10th Grade Sine Law (Nikoleta Simeonova)

1.

Given a triangle ABC with side BC=25 cm, the altitude BD is 15 cm, the circumradius is 32,5 cm. Find the other sides, and .
- A.
- B.
- C.
- D.
Correct Answer(s)
A.
D.
2.

Given a triangle ABC, the shortest side and the circumradius are in the ratio 6:5. The other two sides are 20 cm and 21 cm. Find the shortest side of the triangle.

Correct Answer(s)
13 cm

Explanation
In a triangle, the circumradius is the radius of the circumscribed circle, which passes through all three vertices of the triangle. The shortest side of the triangle is opposite the smallest angle. In this case, since the other two sides are 20 cm and 21 cm, the shortest side must be less than or equal to 20 cm. Since the ratio of the shortest side to the circumradius is given as 6:5, the shortest side can be calculated as (6/5) * circumradius. Substituting the given values, we get (6/5) * circumradius = 20 cm. Solving for circumradius, we find that it is equal to 25 cm. Therefore, the shortest side of the triangle is (6/5) * 25 cm = 30 cm * 6/5 = 36 cm / 5 = 7.2 cm, which is rounded to 7 cm. However, since the shortest side must be less than or equal to 20 cm, the correct answer is 13 cm.

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4
3.

Given an inscribed trapezoid ABCD with a leg m. The angle between the diagonals, lying opposite to the bases is . Find the radius of the circumcircle of ABCD.
- A.
- B.
- C.
- D.
Correct Answer
C.

Explanation
The radius of the circumcircle of an inscribed trapezoid can be found using the formula R = (m/2) * cot(angle/2), where R is the radius, m is the length of the leg, and angle is the angle between the diagonals opposite to the bases.

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4.

Given a right triangle ABC with a hypotenuse AB=2 cm. A, B and the midpoints of the legs (BC and AC) lie on a circle. Find the radius of that circle.
- A.
- B.
- C.
- D.
Correct Answer
B.

Explanation
The radius of the circle can be found by using the property that the circumradius of a right triangle is equal to half the length of the hypotenuse. In this case, the hypotenuse AB is given as 2 cm, so the radius of the circle is 1 cm.

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5.

Given a triangle ABC, the interior and exterior bisectors of angle C are equal. Find the number .

Correct Answer
4

Explanation
The number 4 represents the measure of angle C in degrees. The interior and exterior bisectors of angle C are equal, which means that the angle is divided into two equal angles. Since the sum of the angles in a triangle is 180 degrees, each of the two equal angles will measure 90 degrees. Therefore, angle C measures 90 degrees, and the number representing it is 4.

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4
6.

Given an inscribed trapezoid ABCD with leg 6 cm. The diagonals are perpendicular to each other. Find the radius of the circumcircle of ABCD.
- A.
- B.
- C.
  
  6 cm
- D.
  
  3 cm
Correct Answer
A.

Explanation
In an inscribed trapezoid with perpendicular diagonals, the diagonals are also the diameters of the circumcircle. Since the length of one leg of the trapezoid is given as 6 cm, it is also the diameter of the circumcircle. Therefore, the radius of the circumcircle is half of the diameter, which is 3 cm.

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7.

Given a rectangle ABCD with an angle between the diagonals . ABCD is inscribed in a circle with radius R. Find the area of ABCD.
- A.
- B.
- C.
- D.
Correct Answer
B.

Explanation
The area of a rectangle can be calculated by multiplying the length of one of its sides by the length of an adjacent side. In this case, since ABCD is inscribed in a circle, the diagonals are equal in length and bisect each other at right angles. This means that the length of one diagonal is equal to the radius of the circle, which is R. Therefore, the area of ABCD is equal to R * R, or R^2.

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8.

Given the isosceles inscribed triangle ABC with a radius of the circumcircle R and a leg with length R. Find the length of the base of ABC.
- A.
  
  4R
- B.
- C.
  
