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secondary 3 | E Maths
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(3, 1) is directly below (3, 4).
This means that the triangle will contain one horizontal line and one vertical line, and as we know it too well, any horizontal line meets any vertical line at right angles.
The position of the right angle is at (3, 4), which is point P.
For the shortest distance from P to QR, the line drawn from P must touch line QR at a random point S such that PS is perpendicular to QR.
Now, the area of the triangle can be calculated in two different ways, but both obviously give the same area.
Firstly, we can use 0.5 * PQ * PR, which is 0.5 * 3 * 5 = 7.5 units2.
Secondly, we can use 0.5 * QR * PS, which will also give 7.5 units2.
As such,
0.5 * QR * PS = 7.5
QR * PS = 15
Since QR is the hypotenuse of the right angled triangle PQR, we use the Pythagorean Theorem
“a2 + b2 = c2”
to find the length QR. Of course, QR is the square root of (3^2 + 5^2), or the square root of 34.
As such,
PS = 15 / QR
PS = 15 / sqrt 34
I have no calculator for this.
“^” is the symbol for power, as I do not have the specialised tools to write an actual power. “3^2” means 3 raised to the power of 2 (which equals 9), while “4^3” means 4 raised to the power of 3 (which equals 64).
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