}}}}}}{\blue{\bigstar}}$

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${\blue{✇}}$Therefore :

❝ Area occupied by the park is 65.5 m² (Approx.) . ❞

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