✪ Sanya has a piece of land which is in the shape of a rhombus. She wants her one daughter and one son to work on the land and produce different crops. She divided the land is two equal parts. If the perimeter of the land is 400m and one of the diagonals is 160m, how much area each of them will get for their crops ?
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Answers
Let ABCD be the field.
Given perimeter = 400 m
So, each side = 400/4 = 100 m
Diagonal BD = 160 m
Let a = 100 m
Let B = 100 m
Let C = 160 m
∴ s = (a + b + c)/2 = (100 + 100 + 160)/2 = 180 m.
Therefore, Area of △ABD = root under (√s(s - a)(s - b)(s - c))
= root under (√180(180 - 100)(180 - 100)(180 - 160))
= root under
(√180x80x80x20) =
Alternative method:
As the diagonals of rhombus bisect each other: Therefore
OD = 1/2.BD = 1/2 x 160 = 80 m
OC = 1/2.AC
In △OCD, we have,
Therefore, area of △B CD = 1/2(BD x OC) = 1/2 x 160 x 60 = 4800 m2
∴ Each of them will get 4800 m2 of area for their crops.
Answer:
Let ABCD be the field.
Given perimeter = 400 m
So, each side = 400/4 = 100 m
Diagonal BD = 160 m
Let a = 100 m
Let B = 100 m
Let C = 160 m
∴ s = (a + b + c)/2 = (100 + 100 + 160)/2 = 180 m.
Therefore, Area of △ABD = root under (√s(s - a)(s - b)(s - c))
= root under (√180(180 - 100)(180 - 100)(180 - 160))
= root under
(√180x80x80x20) = 4800 m {}^{2}4800m
2
Alternative method:
As the diagonals of rhombus bisect each other: Therefore
OD = 1/2.BD = 1/2 x 160 = 80 m
OC = 1/2.AC
In △OCD, we have,
CD {}^{2} = OC {}^{2} + OD {}^{2}CD
2
=OC
2
+OD
2
100 {}^{2} = OC {}^{2} + 802 {}^{2} ⇒ OC {}^{2} = 10000 - 6400100
2
=OC
2
+802
2
⇒OC
2
=10000−6400
⇒ OC {}^{2} = 3600 ⇒ OC = 60 m⇒OC
2
=3600⇒OC=60m
Therefore, area of △B CD = 1/2(BD x OC) = 1/2 x 160 x 60 = 4800 m2
∴ Each of them will get 4800 m2 of area for their crops.