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EFRENSALAC32IN
Answered

What I Can Do

A. Solve the value of x. What are the dimensions of the field?

B. What is the length of the wire required to put around the field to complete it's boundary?

C. How much area is required to be irrigated in his field?

GUYS PAHELP NAMAN PO PARA PO SA MATH JOURNAL KO PO. THANKS PO​

What I Can DoA Solve The Value Of X What Are The Dimensions Of The FieldB What Is The Length Of The Wire Required To Put Around The Field To Complete Its Bounda class=

Sagot :

C41992IN

Answer:

A. x=8

B. 136m

c. 832 m²

Step-by-step explanation:

  • The figure is a rectangle
  • Rectangle has 4 right triangles.
  • Rectangle has 2 equal parallel sides

A. any of the given may do.

in width

2x = x+ 8

x= 8

in length

7x-4 = 6x+4

x=8

B. Perimeter = 2L + 2W = 2 (L+W)

L = 7(8)-4=52m

W = 2(8) = 16m

total length of wire is the Perimeter of entire lot

P = 2(52+16) = 136m

C. Area = LW

A = (52)(16) = 832m²

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