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mestny [16]
3 years ago
11

The side of rhombus is 14 in. The obtuse angle is 160º. Find the longer diagonal of the rhombus.

Mathematics
1 answer:
Mandarinka [93]3 years ago
3 0

Answer:

x = 27.57 in to the nearest hundredth.

Step-by-step explanation:

The longer diagonal  is opposite the obtuse angle.

Also as it is a rhombus all sides = 14 in.

Use the Cosine Rule:

x^2 = 14^2 + 14^2 - 2*14^2 cos 160

x^2 =  760.36

x = sqrt 760.36

x = 27.57 in.

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<u>ANSWER:  </u>

The area of the triangle bounded by the y-axis is  \frac{7938}{4225} \sqrt{65} \text { unit }^{2}

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Given, f(x)=9-\frac{-4}{7} x

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Slope of f(x) = \frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-4}{7}

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Perpendicular line is passing through origin(0,0).So by using point slope formula,

y-y_{1}=m\left(x-x_{1}\right)

Where m is the slope and \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)

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7x – 4y = 0  

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for points on y-axis x will be zero, to get y value, put x =0 int f(x)

0 + 7y – 63 = 0

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Hence, the point of intersection is (0, 9)

Third vertex is point of intersection of f(x) and its perpendicular line.

So, solve (1) and (2)

\begin{array}{l}{9-\frac{4}{7} x=\frac{7}{4} x} \\\\ {9 \times 4-\frac{4 \times 4}{7} x=7 x} \\\\ {36 \times 7-16 x=7 \times 7 x} \\\\ {252-16 x=49 x} \\\\ {49 x+16 x=252} \\\\ {65 x=252} \\\\ {x=\frac{252}{65}}\end{array}

Put x value in (2)

\begin{array}{l}{y=\frac{7}{4} \times \frac{252}{65}} \\\\ {y=\frac{441}{65}}\end{array}

So, the point of intersection is \left(\frac{252}{65}, \frac{441}{65}\right)

Length of f(x) is distance between \left(\frac{252}{65}, \frac{441}{65}\right) and (0,9)

\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(9-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+0} \\ &=\frac{252}{65} \end{aligned}

Now, length of perpendicular of f(x) is distance between \left(\frac{252}{65}, \frac{441}{65}\right) \text { and }(0,0)

\begin{aligned} \text { Length } &=\sqrt{\left(0-\frac{252}{65}\right)^{2}+\left(0-\frac{441}{65}\right)^{2}} \\ &=\sqrt{\left(\frac{252}{65}\right)^{2}+\left(\frac{441}{65}\right)^{2}} \\ &=\frac{\sqrt{(12 \times 21)^{2}+(21 \times 21)^{2}}}{65} \\ &=\frac{63}{65} \sqrt{65} \end{aligned}

Now, area of right angle triangle = \frac{1}{2} \times \frac{252}{65} \times \frac{63}{65} \sqrt{65}

=\frac{7938}{4225} \sqrt{65} \text { unit }^{2}

Hence, the area of the triangle is \frac{7938}{4225} \sqrt{65} \text { unit }^{2}

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