The sides of a rhombus with angle of 60° are 6 inches. Find the area of the rhombus.

Mathematics

1 answer:

ehidna [41]3 years ago

6 0

1. Check the drawing of the rhombus ABCD in the picture attached.

2. m(CDA)=60°, and AC and BD be the diagonals and let their intersection point be O.

3. The diagonals:
i) bisect the angles so m(ODC)=60°/2=30°

ii) are perpendicular to each other, so m(DOC)=90°

4. In a right triangle, the length of the side opposite to the 30° angle is half of the hypothenuse, so OC=3 in.

5. By the pythagorean theorem, $DO= \sqrt{ DC^{2}- OC^{2} }= \sqrt{ 6^{2}- 3^{2} }=\sqrt{ 36- 9 }=\sqrt{ 27 }= \sqrt{9*3}=$
$=\sqrt{9}* \sqrt{3} =3\sqrt{3} (in)$

6. The 4 triangles formed by the diagonal are congruent, so the area of the rhombus ABCD = 4 Area (triangle DOC)=4* $\frac{DO*OC}{2}=4 \frac{3 \sqrt{3} *3}{2}$ = $=2*9 \sqrt{3}=18 \sqrt{3}$ ( $in^{2}$ )

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DENIUS [597]

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4 or 7 (Check the explanation to see which one your equation actually is.

Step-by-step explanation:

One thing I want to clarify for you, It's Evaluate, not Avaluate.

Okay so, we want to find the value of 2^3x-1 and we know x = 1.

You didn't really clarify if the expression was $2^{3x-1}$ or $2^{3x}-1$ , so I'll be doing both:

For $2^{3x-1}$ , we should plug in x to get $2^{3(1)-1}$ and then simplify to get $2^{2}$ .

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