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secondary 3 | E Maths
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Hi, i need help with 6ii) Thank you so much!
Date Posted: 3 years ago
Views: 166
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Draw a perpendicular line from Q to PS. Call the intersection point X.
So QRSX is a rectangle.
QX = RS = 12cm , XS = QR and opposite sides are parallel to each other.
PX = PS - XS
= PS - QR (since XS = QR)
= (3r + 1) cm - r cm
= (2r + 1) cm
Since QX is perpendicular to PS, ∠PXQ is a right angle and so△PQX is right angled.
Using Pythagoras' Theorem,
PQ² = PX² + QX²
PQ² = (2r + 1)² cm² + 12² cm²
= (4r² + 4r + 1 + 144) cm²
= (4r² + 4r + 145) cm²
But realise that PQ = PA + AB + BQ and that PA and BQ are the radii of the respective circles.
So PQ = (3r + 1 + 6 + r) cm
= (4r + 7) cm
PQ² = (4r + 7)² cm²
= (16r² + 56r + 49) cm²
Now we can equate the two.
4r² + 4r + 145 = 16r² + 56r + 49
12r² + 52r - 96 = 0
Divide by 4,
3r² + 13r - 24 = 0 (shown)
So QRSX is a rectangle.
QX = RS = 12cm , XS = QR and opposite sides are parallel to each other.
PX = PS - XS
= PS - QR (since XS = QR)
= (3r + 1) cm - r cm
= (2r + 1) cm
Since QX is perpendicular to PS, ∠PXQ is a right angle and so△PQX is right angled.
Using Pythagoras' Theorem,
PQ² = PX² + QX²
PQ² = (2r + 1)² cm² + 12² cm²
= (4r² + 4r + 1 + 144) cm²
= (4r² + 4r + 145) cm²
But realise that PQ = PA + AB + BQ and that PA and BQ are the radii of the respective circles.
So PQ = (3r + 1 + 6 + r) cm
= (4r + 7) cm
PQ² = (4r + 7)² cm²
= (16r² + 56r + 49) cm²
Now we can equate the two.
4r² + 4r + 145 = 16r² + 56r + 49
12r² + 52r - 96 = 0
Divide by 4,
3r² + 13r - 24 = 0 (shown)
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