Could You Actually Pass This Geometry Trivia Test?

1) The sum of all the angles of a triangle is:

90o

180o

270o

360o

The sum of all the angles of a triangle is 180 degrees. This is a well-known property of triangles in Euclidean geometry. The sum of the interior angles of any triangle is always equal to 180 degrees. This property can be proven using various methods, such as the angle sum property of a straight line or the exterior angle theorem. Therefore, the correct answer is 180o.

Explanation

The sum of all the angles of a triangle is 180 degrees. This is a well-known property of triangles in Euclidean geometry. The sum of the interior angles of any triangle is always equal to 180 degrees. This property can be proven using various methods, such as the angle sum property of a straight line or the exterior angle theorem. Therefore, the correct answer is 180o.

2) What would be the distance between the point of intersection of the two tangents and the centre of the circle, if a circle of radius 6 cm is constructed from which two tangents are arise which are inclined at an angle of 60^o.

15

14

13

12

The distance between the point of intersection of the two tangents and the center of the circle can be found using the formula for the length of a radius of a circle inscribed in an equilateral triangle. In this case, the angle between the two tangents is 60 degrees, which means that the triangle formed by the center of the circle and the two points of intersection of the tangents is an equilateral triangle. The formula for the length of the radius of an inscribed equilateral triangle is r = s√3, where r is the radius of the circle and s is the length of each side of the equilateral triangle. In this case, the radius of the circle is 6 cm, so the length of each side of the equilateral triangle is 6 cm. Therefore, the distance between the point of intersection of the tangents and the center of the circle is 6√3 cm, which is approximately equal to 10.39 cm.

Explanation

The distance between the point of intersection of the two tangents and the center of the circle can be found using the formula for the length of a radius of a circle inscribed in an equilateral triangle. In this case, the angle between the two tangents is 60 degrees, which means that the triangle formed by the center of the circle and the two points of intersection of the tangents is an equilateral triangle. The formula for the length of the radius of an inscribed equilateral triangle is r = s√3, where r is the radius of the circle and s is the length of each side of the equilateral triangle. In this case, the radius of the circle is 6 cm, so the length of each side of the equilateral triangle is 6 cm. Therefore, the distance between the point of intersection of the tangents and the center of the circle is 6√3 cm, which is approximately equal to 10.39 cm.

3) To draw a pair of tangents to a circle which is at right angles to each other, it is required to draw tangents at endpoints of the two radii of the circle, which are inclined at an angle of:

60o

80o

90o

120o

To draw a pair of tangents to a circle that are at right angles to each other, it is necessary to draw tangents at the endpoints of the two radii of the circle. Since the radii of a circle are always perpendicular to the tangent at the point of contact, the angle between the two radii is 90 degrees. Therefore, the correct answer is 90o.

Explanation

To draw a pair of tangents to a circle that are at right angles to each other, it is necessary to draw tangents at the endpoints of the two radii of the circle. Since the radii of a circle are always perpendicular to the tangent at the point of contact, the angle between the two radii is 90 degrees. Therefore, the correct answer is 90o.

4) P₃ divides the line segment AP in the ratio:

2 : 3

4 : 1

1 : 4

3 : 2

P3 divides the line segment AP in the ratio 3:2. This means that P3 divides the line segment into two parts, where the length of the first part is 3 times the length of the second part. In other words, if the total length of the line segment AP is 'x', then the length of the first part is (3/5)x and the length of the second part is (2/5)x.

Explanation

P3 divides the line segment AP in the ratio 3:2. This means that P3 divides the line segment into two parts, where the length of the first part is 3 times the length of the second part. In other words, if the total length of the line segment AP is 'x', then the length of the first part is (3/5)x and the length of the second part is (2/5)x.

5) To divide a line segment AB in the ratio 4 : 5, draw a ray AX such that angle BAX is an acute angle, then draw a ray BY parallel to AX and the points A₁, A₂, A₃, and B_1, B₂, B₃.... are located at equal distances n ray AX and BY, respectively. Then the points joined are:

A4 and B5

A5 and B4

A5 and B6

A6 and B6

The question states that to divide the line segment AB in the ratio 4:5, we need to draw a ray AX such that angle BAX is an acute angle. Then, a ray BY parallel to AX is drawn. The points A1, A2, A3... are located at equal distances on ray AX, and the points B1, B2, B3... are located at equal distances on ray BY. According to the given information, the points A4 and B5 are joined. Therefore, the correct answer is A4 and B5.

