Need help on maths questions

[davidle_10]

I need help on a question. Please show working out with your solution.
Express the area of an equilateral triangle as a function of:
a) the length "s" of each side.
b) the altitude "h".

[brightsky]

a) Split the equilateral triangle of side "s" in half (i.e. into two right angled triangles).

Hence, the length of the base of each right-angled triangle is .

The height of the equilateral triangle (the line that cuts the two in half) would therefore be given by:













Because the area of the equilateral triangle is given by: height x side x 1/2,

Hence, the area of the triangle is expressed as a function of s is:

[brightsky]

b) From the previous question, we already know that:



Hence:





Hence, because the area of the triangle is: height x side x 1/2,

The area of the triangle expressed as a function of h is:

[davidle_10]

Brightsky. Shouldn't the pythagoras equation be s^2 - (s/2)^2 = h^2 ? Since s is the longest side and h and (s/2) are the two shortest lengths. I have already checked the solutions for the Essential Maths Methods book, but the answers are different from what you guys have stated.

[brightsky]

Brightsky. Shouldn't the pythagoras equation be s^2 - (s/2)^2 = h^2 ? Since s is the longest side and h and (s/2) are the two shortest lengths. I have already checked the solutions for the Essential Maths Methods book, but the answers are different from what you guys have stated.

Oh yes, hehe...my bad..

[brightsky]

Is it right now? Or have I made another flaw...

[davidle_10]

Dw about it, I  already figured the answer out. There is one last question I need help with.

A man walks at a speed of 2km/h for 45 minutes and then runs at 4km/h for 30 minutes. Let "S" km be the distance the man has run after t minutes. The distance travelled is described by:

S(t)={at if 0 < t < c , bt+d if c < t < e}

Find the values of a,b,c,d,e.

[brightsky]

Are these right?

[brightsky]

The equation you've given me as well as the worded information is split in two ways:

1: The man is moving at a speed of 2km/h (which is 2/60 = 1/30 km per minute) for 45 mins.
2: The man is moving at a speed of 4km/h (which is 4/60 = 1/15 km per minute) for 30 mins.

Hence, logically, the first section of the equation would be about the first 45 mins of the man's travel, and the second section of the equation would be about the next 30 mins.

That is to say, the first section is 0 < t < 45, as t is dealing with minutes, and the next section is 45 < t < 45 + 30 = 45 < t < 75.

So we already know that c = 45 and e = 75.

The general formula for distance relative to speed and time is d = st. t in this case is the minutes elapsed and the distance is S(t), with it being expressed as km.

In the first section, we have S(t) = at. Logically we can see that S(t) is the "d" (distance), and t is the "t" (time) so a would then be the speed. Because S(t) is expressed in km while t is expressed in mins, then a has to be expressed in km/min (the conversion has already been done above. Hence, a = 1/30.

The next section, we have S(t) = bt +d. Remember that this is where 45 < t < 75 (i.e. when the guy is running). We know that S(t) is the speed in km and t is the time in minutes. Because the man, in this section, has already travelled for 45 mins under a speed of 1/30 km/min (i.e. has already completed the walking), hence we need to add the distance he travelled whilst walking to the equation, which is 1/30 * 45 = 3/2. So d = 3/2. That leaves b to be the speed he is travelling in the second section, which has to be in km/min (converted above). Hence b = 1/15.

[davidle_10]

Thank you brightsky. I really appreciate your effort in helping me.