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3 years ago

Show that the area of triangle DEF is 11 cm2.

Mathematics

1 answer:

dedylja [7]3 years ago

5 0

9514 1404 393

Explanation:

The length CD can be found from the Pythagorean theorem.

CF² + CD² = DF²

2² + CD² = (2√10)²

CD² = 40 -4 = 36

CD = √36 = 6

From here, any of several methods could be used to find the area of ΔDEF.

One of them is to subtract the areas of the triangles that are not ΔDEF from the area of the rectangle ABCD.

area ABCD = bh = (6 cm)(5 cm) = 30 cm²

area ΔADE = 1/2bh = 1/2(4 cm)(5 cm) = 10 cm²

area ΔBEF = 1/2bh = 1/2(2 cm)(3 cm) = 3 cm²

area ΔCDF = 1/2bh = 1/2(6 cm)(2 cm) = 6 cm²

Then the area of interest is ...

area ΔDEF = area ABCD -area ΔADE -area ΔBEF -area ΔCDF

area ΔDEF = (30 -10 -3 -6) cm²

area ΔDEF = 11 cm²

<em>Alternate solution</em>

If we define A as the origin of a coordinate system, then the coordinates of the vertices of ΔDEF are ...

D(0, 5), E(4, 0), F(6, 3)

The area can be computed as half the absolute value of the sum of the determinants of 2×2 matrices consisting of adjacent points.

$A=\dfrac{1}{2}\left|\left|\begin{array}{cc}0&5\\4&0\end{array}\right|+\left|\begin{array}{cc}4&0\\6&3\end{array}\right|+\left|\begin{array}{cc}6&3\\0&5\end{array}\right| \right|\\\\=\dfrac{1}{2}|(0-20)+(12-0)+(30-0)|=\dfrac{1}{2}(22) = \boxed{11}$

_____

<em>Additional comment</em>

The solution using coordinates is effectively equivalent to the computation of the area of trapezoid ABFD, less the areas of triangles EBF and AED.

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3 years ago

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Step-by-step explanation:

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1 year ago

There are 140 legs on a planet zids have 5 legs and zods have 7 legs h=how many soulutions are there

abruzzese [7]

Number of zids=x
number of zods=y
number of zid legs=5x
number of zod legs=7y

so
5x+7y=140
try to get it into slope intercept from so you can graph is (y=mx+b)
5x+7y=140
subtract 5x from both sides
7y=140-5x
divide both sides by 7
y=20-5/7x
y=-5/7x+20

plug in numbers for x and get numbers for y (you can only plug in multiples of 7 for x so that there are a whole number of zids and since you are counting, x and y must never be negative)

lets try 0 for x

y=-5/7(0)+20
y=20
so x=0 and y=20 is an answer (if you can have only one of that species)

lets try 7 for x
y=-5/7(7)+20
y=-5+20
y=15

so x=7 and y=15 is an answer

lets try 14 for x

y=-5/7(14)+20
y=-10+20
y=10

so x=14 and y=10 is another answer

lets try 21 for x
y=-5/7(21)+20
y=-15+20
y=5

so x=21 and y=5 is another answer

lets try 28 for x
y=-5/7(28)-20
y=-20+20
y=0

so x=28 and y=0 is an answer (if there can be only one of a species)

if we go further, then y will be negative so the answers are

(x,y)
(0,20)
(7,15)
(14,10)
(21,5)
(28,0)

if it is allowed that only one species exists then there are 5 possible answers
if both must exist simultaneously, then there are only 3 answers
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3 years ago

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3 years ago

4sin²<img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7B2%7D" id="TexFormula1" title="\frac{x}{2}" alt="\frac{x}{2}" align="absm

raketka [301]

Answer:

$\displaystyle x=\left \{\frac{2\pi}{3}+2\pi k,\frac{4\pi}{3}+2\pi k, \frac{8\pi}{3}+2\pi k, \frac{10\pi}{3}+2\pi k\right \}k\in \mathbb{Z}$

Step-by-step explanation:

Hi there!

We want to solve for $x$ in:

$4\sin^2(\frac{x}{2})=3$

Since $x$ is in the argument of $\sin^2$ , let's first isolate $\sin^2$ by dividing both sides by 4:

$\displaystyle \sin^2\left(\frac{x}{2}\right)=\frac{3}{4}$

Next, recall that $\sin^2x$ is just shorthand notation for $(\sin x)^2$ . Therefore, take the square root of both sides:

$\displaystyle \sqrt{\sin^2\left(\frac{x}{2}\right)}=\sqrt{\frac{3}{4}},\\\sin\left(\frac{x}{2}\right)=\pm \sqrt{\frac{3}{4}}$

Simplify using $\displaystyle \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$ :

$\displaystyle \sin\left(\frac{x}{2}\right)=\pm \sqrt{\frac{3}{4}},\\\sin\left(\frac{x}{2}\right)=\pm \frac{\sqrt{3}}{\sqrt{4}}=\pm \frac{\sqrt{3}}{2}$

Let $\phi = \frac{x}{2}$ .

<h3><u>Case 1 (positive root):</u></h3>

$\displaystyle \sin(\phi)=\frac{\sqrt{3}}{2},\\\phi = \frac{\pi}{3}+2\pi k, k\in \mathbb{Z}, \\\\\phi =\frac{2\pi}{3}+2\pi k, k\in \mathbb{Z}$

Therefore, we have:

$\displaystyle \frac{x}{2}=\phi = \frac{\pi}{3}+2\pi k, k\in \mathbb{Z}, \\\\\frac{x}{2}=\phi =\frac{2\pi}{3}+2\pi k, k\in \mathbb{Z},\\\\\begin{cases}x=\boxed{\frac{2\pi}{3}+2\pi k, k\in \mathbb{Z}},\\x=\boxed{\frac{4\pi}{3}+2\pi k , k \in \mathbb{Z}}\end{cases}$

<h3><u>Case 2 (negative root):</u></h3>

$\displaystyle \sin(\phi)=-\frac{\sqrt{3}}{2},\\\phi = \frac{4\pi}{3}+2\pi k, k\in \mathbb{Z}, \\\\\phi =\frac{5\pi}{3}+2\pi k, k\in \mathbb{Z},\\\begin{cases}x=\boxed{\frac{8\pi}{3}+2\pi k, k\in \mathbb{Z}},\\x=\boxed{\frac{10\pi}{3}+2\pi k , k \in \mathbb{Z}}\end{cases}$

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3 years ago

Show that the area of triangle DEF is 11 cm2.​

Show that the area of triangle DEF is 11 cm2.