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2 years ago

Please help me with this question

Mathematics

1 answer:

Trava [24]2 years ago

4 0

Applying the formula for the area of a sector and length of an arc, the value of k is calculated as: 27.

<h3>What is the Area of a Sector?</h3>

Area of a sector of a circle = ∅/360 × πr²

<h3>What is the Length of an Arc?</h3>

Length of arc = ∅/360 × 2πr

Given the following:

Radius (r) = 9 cm
Length of arc = 6π cm
Area of sector = kπ cm²

Find ∅ of the sector using the formula for length of acr:

∅/360 × 2πr = 6π

Plug in the value of r

∅/360 × 2π(9) = 6π

∅/360 × 18π = 6π

Divide both sides by 18π

∅/360 = 6π/18π

∅/360 = 1/3

Multiply both sides by 360

∅ = 1/3 × 360

∅ = 120°

Find the area of the sector:

Area = ∅/360 × πr² = 120/360 × π(9²)

Area = 1/3 × π81

Area = 27π

Therefore, the value of k is 27.

Learn more about area of sector on:

brainly.com/question/22972014

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Surface area is just the area of all these 4 triangles plus the rectangle.

First we can find the area of the rectangle.

$\sf A=lw$

Half of the length is 28 cm, so the full length must be 28 * 2 = 56 cm.

$\sf A=(56)(27)$

$\sf A=1512cm^2$

The base for the left and right triangles are 27. The heights would be the net length minus half the length of the rectangle:

$\sf 82.8-28=54.8\? cm$

Calculate the area:

$\sf A=\dfrac{1}{2}bh$

$\sf A=\dfrac{1}{2}(27)(54.8)$

$\sf A=739.8\?cm^2$

We have two of these triangles.

$\sf 739.8\times 2=1479.6\? cm^2$

Now do the other two pair of triangles. The bases for them are 28 + 28 = 56 cm. The heights would be the net width minus the width of the rectangle:

$\sf 81.2-27=54.2\?cm$

Now find the area:

$\sf A=\dfrac{1}{2}bh$

$\sf A=\dfrac{1}{2}(56)(54.2)$

$\sf A=1517.6\? cm^2$

We have two of these triangles.

$\sf 1517.6\times 2=3035.2\?cm^2$

Add all the areas together:

$\sf 1512+1479.6+3035.2=\boxed{\sf 6026.80\? cm^2}$

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3 years ago

Please help me with this question​

Please help me with this question