Die Durchschnittsgeschwindigkeit eines Zuges, dann die Fläche eines Dreiecks

In this video, we solve two questions on average speed and the area of a triangle.

Question one. A train travels one hundred and eighty kilometres in two hours fifteen minutes. What is its average speed in kilometres per hour? Option A, seventy-two. Option B, eighty. Option C, ninety. Option D, eighty-five.

Let us solve it. Average speed equals total distance divided by total time, so you must first express the time in hours.

First, convert two hours fifteen minutes to hours: two plus fifteen over sixty equals two point two five hours.

Now divide one hundred and eighty by two point two five to get eighty kilometres per hour.

As a check, eighty multiplied by two point two five gives one hundred and eighty kilometres, matching the distance travelled.

Hence, the correct answer is option B, eighty kilometres per hour.

Question two. Find the area of a triangle with sides thirteen centimetres, fourteen centimetres and fifteen centimetres.

Let us solve it. Use Heron's formula: with semiperimeter s, the area equals the square root of s times s minus a times s minus b times s minus c.

First, the semiperimeter is thirteen plus fourteen plus fifteen, divided by two, which equals twenty-one centimetres.

Subtracting each side from twenty-one gives eight, seven, and six centimetres.

So the area is the square root of twenty-one times eight times seven times six.

Multiplying inside the square root gives seven thousand and fifty-six.

Taking the square root gives eighty-four square centimetres. As a check, eighty-four squared is seven thousand and fifty-six, matching the product under the square root.

Hence, the area of the triangle is eighty-four square centimetres.