How to find the length of an edge of a tetrahedron - PSAT Math

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Question

Refer to the above tetrahedron, or four-faced solid.The surface area of the tetrahedron is 444. Evaluate to the nearest tenth.

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength . Since the total surface area is 444, each triangle has area one fourth of this, or 111. To find , set in the formula for the area of an equilateral triangle:

$\frac{s^{2} \sqrt{3} }{4} =A$

$\frac{s^{2} \sqrt{3} }{4} =111$

$s^{2} \sqrt{3} =444$

$s^{2}=\frac{444}{ \sqrt{3} } \approx \frac{444}{ 1.732 } \approx 256.34$

$s\approx \sqrt{256.34} \approx 16.0$

Compare your answer with the correct one above

Question

Refer to the above tetrahedron, or four-faced solid.The surface area of the tetrahedron is 444. Evaluate to the nearest tenth.

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength . Since the total surface area is 444, each triangle has area one fourth of this, or 111. To find , set in the formula for the area of an equilateral triangle:

$\frac{s^{2} \sqrt{3} }{4} =A$

$\frac{s^{2} \sqrt{3} }{4} =111$

$s^{2} \sqrt{3} =444$

$s^{2}=\frac{444}{ \sqrt{3} } \approx \frac{444}{ 1.732 } \approx 256.34$

$s\approx \sqrt{256.34} \approx 16.0$

Compare your answer with the correct one above

Question

Refer to the above tetrahedron, or four-faced solid.The surface area of the tetrahedron is 444. Evaluate to the nearest tenth.

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength . Since the total surface area is 444, each triangle has area one fourth of this, or 111. To find , set in the formula for the area of an equilateral triangle:

$\frac{s^{2} \sqrt{3} }{4} =A$

$\frac{s^{2} \sqrt{3} }{4} =111$

$s^{2} \sqrt{3} =444$

$s^{2}=\frac{444}{ \sqrt{3} } \approx \frac{444}{ 1.732 } \approx 256.34$

$s\approx \sqrt{256.34} \approx 16.0$

Compare your answer with the correct one above

Question

Refer to the above tetrahedron, or four-faced solid.The surface area of the tetrahedron is 444. Evaluate to the nearest tenth.

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength . Since the total surface area is 444, each triangle has area one fourth of this, or 111. To find , set in the formula for the area of an equilateral triangle:

$\frac{s^{2} \sqrt{3} }{4} =A$

$\frac{s^{2} \sqrt{3} }{4} =111$

$s^{2} \sqrt{3} =444$

$s^{2}=\frac{444}{ \sqrt{3} } \approx \frac{444}{ 1.732 } \approx 256.34$

$s\approx \sqrt{256.34} \approx 16.0$

Compare your answer with the correct one above

Question

Refer to the above tetrahedron, or four-faced solid.The surface area of the tetrahedron is 444. Evaluate to the nearest tenth.

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength . Since the total surface area is 444, each triangle has area one fourth of this, or 111. To find , set in the formula for the area of an equilateral triangle:

$\frac{s^{2} \sqrt{3} }{4} =A$

$\frac{s^{2} \sqrt{3} }{4} =111$

$s^{2} \sqrt{3} =444$

$s^{2}=\frac{444}{ \sqrt{3} } \approx \frac{444}{ 1.732 } \approx 256.34$

$s\approx \sqrt{256.34} \approx 16.0$

Compare your answer with the correct one above

Question

Refer to the above tetrahedron, or four-faced solid.The surface area of the tetrahedron is 444. Evaluate to the nearest tenth.

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength . Since the total surface area is 444, each triangle has area one fourth of this, or 111. To find , set in the formula for the area of an equilateral triangle:

$\frac{s^{2} \sqrt{3} }{4} =A$

$\frac{s^{2} \sqrt{3} }{4} =111$

$s^{2} \sqrt{3} =444$

$s^{2}=\frac{444}{ \sqrt{3} } \approx \frac{444}{ 1.732 } \approx 256.34$

$s\approx \sqrt{256.34} \approx 16.0$

Compare your answer with the correct one above

Question

Refer to the above tetrahedron, or four-faced solid.The surface area of the tetrahedron is 444. Evaluate to the nearest tenth.

Answer

The tetrahedron has four faces, each of which is an equilateral triangle with sidelength . Since the total surface area is 444, each triangle has area one fourth of this, or 111. To find , set in the formula for the area of an equilateral triangle:

$\frac{s^{2} \sqrt{3} }{4} =A$

$\frac{s^{2} \sqrt{3} }{4} =111$

$s^{2} \sqrt{3} =444$

$s^{2}=\frac{444}{ \sqrt{3} } \approx \frac{444}{ 1.732 } \approx 256.34$

$s\approx \sqrt{256.34} \approx 16.0$

Compare your answer with the correct one above

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