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3 years ago

The square in the middle has an

Mathematics

1 answer:

eduard3 years ago

3 0

Answer:

4 square units

Step-by-step explanation:

Given that the square we have in the middle of the diagram has an area of 1 square unit, to find the area of the entire angle, all we need to do is decompose and reaarange the other shapes where necessary to form the square in the middle.

See the attachment below showing how we can break and arrange the other shapes to derive their area.

Thus, the 2 smaller triangles can be reaaranged to form the middle square. Since the middle square = 1 unit², each smaller triangle = ½ unit².

The other two medium triangles can be decomposed into 2 smaller triangles, which will give us the middle square.

Therefore, the area if the entire rectangle is:

Middle square = 1 unit²

Two smaller triangles = 1 middle square = 1 unit²

2 medium triangles = 4 smaller triangles = 2 middle squares = 2 units

Area of the entire rectangle = 1 + 1 + 2 = 4 square units

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3 years ago

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givi [52]

Answer:

70/9

Step-by-step explanation:

We have the quadratic:

$3x^2+4x-9$

So, let’s find the roots of the quadratic. We will set the expression equal to 0:

$3x^2+4x-9=0$

Testing for factors, we can see that our quadratic isn’t factorable.

So, we can use the Quadratic Formula. The quadratic formula is given by:

$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case:

$a=3, b=4,\text{ and } c=-9$

Therefore, by substitution:

$\displaystyle x=\frac{-(4)\pm\sqrt{(4)^2-4(3)(-9)}}{2(3)}$

Evaluate:

$\displaystyle x=\frac{-4\pm\sqrt{124}}{6}$

Simplify the square root:

$\sqrt{124}=\sqrt{4\cdot31}=2\sqrt{31}$

Hence:

$\displaystyle x=\frac{-4\pm2\sqrt{31}}{6}$

Reduce:

$\displaystyle x=\frac{-2\pm\sqrt{31}}{3}$

So, our roots are:

$\displaystyle x_1=\frac{-2+\sqrt{31}}{3}, x_2=\frac{-2-\sqrt{31}}{3}$

We want to find the sum of the squares of our two roots. So, let’s square each term:

$\displaystyle (x_1)^2=\Big(\frac{-2+\sqrt{31}}{3}\Big)^2$

Square. For the numerator, we can use the perfect square trinomial patten where:

$(a+b)^2=(a^2+2ab+b^2)$

Therefore:

$\displaystyle (x_1)^2=\Big(\frac{(-2)^2+2(-2)(\sqrt{31})+(\sqrt{31})^2}{9}\Big)$

Simplify:

$\displaystyle (x_1)^2=\frac{35-4\sqrt{31}}{9}$

Similarly, for the second root, we will have:

$\displaystyle (x_2)^2=\Big(\frac{-2-\sqrt{31}}{3}\Big)^2$

So:

$\displaystyle (x_2)^2=\Big(\frac{(-2)^2+2(-2)(-\sqrt{31})+(-\sqrt{31})^2}{9}\Big)$

Simplify:

$\displaystyle (x_2)^2=\frac{35+4\sqrt{31}}{9}$

Therefore, our sum will be:

$\displaystyle (x_1)^2+(x_2)^2\\\\ \begin{aligned} &=\frac{35-4\sqrt{31}}{9}+\frac{35+4\sqrt{31}}{9}\\&=\frac{35-4\sqrt{31}+35+4\sqrt{31}}{9}\\&=\frac{70}{9}\end{aligned}$