Questions
A park, in the shape of a quadrilateral ABCD, has ∠ C = 90º, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
detailed solution
Correct option is A
Let us given that,
∠ C = 90º, AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m
From above information we draw the figure,
we connected B and D such the the triangle is formed
So, the triangle BDC is a right angled triangle
By using the Pythagoras theorem
we get,
BD2 = BC2 CD2
BD2 = 122 52
BD2 = 144 25
BD =
BD = 13 m
Area of quadrilateral ABCD = area of ∆BCD area of ∆ABD
Now, Area of ∆BCD = 1/2 base height
= 1/2 12 m 5 m
= 30 m2
Now, in ∆ABD, AB = a = 9 m, AD = b = 8 m, BD = c = 13 m
Semi Perimeter of ΔABD
s = (a b c)/2
= (9 8 13)/2
= 30/2
= 15 m
We know that, Heron's formula
Area of triangle =
where, s is the semi-perimeter = half of the perimeter
and a, b and c are the sides of the triangle
Area of triangle ABD =
=
=
= 6
= 35.5 m2
Area of triangle ABD = 35.5 m2
Therefore,
Area of park ABCD = 30 m2 35.5 m2 = 65.5 m2
Hence, the park ABCD occupies an area of 65.5 m2.
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detailed solution
Correct answer is 1
Let us given that,
∠ C = 90º, AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m
From above information we draw the figure,
we connected B and D such the the triangle is formed
So, the triangle BDC is a right angled triangle
By using the Pythagoras theorem
we get,
BD2 = BC2 CD2
BD2 = 122 52
BD2 = 144 25
BD =
BD = 13 m
Area of quadrilateral ABCD = area of ∆BCD area of ∆ABD
Now, Area of ∆BCD = 1/2 base height
= 1/2 12 m 5 m
= 30 m2
Now, in ∆ABD, AB = a = 9 m, AD = b = 8 m, BD = c = 13 m
Semi Perimeter of ΔABD
s = (a b c)/2
= (9 8 13)/2
= 30/2
= 15 m
We know that, Heron's formula
Area of triangle =
where, s is the semi-perimeter = half of the perimeter
and a, b and c are the sides of the triangle
Area of triangle ABD =
=
=
= 6
= 35.5 m2
Area of triangle ABD = 35.5 m2
Therefore,
Area of park ABCD = 30 m2 35.5 m2 = 65.5 m2
Hence, the park ABCD occupies an area of 65.5 m2.
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