# Find the lengths of the medians of the triangle with vertices A (0,0,6), B (0,4,0) and C (6, 0, 0).

Last updated date: 25th Mar 2023

•

Total views: 306.3k

•

Views today: 2.83k

Answer

Verified

306.3k+ views

Hint: For solving the problem, we should know about the basics of finding a median from each of the vertices of the triangle. Finally, we can find the length of the medians by using distance formula on the vertices joining two points of a median. Distance formula is given by \[\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}}\]. Here, $({{x}_{1}},{{y}_{1}},{{z}_{1}})\text{ }and\text{ }({{x}_{2}},{{y}_{2}},{{z}_{2}})$ are the respective vertices between which we want to find the distance.

Complete step-by-step answer:

Basically, before starting to solve the problem, we first try to understand the definition of median which would be useful for doing this question. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side, thus bisecting that side. Thus, in the below figure of triangle, AE, CD, and BF are the three medians of the triangle. We first start by plotting the vertices A, B and C and then finding the respective mid-points of the sides AB, BC and CA.

Here, E, D and F are the respective mid-points of CB, AB and AC. Let A (0,0,6), B (0,4,0) and C (6,0,0).

Now, to find the midpoint between $({{x}_{1}},{{y}_{1}},{{z}_{1}})\text{ }and\text{ }({{x}_{2}},{{y}_{2}},{{z}_{2}})$, the formula is \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2}\text{ } \right)\] for $({{x}_{1}},{{y}_{1}},{{z}_{1}})\text{ }and\text{ }({{x}_{2}},{{y}_{2}},{{z}_{2}})$ to be the respective vertices of the side of a triangle. Thus, we use this to find E, D and F. Thus, we get,

E = \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2}\text{ } \right)=\left( \dfrac{0+6}{2},\dfrac{4+0}{2},\dfrac{0+0}{2} \right)\]= (3,2,0)

D=\[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2}\text{ } \right)=\left( \dfrac{0+0}{2},\dfrac{0+4}{2},\dfrac{6+0}{2} \right)\]= (0,2,3)

F=\[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2}\text{ } \right)=\left( \dfrac{0+6}{2},\dfrac{0+0}{2},\dfrac{6+0}{2} \right)\]= (3,0,3)

Now, we try to find median lengths using the distance formula. We have distance formula as \[\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}}\]. Thus,

AD = \[\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}}=\sqrt{{{(3-0)}^{2}}+{{(2-0)}^{2}}+{{(0-6)}^{2}}}=\sqrt{49}\]= 7

BE =\[\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}}=\sqrt{{{(3-0)}^{2}}+{{(0-4)}^{2}}+{{(3-0)}^{2}}}=\sqrt{34}\]

CF = \[\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}}=\sqrt{{{(0-6)}^{2}}+{{(2-0)}^{2}}+{{(3-0)}^{2}}}=\sqrt{49}\]= 7

Hence, the length of the medians are 7, $\sqrt{34}$ and 7.

Note: Another alternative to finding the lengths of the median of the triangle is to use the formula by Apollonius’ theorem –

$\sqrt{\dfrac{2{{b}^{2}}+2{{c}^{2}}-{{a}^{2}}}{4}}$, $\sqrt{\dfrac{2{{a}^{2}}+2{{c}^{2}}-{{b}^{2}}}{4}}$, $\sqrt{\dfrac{2{{a}^{2}}+2{{b}^{2}}-{{c}^{2}}}{4}}$. Here; a, b and c are the lengths of the sides of the triangle opposite to angles A, B and C.

Complete step-by-step answer:

Basically, before starting to solve the problem, we first try to understand the definition of median which would be useful for doing this question. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side, thus bisecting that side. Thus, in the below figure of triangle, AE, CD, and BF are the three medians of the triangle. We first start by plotting the vertices A, B and C and then finding the respective mid-points of the sides AB, BC and CA.

Here, E, D and F are the respective mid-points of CB, AB and AC. Let A (0,0,6), B (0,4,0) and C (6,0,0).

Now, to find the midpoint between $({{x}_{1}},{{y}_{1}},{{z}_{1}})\text{ }and\text{ }({{x}_{2}},{{y}_{2}},{{z}_{2}})$, the formula is \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2}\text{ } \right)\] for $({{x}_{1}},{{y}_{1}},{{z}_{1}})\text{ }and\text{ }({{x}_{2}},{{y}_{2}},{{z}_{2}})$ to be the respective vertices of the side of a triangle. Thus, we use this to find E, D and F. Thus, we get,

E = \[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2}\text{ } \right)=\left( \dfrac{0+6}{2},\dfrac{4+0}{2},\dfrac{0+0}{2} \right)\]= (3,2,0)

D=\[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2}\text{ } \right)=\left( \dfrac{0+0}{2},\dfrac{0+4}{2},\dfrac{6+0}{2} \right)\]= (0,2,3)

F=\[\left( \dfrac{{{x}_{1}}+{{x}_{2}}}{2},\dfrac{{{y}_{1}}+{{y}_{2}}}{2},\dfrac{{{z}_{1}}+{{z}_{2}}}{2}\text{ } \right)=\left( \dfrac{0+6}{2},\dfrac{0+0}{2},\dfrac{6+0}{2} \right)\]= (3,0,3)

Now, we try to find median lengths using the distance formula. We have distance formula as \[\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}}\]. Thus,

AD = \[\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}}=\sqrt{{{(3-0)}^{2}}+{{(2-0)}^{2}}+{{(0-6)}^{2}}}=\sqrt{49}\]= 7

BE =\[\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}}=\sqrt{{{(3-0)}^{2}}+{{(0-4)}^{2}}+{{(3-0)}^{2}}}=\sqrt{34}\]

CF = \[\sqrt{{{({{x}_{1}}-{{x}_{2}})}^{2}}+{{({{y}_{1}}-{{y}_{2}})}^{2}}+{{({{z}_{1}}-{{z}_{2}})}^{2}}}=\sqrt{{{(0-6)}^{2}}+{{(2-0)}^{2}}+{{(3-0)}^{2}}}=\sqrt{49}\]= 7

Hence, the length of the medians are 7, $\sqrt{34}$ and 7.

Note: Another alternative to finding the lengths of the median of the triangle is to use the formula by Apollonius’ theorem –

$\sqrt{\dfrac{2{{b}^{2}}+2{{c}^{2}}-{{a}^{2}}}{4}}$, $\sqrt{\dfrac{2{{a}^{2}}+2{{c}^{2}}-{{b}^{2}}}{4}}$, $\sqrt{\dfrac{2{{a}^{2}}+2{{b}^{2}}-{{c}^{2}}}{4}}$. Here; a, b and c are the lengths of the sides of the triangle opposite to angles A, B and C.

Recently Updated Pages

If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE