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secondary 4 | E Maths
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Solution
Date Posted: 3 years ago
Views: 498
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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.
Could you maybe do on paper? Need visualize
You can easily draw it on your paper.
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.
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.
From this, use Pythagoras Theorem to find BC² and AC² respectively.
Then substitute it into the equation you've worked out at the start.
(Follow my working)
Then substitute it into the equation you've worked out at the start.
(Follow my working)
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