Question

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

Answer 1

Step-by-step explanation:

Join BD in ΔBCD, BC and DC are given.

So, we can calculate BD by applying Pythagoras theorem

⇒BD=

BC

2

+CD

2

=

12

2

+5

2

=

144+25

=13 m=BD

⇒Area of □ABCD= Area of ΔABD+ Area of ΔBCD

⇒Area of ΔBCD

=

2

1

×b×h=

2

1

×12×5

=30 m

2

⇒Area of ΔABD

=

s(s−a)(s−b)(s−c)

(Heron's formula)

⇒2S=9+8+13, S=

2

30

⇒S=15 m

⇒Area of ΔABD

=

15(15−9)(15−8)(15−13)

=

15×6×7×2

=

630×2

=6

1260

=35.49m

2

⇒Area of Park = Quad ABCD

=30+35.49

=65.49 m

2

≈65.5 m

2

Answer 2

Answer:

65.5m² is the area of quadrilateral ABCD

Step-by-step explanation:

First we have to construct a quadrilateral ABCD and join BD as shown in the image.

To Find :-

• Area of the quadrilateral.

Given:-

• Angle c is 90°
• AB is 9 m
• BC is 12 m
• CD is 5 m
• AD is 8 m

First we will find the value of BD by using Pythagoras theorem.

Now playing Pythagoras theorem in triangle BCD

BD² = BC² + CD²

BD² = 12² + 5²

BD² = 144 + 25

BD² = 169

BD= √169

BD = 13m

NOW AREA OF TRIANGLE BCD IS

$\red{ \implies \frac{1}{2} \times 12 \times 5 } \\ \\ \blue{ \implies \: \therefore \: \sf30 {m}^{2} }$

The semi perimeter of triangle ABD is

$\sf \green{semi \: perimeter = \frac{perimeter}{2} } \\ \\ \sf \green{semi \: perimeter = \frac{8 + 9 + 13}{2} } \\ \\ \sf\pink{{semi \: perimeter = \frac{30}{2} = 15m}}$

Now area of ∆ABD

$\red{\sqrt{s(s - a)(s - b)(s - c)} } \\ \\ \implies{\sqrt{15(15 - 13)(15 - 9)(15 - 8)}} \\ \\ \implies \: \sqrt{15 \times 2 \times 6 \times 7 {m}^{2} } \\ \\ \sf \pink{\boxed {\implies65.5 {m}^{2} }}$

$\large \star \mathfrak{ \: \: main \: \: answer \: \: is}\\ \tt65.5m² \: is \: the \: area$