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makvit [3.9K]
2 years ago
10

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]2 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]2 years ago
5 0

Answer:

hiiii your answer is 81!!

Step-by-step explanation:

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On a coordinate plane, kite W X Y Z is shown. Point W is at (negative 3, 3), point X is at (2, 3), point Y is at (4, negative 4)
xz_007 [3.2K]

Answer:

P = 10 + 2\sqrt{53} units

Step-by-step explanation:

Given

Shape: Kite WXYZ

W (-3, 3),  X (2, 3),

Y (4, -4),  Z (-3, -2)

Required

Determine perimeter of the kite

First, we need to determine lengths of sides WX, XY, YZ and ZW using distance formula;

d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}

For WX:

(x_1, y_1)\ (x_2,y_2) = (-3, 3),\ (2, 3)

WX = \sqrt{(-3 - 2)^2 + (3 - 3)^2}

WX = \sqrt{(-5)^2 + (0)^2}

WX = \sqrt{25}

WX = 5

For XY:

(x_1, y_1)\ (x_2,y_2) = (2, 3)\ (4,-4)

XY = \sqrt{(2 - 4)^2 + (3 - (-4))^2}

XY = \sqrt{-2^2 + (3 +4)^2}

XY = \sqrt{-2^2 + 7^2}

XY = \sqrt{4 + 49}

XY = \sqrt{53}

For YZ:

(x_1, y_1)\ (x_2,y_2) = (4,-4)\ (-3, -2)

YZ = \sqrt{(4 - (-3))^2 + (-4 - (-2))^2}

YZ = \sqrt{(4 +3)^2 + (-4 +2)^2}

YZ = \sqrt{7^2 + (-2)^2}

YZ = \sqrt{49 + 4}

YZ = \sqrt{53}

For ZW:

(x_1, y_1)\ (x_2,y_2) = (-3, -2)\ (-3, 3)

ZW = \sqrt{(-3 - (-3))^2 + (-2 - 3)^2}

ZW = \sqrt{(-3 +3)^2 + (-2 - 3)^2}

ZW = \sqrt{0^2 + (-5)^2}

ZW = \sqrt{0 + 25}

ZW = \sqrt{25}

ZW = 5

The Perimeter (P) is as follows:

P = WX + XY + YZ + ZW

P = 5 + \sqrt{53} + \sqrt{53} + 5

P = 5 + 5 + \sqrt{53} + \sqrt{53}

P = 10 + 2\sqrt{53} units

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3 years ago
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