  R
- D.
Correct Answer
D.

Explanation
In an isosceles inscribed triangle, the two legs are equal in length. Since one of the legs has a length of R, the other leg will also have a length of R. Therefore, the length of the base of triangle ABC is equal to R.

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9.

Given the isosceles triangle ABC with angle ACB=120 degrees. ABC is inscribed in a circle with radius R. Find the perimeter P and the area S of ABC.
- A.
- B.
- C.
- D.
- E.
- F.
Correct Answer(s)
A.
F.

Explanation
Since triangle ABC is isosceles, we know that angle BAC = angle BCA. Since angle BAC + angle BCA + angle ACB = 180 degrees, we can conclude that angle BAC = angle BCA = (180 - 120) / 2 = 30 degrees.
Since the sum of angles in a triangle is 180 degrees, angle ABC = 180 - 30 - 30 = 120 degrees.
Since angle ABC is an inscribed angle, it intercepts an arc of 120 degrees on the circle.
Since the measure of an inscribed angle is half the measure of its intercepted arc, the intercepted arc measures 2 * 120 = 240 degrees.
Since the sum of the intercepted arcs of a triangle inscribed in a circle is 360 degrees, the remaining intercepted arcs each measure (360 - 240) / 2 = 60 degrees.
Since each intercepted arc measures 60 degrees, the length of each side of the triangle is R * 60 / 360 = R / 6.
Therefore, the perimeter P = 3 * (R / 6) = R / 2 and the area S = (R / 6)^2 * sqrt(3) / 4 = R^2 * sqrt(3) / 72.

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10.

The distance between the chord AB and the center of the circle k is 1 cm. The radius of the circle is cm. Find the length of AB. If C is a random point, lying on the circle k, find angle ACB.
- A.
  
  AB=2 cm, angle ACB=45 degrees
- B.
  
  AB=2 cm, angle ACB=135 degrees
- C.
  
  AB=2 cm, angle ACB=45 or 135 degrees
- D.
  
  AB=2 cm, angle ACB cannot be determined
Correct Answer
C. AB=2 cm, angle ACB=45 or 135 degrees

Explanation
The length of AB is given as 2 cm, which means the distance between the chord AB and the center of the circle is equal to the radius of the circle. This implies that AB is a diameter of the circle.

Since C is a random point lying on the circle, angle ACB can be any angle formed by two radii of the circle. Therefore, angle ACB can be either 45 degrees or 135 degrees, depending on the specific position of point C on the circle.

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10th Grade Sine Law (Nikoleta Simeonova)

Given a triangle ABC with side BC=25 cm, the altitude BD is 15 cm, the circumradius is 32,5 cm. Find the other sides, and .

Given a triangle ABC, the shortest side and the circumradius are in the ratio 6:5. The other two sides are 20 cm and 21 cm. Find the shortest side of the triangle.

Given an inscribed trapezoid ABCD with a leg m. The angle between the diagonals, lying opposite to the bases is . Find the radius of the circumcircle of ABCD.

Given a right triangle ABC with a hypotenuse AB=2 cm. A, B and the midpoints of the legs (BC and AC) lie on a circle. Find the radius of that circle.

Given a triangle ABC, the interior and exterior bisectors of angle C are equal. Find the number .

Given an inscribed trapezoid ABCD with leg 6 cm. The diagonals are perpendicular to each other. Find the radius of the circumcircle of ABCD.

Given a rectangle ABCD with an angle between the diagonals . ABCD is inscribed in a circle with radius R. Find the area of ABCD.

Given the isosceles inscribed triangle ABC with a radius of the circumcircle R and a leg with length R. Find the length of the base of ABC.

Given the isosceles triangle ABC with angle ACB=120 degrees. ABC is inscribed in a circle with radius R. Find the perimeter P and the area S of ABC.

The distance between the chord AB and the center of the circle k is 1 cm. The radius of the circle is cm. Find the length of AB. If C is a random point, lying on the circle k, find angle ACB.

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