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= 9000 cm
2
Now, it can be seen that the quadrilateral ABED is a parallelogram. So,
AB = ED = 10 m
AD = BE = 13 m
EC = 25 – ED = 25 – 10 = 15 m
Now, consider the triangle BEC,
Its semi perimeter (s) = (13+ 14 + 15)/2 = 21 m
By using Heron’s formula,
Area of ΔBEC =
= 84 m
2
We also know that the area of ΔBEC = (½) × CE × BF
84 cm
2
= (½) × 15 × BF
=> BF = (168/15) cm = 11.2 cm
So, the total area of ABED will be BF × DE, i.e. 11.2 × 10 = 112 m
2
∴ Area of the field = 84 + 112 = 196 m
2
Consider the triangle BCD,
Its semi-perimeter = (48 + 30 + 30)/2 m = 54 m
Using Heron’s formula,
Area of the ΔBCD =
= 432 m
2
∴ Area of field = 2 × area of the ΔBCD = (2 × 432) m
2
= 864 m
2
Thus, the area of the grass field that each cow will be getting = (864/18) m
2
= 48 m
2
Solution:
According to the figure,
AC = BD = 44cm
AO = 44/2 = 22cm
BO = 44/2 = 22cm
From ΔAOB,
AB
2
= AO
2
+ BO
2
⇒ AB
2
= 22
2
+ 22
2
⇒ AB
2
= 2 × 22
2
⇒ AB = 22√2 cm
Area of square = (Side)
2
= (22√2)
2
= 968 cm
2
Area of each triangle (I, II, III, IV) = Area of square /4
= 968 /4
= 242 cm
2
To find area of lower triangle,
Let a = 20, b = 20, c = 14
s = (a + b + c)/2
⇒ s = (20 + 20 + 14)/2 = 54/2 = 27.
Area of the triangle = √[s(s-a)(s-b)(s-c)]
= √[27(27-20)(27-20)(27-14)]
= √[27×7×7×13]
= 131.14 cm
2
Therefore,
We get,
Area of Red = Area of IV
= 242 cm
2
Area of Yellow = Area of I + Area of II
= 242 + 242
= 484 cm
2
Area of Green = Area of III + Area of the lower triangle
= 242 + 131.14
= 373.14 cm
2
Let the given rectangle be rectangle PQRS,
According to the question,
PQ = 40m and QR = 15m
As 3m is left in both front and back,
AB = PQ -3 -3
⇒ AB = 40 -6
⇒ AB = 34m
Also,
Given that 2m has to be left at both the sides,
BC = QR -2 – 2
⇒ BC = 15 -4
⇒ BC = 11m
Now, Area left for house construction is the area of ABCD.
Hence,
Area(ABCD) = AB × CD
= 34 × 11
= 374 m
2
