Q. 5

# A rectangular yar

Let us consider a Rectangular yard ABCD,

According to the problem, it is given that

ΔADF, ΔBCE are a congruent isosceles right triangle.

It is also given that,

Length of AB is 25m.

Length of BC is 15m.

From the figure, we can clearly say that,

AB = DF + FE + EC and

DF = EC.

From the above equations and data are given in the problem,

25 = (2 × DF) + 15

2 × DF = 25 - 15

2 × DF = 10 DF = 5m

We can see that AD = DF,

So breath of Rectangle ABCD is 5m.

We know that Area of Rectangle = Length × Breadth From the figure and calculations, we can clearly say,

Length of Rectangle ABCD is 25m.

The breadth of Rectangle ABCD is 5m.

The base of Triangle ADF is 5m.

The height of Triangle ADF is 5m.

Area of Rectangle ABCD = 25 × 5

Area of Rectangle ABCD = 125m2.

Area of Triangle Area of Triangle Area of Triangle ADF = 12.5m2

The area in the Rectangular yard that containing flower beds is the area of two Triangles ADF and BCE.

But according to the problem, it is told that ΔADF and ΔBCE are congruent.

So, the area of both the triangles is the same.

Area of ΔADF = Area of ΔBCE = 12.5m2.

The area occupies by flower beds = Area of ΔADF + Area of ΔBCE.

The area occupies by flower beds = 12.5 + 12.5

The area occupies by flower beds = 25m2.

The fraction(f) of the yard that is occupied by the flower bed is given by   The fraction is .

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