А (2; 5) В (3; 3) С (-1; 4), in the triangle ABC find the length of the median drawn

| September 29, 2021 | education

А (2; 5) В (3; 3) С (-1; 4), in the triangle ABC find the length of the median drawn from the vertex A, the perimeter and area of the triangle ABC, as well as the angles?

I: 1) Let us denote the base of the median drawn from point A to the side BC (midpoint of segment BC), behind M. Find its coordinates using the formula for coordinates of the midpoint of the segment: xM = (xB + xC) / 2 = (3 – 1) / 2 = 1, yM = (yB + yC) / 2 = (3 + 4) / 2 = 3.5. Thus, M (1; 3.5).
2) Find the length of the segment AM by the formula: AM2 = (xM – xA) 2 + (yM – yA) 2 = (1 – 2) 2 + (3.5 – 5) 2 = 1 + 2.25 = 3.25 , whence AM = √13 / 2 is the required median length.

II: 1) Find the length of the segment AB by the formula: AB2 = (xB – xA) 2 + (yB – yA) 2 = (3 – 2) 2 + (3 – 5) 2 = 1 + 4 = 5, whence AB = √5.
2) Find the length of the segment AC by the formula: AC2 = (xC – xA) 2 + (yC – yA) 2 = (- 1 – 2) 2 + (4 – 5) 2 = 9 + 1 = 10, whence AC = √ ten.
3) Find the length of the segment BC by the formula: BC2 = (xC – xB) 2 + (yC – yB) 2 = (- 1 – 3) 2 + (4 – 3) 2 = 16 + 1 = 17, whence BC = √ 17.
4) Then we have: PABC = AB + AC + BC = √5 + √10 + √17.

III: 1) Find the semiperimeter of triangle ABC: p = PABC / 2 = (√5 + √10 + √17) / 2. Then:
p – AB = (√10 + √17 – √5) / 2;
p – AC = (√5 + √17 – √10) / 2;
p – BC = (√5 + √10 – √17) / 2;
2) Then, according to Heron’s formula, we have: SABC = √ (p * (p – AB) * (p – AC) * (p – BC)) = 3.5.

IV: 1) We have, by the theorem of cosines in triangle ABC: AC2 = AB2 + BC2 – 2AB * BC * cos (B), 10 = 5 + 17 – 2 * √85 * cos (B), whence B = arccos (6 / √85).
2) We have, by the theorem of cosines in triangle ABC: BC2 = AC2 + AB2 – 2AC * AB * cos (A), 17 = 10 + 5 – 2 * √50 * cos (A), whence A = arccos (- √2 / ten).
3) Since the sum of the angles of the triangle is 180o, then C = 180o – arccos (6 / √85) – arccos (- √2 / 10).

ANSWER: √13 / 2 – the required length of the median, √5 + √10 + √17 – perimeter, 3.5 – area, angle A = arccos (- √2 / 10), angle B = arccos (6 / √85) and the angle C = 180o – arccos (6 / √85) – arccos (- √2 / 10).



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