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secondary 2 | Maths
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Help with b
Date Posted: 2 years ago
Views: 858
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Ai) since the given triangles are similar, the ratios of length are the same:
X/Y = 6/2
The question wanted in terms of y,
So make x the subject:
X/Y = 3
X = 3Y
:.PA = Y + X
= 4Y m //
ii) given similar triangles: (same as part a)
TM / QA = PM / PA
TM / 6 = Y / 4Y
TM = 1/4 * 6
= 1.5cm //
B) redo entire part (a) with h and k instead:
Start by finding PA again: (make x subject)
X/Y = k/h
X = ky/h
PA = y + ky/h
Part (ii):
TM / QA = PM / PA
TM / k = y / (y + ky/h)
TM / k = 1 / (1 + k/h) (remove all y)
TM = k / (1 + k/h)
To simplify more:
TM = kh /(h + k) (multiply all terms by h)
Done
X/Y = 6/2
The question wanted in terms of y,
So make x the subject:
X/Y = 3
X = 3Y
:.PA = Y + X
= 4Y m //
ii) given similar triangles: (same as part a)
TM / QA = PM / PA
TM / 6 = Y / 4Y
TM = 1/4 * 6
= 1.5cm //
B) redo entire part (a) with h and k instead:
Start by finding PA again: (make x subject)
X/Y = k/h
X = ky/h
PA = y + ky/h
Part (ii):
TM / QA = PM / PA
TM / k = y / (y + ky/h)
TM / k = 1 / (1 + k/h) (remove all y)
TM = k / (1 + k/h)
To simplify more:
TM = kh /(h + k) (multiply all terms by h)
Done
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