Question

Дан параллелограмм MPKH. Докажите, что треугольники PMK и PKH имеют равные площади.

Answer 1

Стороны парал-ма: a и b
Spmk = ab sin(mpk) / 2
Spkh = ab sin(180 - mpk) / 2 = ab sin(mpk) / 2
Таким образом, они равны

Answer 2

Proof of Equal Areas of Triangles PMK and PKH in Parallelogram MPKH

To prove that triangles PMK and PKH have equal areas in parallelogram MPKH, we can use the fact that the area of a triangle is half the product of its base and height.

Let's denote the base of triangle PMK as b1 and its corresponding height as h1. Similarly, the base of triangle PKH is b2 and its corresponding height is h2.

The area of triangle PMK is given by: Area(PMK) = (1/2) * b1 * h1

The area of triangle PKH is given by: Area(PKH) = (1/2) * b2 * h2

To prove that the areas of these triangles are equal, we need to show that: (1/2) * b1 * h1 = (1/2) * b2 * h2

Since the parallelogram MPKH has opposite sides that are parallel and equal in length, we can use the fact that the area of a parallelogram is the product of its base and height.

The area of parallelogram MPKH is given by: Area(MPKH) = base * height

Let's denote the base of parallelogram MPKH as b and its corresponding height as h.

The area of parallelogram MPKH is also given by: Area(MPKH) = b * h

Now, we can express the bases and heights of triangles PMK and PKH in terms of the base and height of parallelogram MPKH.

The base of triangle PMK (b1) is the same as the base of parallelogram MPKH (b), and the height of triangle PMK (h1) is the same as the height of parallelogram MPKH (h).

Similarly, the base of triangle PKH (b2) is the same as the base of parallelogram MPKH (b), and the height of triangle PKH (h2) is the same as the height of parallelogram MPKH (h).

Therefore, we can conclude that: Area(PMK) = Area(MPKH) = (1/2) * b * h Area(PKH) = Area(MPKH) = (1/2) * b * h

This shows that the areas of triangles PMK and PKH are indeed equal, as they are both equal to half the area of parallelogram MPKH. Thus, we have proved that triangles PMK and PKH have equal areas in parallelogram MPKH.