Question

EXERCISE 12.2
1. A park, in the shape of a quadrilateral ABCD, has 2 C = 90°, AB =9 m. BC
CD=5 m and AD=8 m. How much area does it occupy?
1.​

Answer 1

Correction

Question 1 A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy.

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${\green{\underline{\underline{\huge \mathfrak \red {Answer:-}}}}}$

Total Area of park = Area of ∆ABD + Area of ∆BCD

Area of ∆BCD

Since DC is equal to 5cm ,BC=12, ∠=90°,

Then ∆BCD is a right triangle

$\blue{\boxed{ \rm{Area \: of \triangle \: BCD = \frac{1}{2} \times base \times height}}}</p><p>$

${ \mathtt{Area \: of \triangle \: BCD = \frac{1}{2} \times 12 \times {5m}^{2} }} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt = \frac{1}{2} \times 12 \times 5m^{2} \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{\boxed{\mathtt{= 30 {m}^{2} }}}</p><p>$

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Area of ∆ABD

$\purple{ \boxed {\mathtt{Area \: of \triangle \: ABD = \sqrt{s(s - a)(s - b)(s - c)}}}}$

Here, s is the semi- perimeter and a,b,c are the side of the triangle ,

Here,a = 8m,b = 9m, c = BD

Since= ∠90°

So, applying the Pythagoras theorem.

${ \orange{\boxed{\implies \mathtt{{ BD }^{2} = {BC}^{2} + CD ^{2} }}}}$

$: {\implies \mathtt{{ BD }^{2} = {12}^{2} + 5 ^{2} }} \\ : {\implies \mathtt{{ BD }^{2} = {144} + 25 } } \\ : {\implies \mathtt{{ BD }^{2} = 169}} \\ \rm So,BD = \sqrt{169} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\pink {\boxed {\rm = \sqrt{ {13}^{2} } = 13m}}}$

Hence,c= BD =13 m

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$: \implies{ \rm s = \frac{a + b + c}{2} } \\ : \implies{ \rm s = \frac{8+ 9 + 13}{2} } \\ : \implies{ \rm s = \frac{30}{2} } \\ : \implies{ \rm s = 15m}$

${\mathtt{Area \: of \triangle \: ABD = \sqrt{s(s - a)(s - b)(s - c)}}}$

putting a= 8m,b=9m,c=13,&s=15m

${\mathtt{Area \: of \triangle \: ABD = \sqrt{15(15 - 8)(15 - 9)(15 - 13)}}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\mathtt= \sqrt{15 \times 7 \times 6 \times 2} {m}^{2} }\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{= \sqrt{(3 \times 5) \times 7 \times 6 \times 2} m ^{2} }\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt {= \sqrt{(3 \times 2) \times 6 \times (7 \times 5)} {m}^{2} }\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt {= \sqrt{6 \times 6 \times (35)} {m}^{2} }\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \mathtt {{\sqrt{ {6}^{2} } \times \sqrt{35} {m}^{2} }}\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{= 6 \sqrt{35m ^{2} } }\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \mathtt {= (6 \times 5.91) {m}^{2} } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \mathtt{35.46 {m}^{2} }$

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Area of the park =Area of ∆ ABD +Area of ∆ BCD

$: \implies \mathtt{35.46 + 30 { m}^{2} } \\ : \implies \mathtt{65.46 { m}^{2} } \\ {\red{\fbox{\green{\boxed{: \implies \mathtt \blue{65.5 \: approx}}}}}}$

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Answer 2

Solution:

Given a quadrilateral ABCD in which ∠C = 90º, AB = 9 m, BC = 12 m, CD = 5 m & AD = 8 m.

Join the diagonal BD which divides quadrilateral ABCD in two triangles i.e ∆BCD & ∆ABD.

In ΔBCD,

By applying Pythagoras Theorem

BD²=BC² +CD²

BD²= 12²+ 5²= 144+25

BD²= 169

BD = √169= 13m

∆BCD is a right angled triangle.

Area of ΔBCD = 1/2 ×base× height

=1/2× 5 × 12= 30 m²

For ∆ABD,

Let a= 9m, b= 8m, c=13m

Now,

Semi perimeter of ΔABD,(s) = (a+b+c) /2

s=(8 + 9 + 13)/2 m

= 30/2 m = 15 m

s = 15m

Using heron’s formula,

Area of ΔABD = √s (s-a) (s-b) (s-c)

= √15(15 – 9) (15 – 9) (15 – 13)

= √15 × 6 × 7× 2

=√5×3×3×2×7×2

=3×2√35

= 6√35= 6× 5.92

[ √6= 5.92..]

= 35.52m² (approx)

Area of quadrilateral ABCD = Area of ΔBCD + Area of ΔABD

= 30+ 35.5= 65.5 m²

Hence, area of the park is 65.5m²