All categories

3 years ago

Regular triangular pyramid has 6 cm long base edge and slant height k=9 cm. Find the lateral area of the pyramid.

Mathematics

2 answers:

gregori [183]3 years ago

5 0

Answer:

81 cm²

Step-by-step explanation:

Since, the lateral face of a triangular pyramid is a triangle,

Given,

The base edge or the base of one lateral face of pyramid, a = 6 cm,

And, the slant height or the height of the face, k = 9 cm,

Thus, the area of one lateral face of the pyramid,

$A=\frac{1}{2}\times a\times k$

$=\frac{1}{2}\times 6\times 9$

$=\frac{54}{2}$

$=27\text{ square cm}$

We know that, a Regular triangular pyramid has 3 lateral faces,

Hence, the total lateral area of the pyramid,

$L.A.=3\times\text{ The area of one lateral face}$

$=3\times 27$

$=81\text{ square cm}$

kobusy [5.1K]3 years ago

5 0

Answer:

hiiii your answer is 81!!

Step-by-step explanation:

You might be interested in

One of the interior angles of a triangle is equal to 30°, and one of the exterior angles is equal to 40°. Find the remaining int

irina1246 [14]

Answer:

The remaining interior angles of this triangle are 140º and 10º

Step-by-step explanation:

The sum of the interior angles of a triangle is always 180º.

A triangle has 3 angles. In this problem, we have one of them, that i am going to call A1 = 30º.

The sum of a interior angle with it's respective exterior angle is also always 180º.

We have that one of the exterior angles is equal to 40°. So it's respective interior angle is

40º + A2 = 180º

A2 = 180º - 40º

A2 = 140º

Now we have two interior angles, and we know that the sum of the 3 interior angles is 180º. So:

A1 + A2 + A3 = 180º

A3 = 180º - A1 - A2

A3 = 180º - 30º - 140º

A3 = 180º - 170º

A3 = 10º

7 0

3 years ago

5x + (x÷3) = 38.4 Which value of x makes the equation true

denpristay [2]

5x + X/3 = 38.4

or, 5x* 3 + X = 38.4*3

or,. 15x + X. = 115.2

or,. 16 X. = 115.2

or,. X. = 115.2/16

X = 7.2

4 0

3 years ago

Read 2 more answers

Line Equation from Two Points

Verdich [7]

Hi!

We can use point-slope form to solve this.

$y-y_{1} =m(x -x_{1})$

$y_{1}$ and $x_{1}$ will be from one of the points.

First, we have to find $m$ , the slope. We can use the slope equation to get this.

$\frac{y_{1} -y_{2} }{x_{1} -x_{2} }$

Plug in your points:

$\frac{4-0}{8-(-8)} =\frac{4-0}{8+8} =\frac{4}{16} =\frac{1}{4}$

Your slope is $\frac{1}{4}$

Now plug points and slope into point-slope equation. We will use (8, 4).

$y-4=\frac{1}{4} (x-8)$

Now, if you want to get it into y = mx + b form, you have to solve for y:

$y-4=\frac{1}{4} (x-8)$

$y-4=\frac{1}{4} x-2$

$y=\frac{1}{4} +2$

Your equation is $y=\frac{1}{4} +2$

For more information on how to get the equation of a line when given two points, see here:

brainly.com/question/986503

7 0

2 years ago

How many 1/2 are there in 7 1/2 ?

My name is Ann [436]

Answer:

There are 15 halves in 7 1/2.

4 0

3 years ago

Read 2 more answers

Given that cos (x) = 1/3, find sin (90 - x)

ddd [48]

Answer:

$\sin(90^{\circ} - x)=\frac{1}{3}$

Step-by-step explanation:

Given: $\cos (x)=\frac{1}{3}$

We have to find the value of $\sin(90^{\circ} - x)$

Since Given $\cos (x)=\frac{1}{3}$

Using trigonometric identity,

$\sin(90^{\circ} - \theta)=\cos\theta$

Thus, for $\sin(90^{\circ} - x)$ comparing , we have,

$\theta=x$

We get,

$\sin(90^{\circ} - x)=\cos x=\frac{1}{3}$

Thus, $\sin(90^{\circ} - x)=\frac{1}{3}$

3 0

3 years ago

Read 2 more answers