How to Calculate Area and Circumference of a Unique Field Shape?

[siyanor]

Hello there,
This is my first post in this forum and I am so excited!
I have got this question as my homework ,but i couldn't understand what exactly been asked in this question.the question says :
A right angled isosceles triangle ( with two equal sides ) is cut from each corner of a rectangular field whose width is 40m and length is 60 m. A semi circular plot whose diameter coincides with the hypotenuse of the cut triangle is added to each corner . Thus the field finally composed of two parts: A which is an octagonal ( 8 sided polygon) part and B consisting of four semicircular parts.
Obtain expressions for the area and circumference of the field in terms of x , the length of the leg of the cut triangle and find the domain of each expression.

It's a bit confusing (at least to me!).
just a few tips to clarify the question.
Any help is appreciated

Worm Regards,

 

[pickslides]

Re: an Interesting question!

Some considerations

1st. Draw a picture of the shape(s)

2nd. Let the short side of the isosceles triangle be 'x'

3rd. Label the lengths of each part. Hint: Some side lengths to consider could be (60-2x), (40-2x) & sqrt(x^2+x^2)

 

[siyanor]

Re: an Interesting question!

Hey Pickslides,

Thank you for your reply,
I have drawn the picture as you said.
Now if i want to calculate the circumference of the polygon we can calculate it by summing all sides.but as a second way can we calculate the circumference of the rectangle and then subtract the circumference of two square with side size of x which willl be 2(40+60)-2(4x)?
i couldn't calculate the area of the polygon?how can i calculate the least area of polygon?shall i break it in parts ?

 

[HallsofIvy]

Re: an Interesting question!

Yes, the perimeter of the original retangle is 2(40+ 60)= 200 m and you then cut off corners taking off 8x. But you add back on the circumference of four semi-circles (so two circles). To find those, you need to know the diameter of the circles. That is the "hypotenuse" of the cut triangles so $$d^2= x^2+ x^2= 2x^2$$. Find the circumferences of those two circles.

The area of the original rectangle is 40(60)= 2400 square meters. The four triangles cut off each have area $$(1/2)x^2$$ so have total area $$2x^2$$. Subtract that off. Then add the areas of the two circles (four semicircles). Each has diameter d given by $$d^2= 2x^2$$. The radius is half that.

 

[siyanor]

Re: an Interesting question!

Thank you for replying Hallsofivy.

I think those semicircle will be located on the outside of polygon(having the very same diameter) .now what if we want to calculate the least possible area of polygon ,is there any other way to calculate the area of the polygon ?

 

[siyanor]

Re: an Interesting question!

if i want to calculate the circumference and area of the field can i do it in this way :

Polygon Area = 60*40-4*(x*x)/2 (area of the rectangle minus area of triangle multiple 4)
semicircular area = Pi x^2
Polygon Perimeter =2(40+60)-4*(x+x)+4*x(sqrt2)=200 +(4(sqrt2)-8) >>>perimeter of rectangle minus 4 multiple perimeter of a triangle plus 4 times length of the hypotenuse
semicircular Perimeter= 2(sqrt2)(Pi +2)x >>>we will calculate the perimeter of two circle with radius x(sqrt2)/2

is the above calculation make sense or i have to follow HallsofIvy and pickslides instructions ?

 

[siyanor]

Re: an Interesting question!

is there anyone who can help me out on this issue?