Explanation

The question states that to divide the line segment AB in the ratio 4:5, we need to draw a ray AX such that angle BAX is an acute angle. Then, a ray BY parallel to AX is drawn. The points A1, A2, A3... are located at equal distances on ray AX, and the points B1, B2, B3... are located at equal distances on ray BY. According to the given information, the points A4 and B5 are joined. Therefore, the correct answer is A4 and B5.

6) If two tangents are drawn at the end points of two radii of a circle which are inclined at 130^o to each other, then the pair of tangents will be inclined to each other at an angle of:

40o

50o

60o

70o

When two tangents are drawn at the end points of two radii of a circle, they intersect at a point on the circumference of the circle. This forms an isosceles triangle with the radii as the equal sides. Since the radii are inclined at 130o to each other, the angle between the tangents is half of the angle between the radii. Therefore, the angle between the tangents is 130o/2 = 65o. However, the question asks for the angle between the tangents, not the angle between the radii. Since the angle between the tangents is the exterior angle of the triangle, the angle between the tangents is 180o - 65o = 115o. However, this is the exterior angle, so the angle between the tangents is 180o - 115o = 65o. Therefore, the correct answer is 50o.

Explanation

When two tangents are drawn at the end points of two radii of a circle, they intersect at a point on the circumference of the circle. This forms an isosceles triangle with the radii as the equal sides. Since the radii are inclined at 130o to each other, the angle between the tangents is half of the angle between the radii. Therefore, the angle between the tangents is 130o/2 = 65o. However, the question asks for the angle between the tangents, not the angle between the radii. Since the angle between the tangents is the exterior angle of the triangle, the angle between the tangents is 180o - 65o = 115o. However, this is the exterior angle, so the angle between the tangents is 180o - 115o = 65o. Therefore, the correct answer is 50o.

7) The scale factor means:

The ratio of angles of the triangle to be constructed with the corresponding angle of the given triangle

The ratio of sides of the triangle to be constructed with the corresponding sides of the given triangle

Both (a) & (b)

Neither (a) & (b)

The correct answer is "Both (a) & (b)". The scale factor refers to the ratio of both the angles and the sides of the triangle to be constructed with the corresponding angles and sides of the given triangle. In other words, it encompasses both the ratio of angles and the ratio of sides.

Explanation

The correct answer is "Both (a) & (b)". The scale factor refers to the ratio of both the angles and the sides of the triangle to be constructed with the corresponding angles and sides of the given triangle. In other words, it encompasses both the ratio of angles and the ratio of sides.

8) To divide a line segment AB in the ratio 3 : 7, first a ray AX is drawn so that angle BAX is an acute angle and then at equal distances points are marked on the ray AX such that the minimum number of these points is

3

7

10

12

To divide a line segment AB in the ratio 3:7, a ray AX is drawn from point A such that angle BAX is an acute angle. Then, points are marked on the ray AX at equal distances. The minimum number of these points required is 10. This is because the ratio 3:7 can be simplified to 3/10:7/10, indicating that the line segment AB needs to be divided into 10 equal parts. Therefore, 10 points are marked on the ray AX to divide the line segment AB in the desired ratio.

Explanation

To divide a line segment AB in the ratio 3:7, a ray AX is drawn from point A such that angle BAX is an acute angle. Then, points are marked on the ray AX at equal distances. The minimum number of these points required is 10. This is because the ratio 3:7 can be simplified to 3/10:7/10, indicating that the line segment AB needs to be divided into 10 equal parts. Therefore, 10 points are marked on the ray AX to divide the line segment AB in the desired ratio.

9) To construct a triangle similar to a given ΔABC with its sides 2/5 of the corresponding sides of ΔABC, first draw a ray BX such that angle CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then locate points B₁, B₂, B₃,..... on BX at equal distances and next step is to join:

B2 to C

B3 to C

B4 to C

B5 to C

In order to construct a triangle similar to triangle ABC with its sides 2/5 of the corresponding sides of triangle ABC, we start by drawing a ray BX such that angle CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then, we locate points B1, B2, B3, ... on BX at equal distances. The next step is to join B5 to C. This step completes the construction of the triangle similar to triangle ABC with its sides 2/5 of the corresponding sides of triangle ABC.

