1.
A park, in the shape of a quadrilateral ABCD, has C = 90°, AB = 9 m. BC = 12.
CD=5 m and AD=8 m. How much area does it occupy?
Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm. CD = 4cm
Answers
Given: The figure is given below:
The dimensions of the park as follows:
Angle C = 90°,
AB = 9m,
BC = 12m,
CD = 5m
And, AD = 8m
Now, BD is joined.
Thus quadrilateral ABCD can now be divided into two triangles ABD and BCD
Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD
Pythagoras Theorem: It states that the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.
In ΔBCD by Pythagoras theorem,
BD2 = BC2 + CD2
BD2 = 122 + 52
BD2 = 169
BD = 13m
As triangle BCD is right angled triangle,
Now, Semi perimeter of triangle (ABD) = (8 + 9 + 13)/2
= 30/2 = 15m
Using Heron’s formula,
ar(ABD) =
where, s = semiperimeter of the triangle and a, b, and c are the sides of the triangle.
=
=
=
= 6√35 m2
= 6 × 5.91= 35.46 m2
Area of quadrilateral ABCD = 35.46 + 30
Area of quadrilateral ABCD = 65.46 m2
Thus, the park acquires an area of 65.5 m2(approx).
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