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Secondary 1 | Maths
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Date Posted: 4 years ago
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I assume that the front face has the length and height as its dimensions, and top face has length and breadth as its dimensions.
Let the height be H cm , length be L cm breadth be B cm.
Area of front and top face combined = L × H + L × B
35 = LH + LB = L(H+B)
L is prime so we have to express 35 as a product of two factors, whereby one factor is a prime number (L) and the other factor is (H+B)
35 = 7 x 5
Now both 7 and 5 are prime. So which one is L?
① If L = 7, then H + B = 5
This can be rexpressed as the sum of two prime numbers. H + B = 2 + 3. So we can have B = 2, H = 3 or B = 3, H = 2.
② If L = 5, then H + B = 7
We can also rexpress this as the sum of 2 numbers. H + B = 2 + 5 . So we can have H = 2, B = 5 or H = 5, B = 2
Now the question doesn't specify that the dimensions must all different prime numbers. If we want all the dimensions to be different and breadth to be strictly smaller than length, then we go with option ①
Volume of cuboid = Lcm x Bcm x Hcm = 7cm × 3cm × 2cm = 42cm³ or 7cm × 2cm × 3cm = 42cm³ (either way, we get 42cm³ so it doesn't matter if B = 2, H = 3 or B = 3, H = 2)
Now front and back faces have same surface area. Likewise for left and right, top and bottom.
Total surface area of cuboid = 2 × (Lcm × Bcm + Lcm × Hcm + Bcm × Hcm)
= 2 × (LBcm² + LHcm² + 2cm × 3cm)
= 2 × (35cm² + 6cm²)
= 2 × 41cm²
= 82cm²
Or
= 2 × (LBcm² + LHcm² + 3cm × 2cm)
= 2 × (35cm² + 6cm²)
= 2 × 41cm²
= 82cm²
Let the height be H cm , length be L cm breadth be B cm.
Area of front and top face combined = L × H + L × B
35 = LH + LB = L(H+B)
L is prime so we have to express 35 as a product of two factors, whereby one factor is a prime number (L) and the other factor is (H+B)
35 = 7 x 5
Now both 7 and 5 are prime. So which one is L?
① If L = 7, then H + B = 5
This can be rexpressed as the sum of two prime numbers. H + B = 2 + 3. So we can have B = 2, H = 3 or B = 3, H = 2.
② If L = 5, then H + B = 7
We can also rexpress this as the sum of 2 numbers. H + B = 2 + 5 . So we can have H = 2, B = 5 or H = 5, B = 2
Now the question doesn't specify that the dimensions must all different prime numbers. If we want all the dimensions to be different and breadth to be strictly smaller than length, then we go with option ①
Volume of cuboid = Lcm x Bcm x Hcm = 7cm × 3cm × 2cm = 42cm³ or 7cm × 2cm × 3cm = 42cm³ (either way, we get 42cm³ so it doesn't matter if B = 2, H = 3 or B = 3, H = 2)
Now front and back faces have same surface area. Likewise for left and right, top and bottom.
Total surface area of cuboid = 2 × (Lcm × Bcm + Lcm × Hcm + Bcm × Hcm)
= 2 × (LBcm² + LHcm² + 2cm × 3cm)
= 2 × (35cm² + 6cm²)
= 2 × 41cm²
= 82cm²
Or
= 2 × (LBcm² + LHcm² + 3cm × 2cm)
= 2 × (35cm² + 6cm²)
= 2 × 41cm²
= 82cm²
Thank you!
done
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Note: if you let L=5 and H+B= 7, it still gives the same ans
Date Posted:
4 years ago
Not really.
If L = 5, then H = 5,B = 2 or H = 2, B = 5 for them to be prime.
Then Volume = 5cm x 5cm x 2cm = 50cm³
Total Surface area = 2 x (5cm x 5cm + 5cm x 2cm + 5cm x 2cm)
= 2 x 45cm²
= 90cm²
If L = 5, then H = 5,B = 2 or H = 2, B = 5 for them to be prime.
Then Volume = 5cm x 5cm x 2cm = 50cm³
Total Surface area = 2 x (5cm x 5cm + 5cm x 2cm + 5cm x 2cm)
= 2 x 45cm²
= 90cm²
I overlooked that part, thanks for pointing out!
Glad to help
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