In rectangle ABCD with sides AB = 10 and BC = 16.5 point L is the middle of AB. On the AD side

| May 1, 2021 | education

In rectangle ABCD with sides AB = 10 and BC = 16.5 point L is the middle of AB. On the AD side, the points A, M, N and D are sequentially located in such a way that AM: MN: ND = 1: 17: 15. Find the area of the triangle MNP, where P is the intersection point of the segments LN and CM.

In a rectangle, the lengths of the opposite sides are equal, AB = SD = 10 cm, AD = BC = 16.5 cm.

Let the length of the segment AM = X cm, then MN = 17 * X cm, ND = 15 * X cm.

AD = AM + MN + ND = X + 17 * X + 15 * X = 16.5.

33 * X = 16.5 cm.

X = 16.5 / 33 = 0.5 cm.

Then AM = 0.5 cm, MN = 17 * 0.5 = 8.5 cm, ND = 15 * 0.5 = 7.5 cm.

AN = AM + MN = 0.5 + 8.5 = 9 cm.

DM = MN + ND = 8.5 + 7.5 = 16 cm.

Let’s draw the height of the PH of the triangle MPN.

Triangles ALN and PHN are similar in acute angle.

Then: AN / HN = AL / PH. HN = AN * PH / AN = 9 * PH / 5 /.

Triangles CDM and MPH are also similar in acute angle.

Then DM / MH = CD / PH. MN = DM * PH / CD = 16 * PH / 10.

HN + MH = MN = 8.5 cm.

Then: 9 * PH / 5 / + 16 * PH / 10 = 8.5 cm.

PH * (9/5 + 16/10) = 8.5.

3.4 * PH = 8.5.

PH = 8.5 / 3.4 = 2.5 cm.

Let us define the area of ​​the triangle MNP.

Smnp = MN * PH / 2 = 8.5 * 2.5 / 2 = 10.625 cm2.

Answer: The area of ​​the MNP triangle is 10.625 cm2.



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