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secondary 2 | Maths
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Please help with (a) and (b) urgent thanks!
Date Posted: 4 years ago
Views: 257
dia. = diagonal of rect. = 10 cm
length: x + 2
breadth: x
diagonal = √[ (x+2)² + x² ]
(x+2)² + x² = 100
x² + 4x + 4 + x² = 100
x² + 2x - 48 = 0
(x+8)(x-6) = 0
x = 6 cm or x = -8 (rejected)
(a)
length is 6+2 = 8 cm
(b)
area rect. = 8x6 = 48 cm²
area circle = π x 25 = 25π cm²
area of shaded region
= 25π - 48 = 30.5 cm²
...
length: x + 2
breadth: x
diagonal = √[ (x+2)² + x² ]
(x+2)² + x² = 100
x² + 4x + 4 + x² = 100
x² + 2x - 48 = 0
(x+8)(x-6) = 0
x = 6 cm or x = -8 (rejected)
(a)
length is 6+2 = 8 cm
(b)
area rect. = 8x6 = 48 cm²
area circle = π x 25 = 25π cm²
area of shaded region
= 25π - 48 = 30.5 cm²
...
The diagonal is 10cm because it's the diameter of the circle. Idea is to use Pythagoras' theorem
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done
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Combine your unknown x into an equation and solve for x.
The answer provide the length of the rect.
The answer provide the length of the rect.
Date Posted:
4 years ago
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