Simplify. 1/3(1−1/4)^2
1 answer:
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Answer:
12.9 yd
Step-by-step explanation:
It helps if you draw a triangle. Draw a horizontal segment. That is the ground. Now at the left end start a new segment that goes up to the right at approximately 15 degrees until it its other endpoint directly above the right endpoint of the horizontal segment. Connect these two endpoints. The vertical side on the right shows the height of the kite. The hypotenuse is the string.
For the 15-deg angle, the height of the triangle is the opposite leg, and the string is the hypotenuse. The trig ratio that relates the opposite leg tot he hypotenuse is the sine.
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,
6. The 4 triangles formed by the diagonal are congruent, so the area of the rhombus ABCD = 4 Area (triangle DOC)=4*
=
(
)
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,
6. The 4 triangles formed by the diagonal are congruent, so the area of the rhombus ABCD = 4 Area (triangle DOC)=4*