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Lisa [10]
2 years ago
8

D is the centroid of triangle abc. what is the value of x when bd=2x+40 and bf=18x

Mathematics
1 answer:
Luda [366]2 years ago
7 0

The value of x= 20 units

BD =  BF

BD =2x + 40

BF = 18x

Equate the expressions thus:

2x + 40 = 2/3(18x)

2x + 40 = 12x

Collect like terms

2x - 12x = - 40

-2x = - 40

Divide both sides by - 2

x = 20

Hence, x =  20 units

The centroid is the center point of the item. The factor in which the 3 medians of the triangle intersect is known as the centroid of a triangle. it is also defined because of the point of intersection of all the 3 medians. The median is a line that joins the midpoint of a facet and the opposite vertex of the triangle.

Then, we can calculate the centroid of the triangle by means of taking the common of the x coordinates and the y coordinates of all of the three vertices. So, the centroid method can be mathematically expressed as G(x, y) = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/three).

The centroid of a triangle is the point at which the 3 medians coincide. The centroid theorem states that the centroid is 23 of the space from each vertex to the midpoint of the opposite aspect.

Learn more about the centroid of the triangle here brainly.com/question/7644338

#SPJ4

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Arrange the cones in order from lease volume to greatest volume
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cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let V_{2}. V_{3}. and\  V_{4}. be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

d_{1} = 20\ and\ h_{1}=12

d_{2} = 18\ and\ h_{2}=10

r_{3} = 10\ and\ h_{3}=9

r_{4} = 11\ and\ h_{14}=9

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

V=\pi r^{2} \frac{h}{3}----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for d_{1} = 20\ and\ h_{1}=12

r_{1} = \frac{d_{1}}{2}

r_{1} = \frac{20}{2}=10\ units

V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}

V_{1}=\pi (10)^{2} \frac{12}{3}

V_{1}=\pi\times 100\times 4

V_{1}=400\pi\ units^{3}

Similarly, for volume of the cone for d_{2} = 18\ and\ h_{2}=10

r_{2} = \frac{d_{2}}{2}

r_{2} = \frac{18}{2}=9\ units

V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}

V_{2}=\pi (9)^{2} \frac{10}{3}

V_{2}=\pi\times 81\times \frac{10}{3}

V_{2}=\pi\times 27\times 10

V_{2}=270\pi\ units^{3}

Similarly, for volume of the cone for r_{3} = 10\ and\ h_{3}=9

V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}

V_{3}=\pi (10)^{2} \frac{9}{3}

V_{3}=\pi\times 100\times 3

V_{3}=\pi\times 300

V_{3}=300\pi\ units^{3}

Similarly, for volume of the cone for r_{4} = 11\ and\ h_{4}=9

V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}

V_{4}=\pi (11)^{2} \frac{9}{3}

V_{4}=\pi\times 121\times 3

V_{4}=\pi\times 363

V_{4}=363\pi\ units^{3}

So, the volume of the cone in ascending order.

V_{2}=270\pi\ units^{3}

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