In an isosceles triangle ABC, the base is AC = 18, and the side is 15. On side AB, point K is selected, and on side BC

univerkov | May 30, 2021 | education

In an isosceles triangle ABC, the base is AC = 18, and the side is 15. On side AB, point K is selected, and on side BC, point M, and AK: KM: MC = 5: 3: 5. Then the area of the quadrilateral AKMC is equal.

Since the ratio of the sides AK / MC = 5/5, then the segment AK = MC, and then the CM is parallel to the AC.

Triangles ABC and KMB are similar in two angles, the angle B of the triangles is common, the angle BKM = BAC as the corresponding angles at the intersection of parallel straight lines.

Let the length of the segment KM = 3 * X cm, then AK = CM = 5 * X cm.

The length of the segment BK = (15 – 5 * X) cm.

Then in similar triangles KM / AC = BK / AB.

3 * X / 18 = (15 – 5 * X) / 15.

45 * X = 270 – 90 * X.

135 * X = 270.

X = 2.

Then KM = 2 * 3 = 6 cm, AK = CM = 2 * 5 = 10 cm.

By Heron’s theorem, we determine the area of the triangle ABC.

The semi-perimeter of the triangle is: p = (AB + BC + AC) / 2 = 48/2 = 24 cm.

Then Sav = √24 * (24 – 15) * (24 – 15) * (24 – 18) = √11664 = 108 cm2.

Similarly, we define the area of the triangle BKM.

P = (BK + KM + BM) = (5 + 6 + 5) / 2 = 16/2 = 8 cm.

Svkm = √8 * (8 – 5) * (8 – 6) * (8 – 5) = √144 = 12 cm2.

Then the area of the trapezium AKMС is equal to: Sacms = Savs – Svkm = 108 – 12 = 96 cm2.

Answer: The area of the AKMC quadrangle is 96 cm2.

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