J's answer to Crystal's Secondary 4 E Maths Singapore question.
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You just need to continue by applying Pythagoras's Theorem on the two smaller right-angled triangles.
AC² = 8² + h² = 64 + h²
BC² = 4² + h² = 16 + h²
So, 144 = 64 + h² + 16 + h²
144 = 80 + 2h²
2h² = 144 - 80 = 64
h² = 32
h = √32 = √(16 × 2) = √16 √2 = 4√2
The height is 4√2 cm.
AC² = 8² + h² = 64 + h²
BC² = 4² + h² = 16 + h²
So, 144 = 64 + h² + 16 + h²
144 = 80 + 2h²
2h² = 144 - 80 = 64
h² = 32
h = √32 = √(16 × 2) = √16 √2 = 4√2
The height is 4√2 cm.
Date Posted:
3 years ago
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Let the intersection point of the height and AB be D.
We are looking at triangles ADC and BDC
You are given that ∠ CDB is 90° (right angle).
So ∠CDA also = 90°
(180° - 90° = 90°, angle sum of a straight line)
This tells you that ADC is also a right-angled triangle. AC is its hypotenuse.
triangle BDC's hypotenuse is BC.
Then substitute it into the equation you've worked out at the start.
(Follow my working)