Question

find the length of the medians of the triangle with vertices A(0,0,6) ,B(0,4,0) and (6,0,0)​

Answer 1

Given:-

• The triangle of the vertices A(0,0,6) ,B(0,4,0) and (6,0,0).

To find:-

• Find the length of the medians of the triangle..?

Solutions:-

• Let AD, BE and CF be the medians of the given triangle ABC.

Since AD is the median, D is the mid point of BC.

Therefore,

Coordinates of point = D = (0 + 6 / 2, 4 + 0 / 2, 0 + 0 / 2)

D = (3, 2, 0)

AD = √(0 - 3)² + (0 - 2)² + (6 - 0)²

= √9 + 4 + 36

= √49

= 7

Since BE is the median, E is the mid point of AC.

Therefore,

Coordinates of point = E = (0 + 6 / 2, 0 + 0 / 2, 6 + 0 / 2)

E = (3, 0, 3)

BE = √(3 - 0)² + (0 - 4)² + (3 - 0)²

= √9 + 16 + 9

= √34

Since CF is the median, F is the mid point of AB.

Therefore,

Coordinates of point = F = (0 + 0 / 2, 0 + 4 / 2, 6 + 0 / 2)

D = (0, 2, 3)

CF = √(6 - 0)² + (0 - 2)² + (0 - 3)²

= √36 + 4 + 9

= √49

= 7

Hence, the length of the median of ∆ABC are 7, √34 and 7.

Answer 2

$\huge\rm { ☆_! Question !_! ☆ }$

find the length of the medians of the triangle with vertices A(0,0,6) ,B(0,4,0) and (6,0,0)?

$\huge\rm { ☆_! solution !_! ☆ }$

$\huge\star{\underline{\mathtt{\red{G}\pink{I}</p><p>\green{V}\blue{E}\purple{N}}}}:\star$

A triangles with vertices A(0,0,6) ,B(0,4,0) and (6,0,0).

$\huge\star{\underline{\mathtt{\red{T}\pink{O} \:</p><p>\green{F}\blue{I}\purple{N}\pink{D}}}}:\star$

find the length of the medians of the triangle with vertices A(0,0,6) ,B(0,4,0) and (6,0,0).

⇒medians of triangle:

$\huge\star{\underline{\mathtt{</u><u>\</u><u>r</u><u>e</u><u>d</u><u>{</u><u>A</u><u>}\pink{</u><u>D</u><u>}}}}:\star$

$\implies \sf AD = \sqrt (0-3)^2+(0-2)^2+(6-0)^2$

$\implies \sf \sqrt{49}$

$\implies \sf \: 7$

$\longrightarrow\underline{\underline{\red{\sf{AD=7}}}}$

$\huge\star{\underline{\mathtt</u><u>{</u><u>\green{</u><u>B</u><u>}</u><u>\blue{</u><u>E</u><u>}</u><u>}</u><u>}</u><u>}</u><u>:</u><u>\</u><u>star$

$\implies \sf BE = (3 - 0) ^{2} + (0-4) ^{2} +(3-0) ^{2}$

$\implies \sf \sqrt{34}$

$\longrightarrow\underline{\underline{\red{\sf{BE=√34}}}}$

$\huge\star{\underline{\mathtt{\purple{N}\pink{D}}}}:\star$

$\implies \sf \: SF = (6 - 0)^{2} + (0 - 2 ) ^{2} + (0 - 3) ^{2}$

$\implies \sf \sqrt{49}$

$\implies \sf \: 7$

$\longrightarrow\underline{\underline{\red{\sf{SF=7}}}}$

∴medians of the triangle=AD= 7, BE= √34,SF = 7 i.e. the length of medians of ∆ABC = 7 , 37 and 7.