# Could anyone help with this question please

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#1
Hi,

This question in a practice paper has me stumped, I think i've done it but wondered if someone could check my answer.

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3 years ago
#2
(Original post by s4rah19)
Hi,

This question in a practice paper has me stumped, I think i've done it but wondered if someone could check my answer.

Not quite. It's way too little.
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3 years ago
#3
(Original post by s4rah19)
Hi,

This question in a practice paper has me stumped, I think i've done it but wondered if someone could check my answer.

I'm nkt exactly sure on the answer but find area of rectangle and area of semi circle then minus area of semi circle from area of rectangle then divide by 2
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#4
I thought about finding the area but didn't see how that would help me find the perimeter.
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3 years ago
#5
(Original post by ussy123)
I'm nkt exactly sure on the answer but find area of rectangle and area of semi circle then minus area of semi circle from area of rectangle then divide by 2
Its asking for the perimeter not the area.
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3 years ago
#6
(Original post by s4rah19)
I thought about finding the area but didn't see how that would help me find the perimeter.
It doesn't.

Firstly, what are the lengths of the straight edges of the shaded region?
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3 years ago
#7
(Original post by s4rah19)
I thought about finding the area but didn't see how that would help me find the perimeter.
Were able to figure out the width of the rectangle?
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#8
I wrote the wrong answer, my scrap sheet of paper is full of scribblings ha ha. Is the answer: 214.247cm
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3 years ago
#9
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3 years ago
#10
(Original post by s4rah19)
I wrote the wrong answer, my scrap sheet of paper is full of scribblings ha ha. Is the answer: 214.247cm
Yes. Round to 3 s.f.
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3 years ago
#11
1.If you split the semicircle into half you know that the angle is 90 degrees.

Spoiler:
Show

90 degrees = pi/2

3. To find the length of the curved edge multiply the angle (in radians) by the radius of the sector.

4. Then add the length of the two straight sides, which are equal to the radius.

Spoiler:
Show

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#12
Thanks for your help, also noticed I wrote cm instead of m 1
3 years ago
#13
(Original post by s4rah19)
Hi,

This question in a practice paper has me stumped, I think i've done it but wondered if someone could check my answer.

assume a circle with a diameter of 120....now you draw the two chords that show the diamter of the circle...one verticle and one horizontal and then you draw tangents to these diameters....you will have a circle inside a square. ...you can't fit a circle in a rectangle so the only possible shape is a circle in a square.

let's divide that shape you assumed into two equal parts by a horizontal line and you will end up with a shape just like in the question.

divide this shape further into two equal parts by a verticle line and you will get that part of the shaded region. now you realise that this part is a 1/4th of the original shape you imagined.

for the perimeter you will need the two sides and the arc length. ..the two sides will be 60 because it was a square initially so the sides remain equal...now you need the arc length.

the arc made there is 1/4 of the original circumference of the whole circle we imagined that means that the length of that arc will be 1/4 * circumference of the circle. and you know circumference = pi*diameter and our diameter was 120 so the length of the arc will be 1/4 * pi*120 = 30pi.

so finally the perimeter is 30pi + 60 + 60 = 30pi + 120.

let me know if I helped 0
3 years ago
#14
(Original post by s4rah19)
Thanks for your help, also noticed I wrote cm instead of m No worries. Just a quick question, is this GCSE or A level maths?
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#15
It's from my nephew's GCSE revision, I think it's from last year's foundation paper.
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3 years ago
#16
Sorry but where did you get this exam paper?
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3 years ago
#17
(Original post by kasacolx)
Sorry but where did you get this exam paper?
On OCR paper 4 higher tier 2017 I believe
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