Embarrassing question to ask

Mattjones12291

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Apr 16, 2019
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Hi guys, I have a totally embarrassing question to ask.

Iv always been hopeless at maths.

Basically I have recently purchased a home and have been doing the garden, the previous owners had a raised flower bed that I've removed, leaving a bare patch of grass in the corner in a quadrant shape.

The two sides that run along the fence are 290cm and 260 cm.

I need to find out the rough area of the (almost) quadrant so I can purchase turf to go over it.

Any help would be greatly appreciated thanks
 

Jomo

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Dec 30, 2014
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The area of a rectangle is length times width.
 

Mattjones12291

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Apr 16, 2019
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But it's like a quarter circle?
 

Dr.Peterson

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How are you using the word "quadrant"?

Jomo is apparently assuming you mean "rectangle" (which I agree is likely); but a quadrant is actually either a quarter of a circle, or a quarter of a plane. Neither would fit your description, but it could be close to a quarter of a circle with radius 290 cm, perhaps.

In any case, very likely the people where you buy the turf can help you not only with finding the area, but also with practicalities like fitting the form in which the sod comes to the actual dimensions of the patch.
 

Harry_the_cat

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The area of a circle is calculated by \(\displaystyle \pi r^2\) where r is the radius and \(\displaystyle \pi\) is about 3.14.

Your shape can't be a quarter circle because the 2 sides that run along the fence are different lengths. But as you said its "like" a quarter circle.

So if you average out the lengths to get (260+290)/2 = 275cm =2.75m, that'll be close enough to the figure you can use for the radius.

\(\displaystyle \pi r^2 = 3.14*2.75^2 = 23.75\) square metres

You only have one quarter of this so, \(\displaystyle 23.75 \div 4 = 5.937\) square metres.

So order 6 square metres. Happy gardening!
 

HallsofIvy

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You can treat this "like a circle figure" as an ellipse. The area of an ellipse with "semi-axes" of length $r_1$ and $r_2$ is simply $\pi r_1r_2$. An ellipse with one axis of length 2.9 m and 2.6 m has area (3.14)(2.9)(2.6)= 23.7 square meters. That is essentially the same as Harry_The_Cat got by treating this as a circle with radius the average of 2.9 and 2.6. Your quadrant is 23.7/4= 5.92 square meters or close enough to 6 square meters that you can use that.
 

Dr.Peterson

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Now that we know it's something like a quarter of an ellipse, we get into the practicalities I mentioned. The 6 m^2 you get would presumably be in a rectangular shape, and it might be hard to cut it into pieces that would fill in the space you need. So you'll have to buy a little extra, probably, and make that decision based on a specific plan for fitting it in.

On the other hand, turf will probably be easier to cut and fit than if it were solid stone or tile!
 

Dr.Peterson

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Nov 12, 2017
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Yes, I intentionally added the word "solid" to eliminate "stone" in the sense of gravel, which doesn't need cutting. That would fit just fine. Maybe I should have said "a slab of stone".

But I suppose liquid stone (lava?) would also be hard to work with.
 
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