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borishaifa [10]
1 year ago
6

Find the height of a triangle that has an area of 54 cm2 with the width 3 cm less than its height.

Mathematics
1 answer:
iren2701 [21]1 year ago
8 0

The height of a triangle is 9 cm

<h3>What is Triangle?</h3>
  • A polygon with three edges and three vertices is called a triangle. It is one of the fundamental geometric shapes. Triangle ABC is the display style for a triangle with vertices A, B, and C.
  • In Euclidean geometry, any three points that are not collinear produce a singular triangle and a singular plane (i.e. a two-dimensional Euclidean space).
  • In other words, every triangle is contained in a plane, and there is only one plane that contains that triangle.
  • All triangles are enclosed in a single plane if all of the geometry is the Euclidean plane, however, this is no longer true in higher-dimensional Euclidean spaces.
  • Except when otherwise specified, this article discusses triangles in Euclidean geometry, namely the Euclidean plane.

given that

let the height of the triangle be x

then the width is x-3

Area = \frac{height \times length}{2}

54 = x(x -3)

54 = x^2 -3x

x = 9

height = 9 cm

width = 6cm

Therefore, the height of a triangle is 9 cm.

To learn more about the triangle with the given link brainly.com/question/11897796

#SPJ4

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A ship leaves port at noon and has a bearing of S29oW. The ship sails at 20 knots. How many nautical miles south and how many na
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Approximately 58.2\; \text{nautical miles} (assuming that the bearing is {\rm S$29^{\circ}$W}.)

Step-by-step explanation:

Let v denote the speed of the ship, and let t denote the duration of the trip. The magnitude of the displacement of this ship would be v\, t.

Refer to the diagram attached. The direction {\rm S$29^{\circ}$W} means 29^{\circ} west of south. Thus, start with the south direction and turn towards west (clockwise) by 29^{\circ} to find the direction of the displacement of the ship.

The hypothenuse of the right triangle in this diagram represents the displacement of the ship, with a length of v\, t. The dashed horizontal line segment represents the distance that the ship has travelled to the west (which this question is asking for.) The angle opposite to that line segment is exactly 29^{\circ}.

Since the hypotenuse is of length v\, t, the dashed line segment opposite to the \theta = 29^{\circ} vertex would have a length of:

\begin{aligned}& \text{opposite (to $\theta$)} \\ =\; & \text{hypotenuse} \times \frac{\text{opposite (to $\theta$)}}{\text{hypotenuse}} \\ =\; & \text{hypotenuse} \times \sin (\theta) \\ =\; & v\, t \, \sin(\theta) \\ =\; & v\, t\, \sin(29^{\circ})\end{aligned}.

Substitute in \begin{aligned} v &= 20\; \frac{\text{nautical mile}}{\text{hour}}\end{aligned} and t = 6\; \text{hour}:

\begin{aligned} & v\, t\, \sin(29^{\circ}) \\ =\; & 20\; \frac{\text{nautical mile}}{\text{hour}} \times 6\; \text{hour} \times \sin(29^{\circ}) \\ \approx\; & 58.2\; \text{nautical mile}\end{aligned}.

7 0
2 years ago
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