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secondary 3 | E Maths
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Hello can someone help with this question? Not sure why I cannot get the answer. Thank u!
Date Posted: 3 years ago
Views: 218
The dotted line bisects chord PQ
(The perpendicular bisector of a chord passes through the centre)
Call the point of intersection X.
So X is the midpoint of the chord
PX = QX = ½ PQ = ½(9.6cm) = 4.8cm
Draw lines to join O to P and O to Q.
Now we use Pythagoras' theorem, since we have two right angled triangles.
PX² + OX² = OP²
4.8² + 3² = OP²
OP² = 32.04
OP = √32.04
OP is a radius of the circle.
Cross sectional area
= πr²
= π(√32.04)²
= 32.04π
≈ 100.6566
≈ 101 (3s.f)
(The perpendicular bisector of a chord passes through the centre)
Call the point of intersection X.
So X is the midpoint of the chord
PX = QX = ½ PQ = ½(9.6cm) = 4.8cm
Draw lines to join O to P and O to Q.
Now we use Pythagoras' theorem, since we have two right angled triangles.
PX² + OX² = OP²
4.8² + 3² = OP²
OP² = 32.04
OP = √32.04
OP is a radius of the circle.
Cross sectional area
= πr²
= π(√32.04)²
= 32.04π
≈ 100.6566
≈ 101 (3s.f)
Yea I did this too and got the same answer but the answer key says it's supposed to be 18.0cm^2
The answer key is actually referring to the cross sectional area of the water.
Find the angle POQ, then use it to find the area of the sector POQ. Subtract area of △POQ from it and you will get ≈ 18.03
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