The median of a triangle is the piece which is carried out from any its top to the opposite side, at the same time it divides it into parts of equal length. The maximum number of medians in a triangle - three, by quantity of tops and the parties.
Instruction
1. Task 1. In any triangle of ABD BE median is carried out. Find its length if it is known that the parties, respectively, are equal to AB = 10 cm, BD = 5 cm and AD = 8 cm.
2. Decision. Apply a median formula with expression through all parties of a triangle. It is a simple task as all lengths of the parties are known: BE = √ ((2*AB^2 + 2*BD^2 - AD^2)/4) = √ ((200 + 50 - 64)/4) = √ (46.5) ≈ 6.8 (cm).
3. Task 2. The parties of AD and BD are equal in an isosceles triangle of ABD. The median from top D on the party of BA is carried out, at the same time it makes a corner about BA equal 90 °. Find DH median length if it is known that BA = 10 cm, and the corner of DBA is equal 60 °.
4. Decision. For finding of a median define one and equal parties of a triangle of AD or BD. For this purpose consider one of rectangular triangles, suppose, of BDH. Follows from definition of a median that BH = BA/2 = 10/2 = 5. Find the party of BD on a trigonometrical formula from property of a rectangular triangle - BD = BH/sin(DBH) = 5/sin60 ° = 5 / (√ 3/2) ≈ 5.8.
5. Now there are two possible options of finding of a median: on the formula used in the first task or of Pythagorean theorem for a rectangular triangle of BDH: DH^2 = BD^2 - BH^2.DH^2 = (5.8) ^2 - 25 ≈ 8.6 (cm).
6. Task 3. In any triangle of BDA three medians are carried out. Find their lengths if it is known that height of DK is equal to 4 cm and divides the basis into pieces BK length = 3 and KA = 6.
7. Decision. Lengths of all parties are necessary for finding of medians. Length of BA can be found from a condition: BA = BH + HA = 3 + 6 = 9. Consider a rectangular triangle of BDK. On Pythagorean theorem find BD hypotenuse length: BD^2 = BK^2 + DK^2; BD = √ (9 + 16) = √25 = 5.
8. Similarly find a hypotenuse of a rectangular triangle of KDA: AD^2 = DK^2 + KA^2; AD = √ (16 + 36) = √52 ≈ 7.2.
9. On an expression formula through the parties find medians: BE^2 = (2*BD^2 + 2*BA^2 - AD^2)/4 = (50 + 162 - 51.8)/4 ≈ 40, from here BE ≈ 6.3 (cm) .DH^2 = (2*BD^2 + 2*AD^2 - BA^2)/4 = (50 + 103.7 - 81)/4 ≈ 18.2, from here DH ≈ 4.3 (cm) .AF^2 = (2*AD^2 + 2*BA^2 - BD^2)/4 = (103.7 + 162 - 25)/4 ≈ 60, from here AF ≈ 7.8 (cm).