r/HomeworkHelp • u/Particular-Cat-5331 GCSE Candidate • 1d ago
High School Math—Pending OP Reply (GCSE Maths-Area) Can someone help me?
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u/___OldUser101 Pre-University Student 1d ago
For the first one, you can break the shape down into 2 rectangles and 2 triangles, you should be able to calculate the area from there. You can do the same for the second, or use the trapezium YXWV.
For the other two, you need to calculate the area of the whole rectangle using the provided side lengths, and subtract the non-shaded areas to find the shaded area.
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u/SinisterSnipes 1d ago
For the first one, a rectangle minus the area of a trapezoid is more efficient.
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u/WishboneHot8050 👋 a fellow Redditor 1d ago edited 1d ago
For the first problem:
See extended picture here for where I added some more referenced points: https://imgur.com/a/O2mY3e7
TLDR: You know there's a core rectangle that is 11*21 and two "triangle" sections to add in.
The trick is to remember that a right triangle's area is half of what it would be of a rectangle diagnolly sliced. And if you have two rectangles you know the total area for, the area of the two triangles sliced out of it will still be half.
area(ABEF) is 21*17 = 357
area(CGHD) is 6*13 = 78
area(IBEF) is 6*21 = 126
Area(AIJF) is 21*11 = 231
area(IBGC) + area(DHEJ) is area(IBEF) minus area(CGHD)
area(IBGC) + area(DHEJ) is 126 - 78 = 48
But we only care about the triangle areas!
area(IBC) + area(DEJ) is half of area(IBGC) + area(DHEJ) = 24
Area(AIJF) is 21*11 = 231
Add the triangle areas back in:
Total hexagon areas = 231 + 24 = 255
We don't need to know the individual areas of IBC and DEJ.
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u/WishboneHot8050 👋 a fellow Redditor 1d ago edited 1d ago
The second problem looks to have a similar solution as the first. You're going to have two triangles that you won't know the individual areas of. But you can determine their total area - just like above.
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u/WishboneHot8050 👋 a fellow Redditor 1d ago
The third problem and fourth problem have a technical issue. The give height of the rectangles are truncated by the margins on the left side. You can probably get them if you switched to using a desktop browser on a PC or Mac. Then these problems become solvable.
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u/SummerEden 1d ago
For the first image it would be easier to take area of the larger rectangle and subtract the trapezium, which has easily determined parallel sides and perpendicular height.
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u/TruckerJay 1d ago
Even simpler, the trapezoid BCDE has parallel lengths of 13 (CD) and 21 (BE) and a width of 6. The area of a parallel trapezoid is (the average of the parallel lengths) x the width. 17x6 =102
Take 102 from the bigger core rectangle
357-102=255
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u/OtherwiseAd2040 1d ago
Pretty straightforward, find area of trapezium BCDE, and substract it from area of rectangle ABEF
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u/Deapsee60 👋 a fellow Redditor 1d ago
1 is a rectangle with a trapezoid cut out
(17)(21) - 1/2(6)(13 + 21) = 255
2 is a rectangle + a trapezoid
(13)(15) + 1/2(6)(8 + 15) 264
Can’t read all measurements on #3 & #4
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u/Particular-Cat-5331 GCSE Candidate 1d ago
4 is the measurement that you cannot see is 4cm. Are you able to do it now?
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u/One_Wishbone_4439 University/College Student 1d ago
How is the length of the rectangle 4 cm longer than 6 cm??
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u/CardinalCountryCub 21h ago
Are you sure it's not 24cm?
We can see the last digit is 4, but 4cm doesn't make sense as a side length that is clearly longer than the given 14cm up top. It also can't be 14 cm because, again, it's longer, and also it would be labelled "square" and not "rectangle."
Also, when assuming it's 24 cm, the answer comes out to a nice, whole number when you find the area of the rectangle and then subtract the areas of the 3 given triangles.
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u/selene_666 👋 a fellow Redditor 1d ago
Split each figure into rectangles, triangles, and trapezoids whose areas you can calculate. Then add or subtract accordingly.
For example, the last problem is the area of a square minus the areas of three triangles.
Where you know enough parallel sides to say that something is a rectangle, the opposite sides must have the same length. In the second problem, YZUV is a rectangle with YV = 15mm.
If you have two triangles whose individual heights you don't know, but you know the sum of their heights ,you may still be able to find the sum of their areas.
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u/ACTSATGuyonReddit 👋 a fellow Redditor 1d ago
1st one: since the heights are the same, treat the two triangles as one triangle. Height is 17-11 = 6. Base is 21-13 = 8. The area of the two triangles is 1/2 * 6 * 8 = 24.
The remaining rectangle is 21 by 11: 21 * 11 = 231
231 + 24 = 255
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u/Jaymac720 👋 a fellow Redditor 1d ago
Cut off the triangles. Get those areas separately from the rectangles. The shapes are fully defined. Just draw some lines to separate them
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u/gerburmar 23h ago edited 23h ago
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A simple "positive" way is seeing a rectangle, and seeing that its area should be 12*11. the difference between 17 and 11 gives the "heights' of two triangles whose areas need summed to that area. Then comes a trick with their bases. (1/2)bh + (1/2)bh = (1/2)(b+b)h where say one base is "b1" and one b is "b2" it doesn't matter what they each are only what the sum is. You can see from the difference between 13 and 21 the total difference should be 8 and that is the sum of their bases.
21*11 + (1/2)*(8)*(6)
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there is a 'negative' way of taking the area of 21*17 and subtracting out a trapezoid area that seems more complex, but it can be viewed as taking that 21*17 and subtracting out (1/2)*(21+13)*(6) and you can still get the same answer.
21*17 - (1/2)*(21+13)*(6)
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u/Paounn 1d ago
What did you try? Have you ever tried anything?
First one is easily done by subtraction, join B and E.
Second one Similarly, join Y and V.
Third, get the surface, again by subtraction, then divide by 17 and round up. That's the number of bags, multiply by 11, price per bags, then by 0,7 for the discount.
Fourth, it's a square you removed the corners, again by subtraction.
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u/tay_tfs 1d ago
Since it's not specified, I'd assume that BC is 6cm long, thus leading to C lying on AB. From there it's simpler.
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u/pogsnacks 1d ago
You can't do that
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u/DerfK 1d ago
You can do that! The trick is that the trapezoid shape's area doesn't actually depend on the angles of the triangle bits on the end. You could have no triangle on one side and an 8x5 triangle on the other and the area is the same as having a 3x5 and 5x5 triangle or whatever.
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u/pogsnacks 1d ago
You can say the height of the trapezoid is 6 cm but you can't say BC is 6 cm
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u/ThunkAsDrinklePeep Educator 1d ago
If you're only finding the area it doesn't matter. If you extend CD, the two triangles have indeterminate individual size, but the sum of those areas is fixed.
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u/One_Wishbone_4439 University/College Student 1d ago
Why would u need to find BC? It's not necessary to do that.
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u/One_Wishbone_4439 University/College Student 1d ago
1st problem: rectangle - trapezoid
2nd problem: rectangle - two triangles
3rd problem: rectangle - triangle
4th problem: rectangle - three triangles