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2 years ago

9

Which solid could have exactly two planes of symmetry?

1 answer:

kodGreya [7K]2 years ago

4 0

Answer:

Triangular Prism

Step-by-step explanation:

A plane of symmetry divides a three dimensional shape into two congruent halves that are mirror images of each other. An isosceles triangular based prism has 2 planes of symmetry. An isosceles triangular based prism has 2 planes of symmetry. An isosceles triangular based prism has 2 planes of symmetry.

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Answer:

The area of the rectangle <em>TOUR</em> is 80.00 unit².

Step-by-step explanation:

The area of a rectangle is computed using the formula:

$Area\ of\ a\ Rectangle=length\times width$

Since the dimensions of the rectangle are not provided we can compute the dimensions using the distance formula for two points.

The distance formula using the two point is:

$distance=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$

Considering the rectangle <em>TOUR</em> the area formula will be:

Area of Rectangle <em>TOUR</em> = <em>TO × OU</em>

The co-ordinates of the four vertices of a triangle are:

T = (-8, 0), O = (4, 4), U = (6, -2) and R = (-6, -6)

Compute the distance between the vertices <em>T</em> and <em>O</em> as:

$TO=\sqrt{(4-(-8))^{2}+(4-0)^{2}}\\=\sqrt{12^{2}+4^{2}} \\=\sqrt{160} \\=4\sqrt{10}$

Compute the distance between the vertices <em>O </em>and <em>U</em> as:

$OU=\sqrt{(6-4)^{2}+(-2-4)^{2}}\\=\sqrt{2^{2}+6^{2}} \\=\sqrt{40} \\=2\sqrt{10}$

Compute the area of rectangle TOUR as follows:

$Area\ of\ TOUR=TO\times OU\\=4\sqrt{10}\times 2\sqrt{10}\\=80\\\approx80.00 unit^{2}$

Thus, the area of the rectangle <em>TOUR</em> is 80.00 unit².

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