The sides of a triangular field are 41m and 9m find the number of rose beds that can be prepared in the field, if each rose bed, on an average needs 900m2
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Solution:-
Given : two sides = 41 m (Hypotenuse or longest side) and 9 m (base) perpendicular = ?
By Pythagoras Theorem,
H² = B² + P²
41² = 9² + P²
P² = 41² - 9²
P² = 1681 - 81
P² = 1600
P = √1600
P = 40 m
Now we will find out the area of triangle by the Heron's formula.
Let a = 41 m, b = 40 m and c = 9 m
s = (a+b+c)/2
s = (41+40+9)/2
s = 90/2
s = 45 m
area of triangle = √s(s-a)(s-b)(s-c)
⇒ √45(45-41)(45-40)(45-9)
⇒ √45×4×5×36
⇒ √32400
Area = 180 m²
There is a mistake in this question. If the area of the triangular field having sides 41 m, 40 m and 9 m is 180 m² then how can a rose bed needs space of 900 m² because it is more than the area of the whole field Instead it must be 900 cm² of space that is needed by a rose bed.
Number of rose beds can be prepared in the field
= Total area of field/area occupied by a rose bed
⇒ 180/0.09 (As 1 m² = 10000 cm²)
= 2000 rose beds can be prepared.
Given : two sides = 41 m (Hypotenuse or longest side) and 9 m (base) perpendicular = ?
By Pythagoras Theorem,
H² = B² + P²
41² = 9² + P²
P² = 41² - 9²
P² = 1681 - 81
P² = 1600
P = √1600
P = 40 m
Now we will find out the area of triangle by the Heron's formula.
Let a = 41 m, b = 40 m and c = 9 m
s = (a+b+c)/2
s = (41+40+9)/2
s = 90/2
s = 45 m
area of triangle = √s(s-a)(s-b)(s-c)
⇒ √45(45-41)(45-40)(45-9)
⇒ √45×4×5×36
⇒ √32400
Area = 180 m²
There is a mistake in this question. If the area of the triangular field having sides 41 m, 40 m and 9 m is 180 m² then how can a rose bed needs space of 900 m² because it is more than the area of the whole field Instead it must be 900 cm² of space that is needed by a rose bed.
Number of rose beds can be prepared in the field
= Total area of field/area occupied by a rose bed
⇒ 180/0.09 (As 1 m² = 10000 cm²)
= 2000 rose beds can be prepared.
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