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3 years ago

Find the area of a regular hexagon if given a side length of 1.9 to the square root of 3.

Mathematics

1 answer:

dalvyx [7]3 years ago

6 0

Answer:

28.137 sq. units.

Step-by-step explanation:

Area of a regular hexagon is

$A=\dfrac{3\sqrt{3}}{2}a^2$

where, a is the side length.

It is given that the side length of the regular hexagon is $1.9\sqrt{3}$ .

Substitute $a=1.9\sqrt{3}$ in the above formula.

$A=\dfrac{3\sqrt{3}}{2}(1.9\sqrt{3})^2$

$A=(1.5\sqrt{3})(10.83)$

$A=16.245\sqrt{3}$

$A=28.137$

Therefore, the area of regular hexagon is 28.137 sq. units.

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Answer:

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Step-by-step explanation:

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iogann1982 [59]

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Answer:

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Step-by-step explanation:

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3 years ago

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3 years ago

Read 2 more answers

Find the least common multiple (LCM) of 8y^6+ 144y^5+ 640y^4 and 2y^4 + 40y^3 + 200y^2.

Mice21 [21]

Answer:

LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

Step-by-step explanation:

Making factors of $8y^{6}+ 144y^{5}+ 640y^{4}$

Taking $8y^{4}$ common:

$\Rightarrow 8y^{4} (y^{2}+ 18y+ 80)$

Using factorization method:

$\Rightarrow 8y^{4} (y^{2}+ 10y + 8y + 80)\\\Rightarrow 8y^{4} (y (y+ 10) + 8(y + 10))\\\Rightarrow 8y^{4} (y+ 10)(y + 8))\\\Rightarrow \underline{2y^{2}} \times 4y^{2} \underline{(y+ 10)}(y + 8)) ..... (1)$

Now, Making factors of $2y^{4} + 40y^{3} + 200y^{2}$

Taking $2y^{2}$ common:

$\Rightarrow 2y^{2} (y^{2}+ 20y+ 100)$

Using factorization method:

$\Rightarrow 2y^{2} (y^{2}+ 10y+ 10y+ 100)\\\Rightarrow 2y^{2} (y (y+ 10) + 10(y + 10))\\\Rightarrow \underline {2y^{2} (y+ 10)}(y + 10) ............ (2)$

The underlined parts show the Highest Common Factor(HCF).

i.e. HCF is $2y^{2} (y+ 10)$ .

We know the relation between LCM, HCF of the two numbers 'p' , 'q' and the numbers themselves as:

$HCF \times LCM = p \times q$

Using equations (1) and (2): $\Rightarrow 2y^{2} (y+ 10) \times LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times 2y^{2} (y+ 10)(y + 10)\\\Rightarrow LCM = 2y^{2} \times 4y^{2}(y+ 10)(y + 8) \times (y + 10)\\\Rightarrow LCM = 8y^{4}(y+ 10)^{2}(y + 8)$

Hence, LCM = $8y^{4}(y+ 10)^{2}(y + 8)$

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3 years ago