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3 years ago

What is the perimeter of a square which has the same area as a circle with circumference of 4π?

Mathematics

1 answer:

enot [183]3 years ago

8 0

Answer:

perimeter = 8 $\sqrt{\pi }$

Step-by-step explanation:

We require to calculate the area (A) of the circle

A = πr² ← r is the radius

circumference = 4π = 2πr, hence

2πr = 4π ( divide both sides by 2π )

r = 2, hence

A = π × 4 = 4π

The area of the square is therefore 4π and area = s² ← s is the side length

s² = 4π ( take the square root of both sides )

s = $\sqrt{4\pi }$ = 2 $\sqrt{\pi }$ , hence

perimeter = 4 × 2 $\sqrt{\pi }$ = 8 $\sqrt{\pi }$

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Hannah notices that segment HI and segment KL are congruent in the image below: Two triangles are shown, GHI and JKL. G is at ne

BlackZzzverrR [31]

Answer:

segment IG ≅ segment LJ

Step-by-step explanation:

Please refer to the attached image as per the triangles as given in the question statement.

$\triangle HGI, \triangle JKL$

$G\left(-3,1\right),\ H\left(-1,1\right),\ I\left(-2,3\right)$

$J\left(3,3\right),K\left(1,3\right),L\left(2,1\right)$

Given that:

$HI\cong KL$ and

$\angle I \cong \angle L$

SAS congruence between two triangles states that two triangles are congruent if two corresponding sides and the angle between the two sides are congruent.

We are given that one angle and one sides are congruent in the given triangles.

We need to prove that other sides that makes this angle are also congruent.

To show the triangles are congruent i.e. $\triangle GHI \cong \triangle JKL$ by SAS congruence we need to prove that

segment IG ≅ segment LJ

Let us use Distance formula to find IG and LJ:

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

$IG =\sqrt{(-2+3)^2+(3-1)^2} =\sqrt5\ units$

$LJ =\sqrt{(2-3)^2+(1-3)^2} =\sqrt5\ units$

Hence, segment IG ≅ segment LJ

$\therefore$ ΔGHI ≅ ΔJKL by SAS

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3 years ago

Rational numbers are __ natural numbers

Brums [2.3K]

Lol that’s why not I just got to tell

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AnnZ [28]

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Mkey [24]

Answer:

$2^{11}$

Step-by-step explanation:

Key indices rule: When dividing powers, subtract the indices and keep the base numbers constant.

$\frac{2^{3} }{2^{-8} }$ Based on the rule above, the indices are determined by 3--8, which is 11.

The base number is kept constant.

Therefore, the answer is $2^{11}$ .

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Sergio [31]

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