Mathematics, 25.07.2019 19:30 IkarosSakurai

Asquare of area 32 cm2 is inscribed into a semi-circle. what is the area of the semi-circle?

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Find Length of the square :
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Given that the area of the square is 32 cm²:

Area = Length²

$\text {Length =} \sqrt{32}$

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Find Radius :
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Using Pythagorus Theorem to find the radius of the circle:

a² + b² = c²

$( \sqrt{32})^2 + (\frac{ \sqrt{32} }{2} )^2 = ( \text {radius} )^2$

$(\text {radius} )^2 = 32 + \dfrac{32}{4}$

$(\text {radius} )^2 = 40$

$\text { Radius = } \sqrt{ 40 }$

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Find Area of the semi-circle :
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$\text {Area of the semi-circle = } \dfrac{1}{2} \pi r^2$

$\text {Area of the semi-circle = } \dfrac{1}{2} \pi ( \sqrt{40}) ^2$

$\text {Area of the semi-circle = } 20 \pi$

$\text {Area of the semi-circle = } 62.83 \ cm^2$

Answer from: lmolmlm5420

*incorect caculations do not read*
we have a formula for that
${r}^{2} = \frac{5 {a}^{2} }{4}$
r being the radius and a being the edge of our square

a is
$\sqrt{32}$
so lets find r
${r}^{2} = \frac{5 \times { \sqrt{32} }^{2} }{4}$
after solving we reach these two answers
$r = - 2 \sqrt{10} \\ r = 2 \sqrt{10}$
we are looking for a radius so negetive numbers are disregarded. now that we have the radius for our semi circle we can calculate the area of our semi circle. with this formula
$\frac{\pi \times {r}^{2} }{2}$
(just the formula for the area of a circle divided by half)
$area = \frac{\pi \times 2\sqrt{10} }{2} \\area = \pi \times 20$

Answer from: Quest

y = 2x

step-by-step explanation:

if you divide each number in the sequence, you'll see that they all have a multiple factor of 2. each number in the series, and any other numbers which may continue, will be multiplied by 2. therefore, y is the next number in the sequence, and x is the current one, so, our formula will be y = 2x.

Answer from: Quest

80° is the measure of gh

step-by-step explanation:

circle f is congruent to circle j

we have two circle with center f and j.

both circles are congruent. the radius of the circle would be equal.

see the attachment for the circle.

we are given ∠efd ≅ ∠gjh

therefore, arc (ed) = arc (gh) (∴ circle f is congruent to circle j)

but ∠dfe=80°=arc (ed)

therefore, arc (gh) = 80°

thus, 80° is the measure of gh

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