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2 years ago

4. ADEF is an equilateral triangle. E 24 20 D F

Mathematics

1 answer:

irinina [24]2 years ago

7 0

By Pythagorean theorem

x²+GF²=EF² ( where G is the point of intersection of perpendicular lines)

now, GF=1/2 EF= 12

x²+12²=24²

x²=24²-12²

x²=432

x=√432

x=12√3

x≈20.78

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What is the following product? (4x square root 5x^2 + 2^2 square root 6)^2

tangare [24]

The product is $104 x^{4}+16 \sqrt{30} x^{4}$

Explanation:

The given expression is $\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}$

We need to determine the product of the given expression.

First, we shall simplify the given expression.

Thus, we have,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x \sqrt{5} x+2 x^{2} \sqrt{6}\right)^2$

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)^2$

Expanding the expression, we have,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)\left(4 x^{2} \sqrt{5}+2 x^{2} \sqrt{6}\right)$

Now, we shall apply FOIL, we get,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=\left(4 x^{2} \sqrt{5}\right)^{2}+2 ( 2 x^{2} \sqrt{6})(4 x^{2} \sqrt{5})+\left(2 x^{2} \sqrt{6}\right)^{2}$

Simplifying the terms, we have,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=16 \cdot 5 x^{4}+16 \sqrt{30} x^{4}+4 \cdot 6 x^{4}$

Multiplying, we get,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=80 x^{4}+16 \sqrt{30} x^{4}+24 x^{4}$

Adding the like terms, we get,

$\left(4 x \sqrt{5 x^{2}}+2 x^{2} \sqrt{6}\right)^{2}=104 x^{4}+16 \sqrt{30} x^{4}$

Thus, the product of the given expression is $104 x^{4}+16 \sqrt{30} x^{4}$

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