# Exercise 6.step three Medians and Altitudes off Triangles

Exercise 6.step three Medians and Altitudes off Triangles

Question 1. Code Label the fresh new five sort of issues out of concurrency. And that lines intersect to form each one of the activities? Answer:

Question 2PLETE The Phrase The size of a segment off an excellent vertex into centroid is actually ______________ the size of new median of one to vertex.

Answer: The size of a section from a vertex towards centroid is the one-third of one’s length of the fresh average of that vertex.

## Explanation: The centroid of the trinagle = ($$\frac < 1> < 3>$$, $$\frac < 4> < 3>$$) = ($$\frac < 10> < 2>$$, 3)

Explanation: PN = $$\frac < 2> < 3>$$QN PN = $$\frac < 2> < 3>$$(21) PN = 14 QP = $$\frac < 1> < 3>$$QN = $$\frac < 1> < 3>$$(21) = 7

Explanation: PN = $$\frac < 2> < 3>$$QN PN = $$\frac < 2> < 3>$$(42) PN = 28 QP = $$\frac < 1> < 3>$$QN = $$\frac < 1> < 3>$$(42) = 14

## Explanation: DE = $$\frac < 1> < 3>$$CE 11 = $$\frac < 1> < 3>$$ CE CE = 33 CD = $$\frac < 2> < 3>$$ CE CD = $$\frac < 2> < 3>$$(33) CD = 22

Explanation: DE = $$\frac < 1> < 3>$$CE 15 = $$\frac < 1> < 3>$$ CE CE = 45 CD = $$\frac < 2> < 3>$$ CE CD = $$\frac < 2> < 3>$$(45) CD = 30

When you look at the Teaching 11-fourteen. area Grams ’s the centroid off ?ABC. BG = six, AF = a dozen, and AE = 15. Find the period of this new part.

Explanation: The centroid of the trinagle = ($$\frac < 1> < 3>$$, $$\frac < 5> < 3>$$) = ($$\frac < -7> < 3>$$, 5)

## Explanation: DE = $$\frac < 1> < 3>$$CE 11 = $$\frac < 1> < 3>$$ CE CE = 33 CD = $$\frac < 2> < 3>$$ CE CD = $$\frac < 2> < 3>$$(33) CD = 22

In the Practise 19-twenty two. give bristlr ekÅŸi perhaps the orthocenter are into the, towards the, otherwise outside of the triangle. Up coming get the coordinates of your own orthocenter.

Explanation: The slope of YZ = $$\frac < 6> < -3>$$ = $$\frac < -1> < 2>$$ The slope of the perpendicular line is 2 The equation of perpendicular line is (y – 2) = 2(x + 3) y – 2 = 2x + 6 2x – y + 8 = 0 The slope of XZ = $$\frac < 6> < -3>$$ = 0 The equation of perpendicular line is (y – 2) = 0 y = 2 Substitute y = 2 in 2x – y + 8 = 0 2x – 2 + 8 = 0 2x + 6 = 0 x = -3 the orthocenter is (-3, 2) The orthocenter lies on the vertex of the triangle.

Explanation: The slope of UV = $$\frac < 4> < 0>$$ = $$\frac < -3> < 2>$$ The slope of the perpendicular line is $$\frac < 2> < 3>$$ The equation of the perpendicular line is (y – 1) = $$\frac < 2> < 3>$$(x + 2) 3(y – 1) = 2(x + 2) 3y – 3 = 2x + 2 2x – 3y + 5 = 0 – (i) The slope of TV = $$\frac < 4> < 0>$$ = $$\frac < 3> < 2>$$ The slope of the perpendicular line is $$\frac < -2> < 3>$$ The equation of the perpendicular line is (y – 1) = $$\frac < -2> < 3>$$(x – 2) 3(y – 1) = -2(x – 2) 3y – 3 = -2x + 4 2x + 3y – 7 = 0 -(ii) Add two equations 2x – 3y + 5 + 2x + 3y – 7 = 0 4x – 2 = 0 x = 0.5 2x – 1.5 + 5 = 0 x = -1.75 So, the orthocenter is (0, 2.33) The orthocenter lies inside the triangle ABC.