Question

A plot is in the shape of quadrilateral abcd where angle c is 90 degree, AB is 7cm, BC is 4cm CD is 3cm and AD is 6cm
How much area does the plot occupy

Answer 1

Given :-

• AB = 7 cm
• BC = 4cm
• CD= 3cm
• AD = 6cm
• <C = 90°

To find :- Area of plot

Construction:- Join Line BD. So that one triangle will form ABD and second as BCD.

SolutioN:-

• First we will find area of ∆ BCD beacuse it is also part of quadrilateral ABCD.

Area of ∆ BCD = 1 /2 × base × height

➟ 1/2 × 4 × 3

➟ 2 × 3

➟ 6 cm²

• We know that right angle triangle have one angle as 90° . So , we can say that ∆BCD is a Right angle triangle.
• Now , we will find area of triangle ∆ ABD , but for it we need need length of BD.
• So, for finding BD we use Pythagoras theorem in ∆BCD because it is also a side of this triangle. And ∆BCD is a Right angle triangle.

Formula of Pythagoras theorem is :

✑ (Base)+(perpendicular)² = (Hypotenuse)²

Now ,

• DC = Base = 3cm
• BC = height = perpendicular = 4cm
• BD = Hypotenuse = ?

➟ (3)² + (4)² = (BD)²

➟ 9 + 16 = (BD)²

➟ 25 = (BD)²

➟ BD = √25

➟ BD = 5cm

Now, area of ∆ABD

We use heron's formula because all sides are not Equal.

Heron's formula is :-

✑ Area of triangle = Whole√s(s - a)(s - b)(s -c)

Then , S = perimeter/2

S = 7 + 6 + 5/2

S = 18/2

S= 9cm

Area of triangle

➟ Whole √ 9× (9 - 7) ( 9 - 6)( 9-5)

➟ Whole √ 9 × 2 × 3× 4

➟ Whole √ 3 × 3 × 2 × 3 × 2 × 2

➟ 3× 2 √ 2 × 3

➟ 6 √6 cm²

So, Area of plot ABCD = area of ∆ABD + area of ∆BCD

➟ 6 + 6√6

⛬ √6 = 2.449 (approx.)

➟ 6 + 6 × 2.449

➟ 6 + 14. 6969

➟ 20. 6969cm² (approx.)

Hence area of plot ABCD is 20.6969cm² ( approximately).

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