Explanation

In order to construct a triangle similar to triangle ABC with its sides 2/5 of the corresponding sides of triangle ABC, we start by drawing a ray BX such that angle CBX is an acute angle and X lies on the opposite side of A with respect to BC. Then, we locate points B1, B2, B3, ... on BX at equal distances. The next step is to join B5 to C. This step completes the construction of the triangle similar to triangle ABC with its sides 2/5 of the corresponding sides of triangle ABC.

10) Two circles touch each other externally at Q and PR is a common tangent to the circles. Then, ∠PQR is:

0o

90o

180o

360o

When two circles touch each other externally, the point of contact is the only point where the circles intersect. In this case, the common tangent PR is perpendicular to the line joining the centers of the circles, since the tangent is always perpendicular to the radius at the point of contact. Therefore, the angle PQR formed at the point of contact Q is a right angle, measuring 90 degrees.

Explanation

When two circles touch each other externally, the point of contact is the only point where the circles intersect. In this case, the common tangent PR is perpendicular to the line joining the centers of the circles, since the tangent is always perpendicular to the radius at the point of contact. Therefore, the angle PQR formed at the point of contact Q is a right angle, measuring 90 degrees.

11) What ratio does the line segment A₃B₂ represent in the following figure:

3 : 2

2 : 3

Both (a) & (b)

Neither (a) & (b)

The line segment A3B2 represents the ratio 3:2. This means that the length of A3B2 is three times the length of A1A2 and two times the length of A2B1.

Explanation

The line segment A3B2 represents the ratio 3:2. This means that the length of A3B2 is three times the length of A1A2 and two times the length of A2B1.

12) To divide a line segment AB in the ratio 4 : 5, a ray AX is drawn first such that angle BAX is an acute angle and then points A₁, A₂, A₃,…. are located at equal distances on the ray AX and the point B is joined to:

A4

A5

A9

A10

In order to divide a line segment AB in the ratio 4:5, a ray AX is drawn first such that angle BAX is an acute angle. Then, points A1, A2, A3,... are located at equal distances on the ray AX. Since the ratio is 4:5, it means that the line segment AB is divided into 4 parts by the points A1, A2, A3,..., and B is joined to the 9th point, which is A9. Therefore, the correct answer is A9.

Explanation

In order to divide a line segment AB in the ratio 4:5, a ray AX is drawn first such that angle BAX is an acute angle. Then, points A1, A2, A3,... are located at equal distances on the ray AX. Since the ratio is 4:5, it means that the line segment AB is divided into 4 parts by the points A1, A2, A3,..., and B is joined to the 9th point, which is A9. Therefore, the correct answer is A9.

13) To draw a pair of tangents to circle which are inclined to each other at an angle of 30°, it is required to draw tangents at end points of those two radii of the circle, the angle between them should be :

60o

90o

120o

150o

To draw a pair of tangents to a circle that are inclined to each other at an angle of 30°, we need to draw tangents at the end points of two radii of the circle. The angle between these two tangents can be found by subtracting the given angle of 30° from 180° (since the sum of angles in a straight line is 180°). Therefore, the angle between the two tangents should be 180° - 30° = 150°.

Explanation

To draw a pair of tangents to a circle that are inclined to each other at an angle of 30°, we need to draw tangents at the end points of two radii of the circle. The angle between these two tangents can be found by subtracting the given angle of 30° from 180° (since the sum of angles in a straight line is 180°). Therefore, the angle between the two tangents should be 180° - 30° = 150°.

14) To construct a triangle similar to a given ΔABC with its sides 5/3 of the corresponding sides of ΔABC draw a ray BX such that CBX is an acute angle and X is on the opposite side of A with respect to BC. The minimum number of points to be located at equal distances on ray BX is

2

3

5

8

To construct a triangle similar to a given ΔABC with its sides 5/3 of the corresponding sides of ΔABC, we need to draw a ray BX such that CBX is an acute angle and X is on the opposite side of A with respect to BC. In order to locate points at equal distances on ray BX, we need at least three points. This is because with two points, we can only determine a line segment, but with three points, we can determine a ray. Therefore, the minimum number of points to be located at equal distances on ray BX is 3.

Explanation

To construct a triangle similar to a given ΔABC with its sides 5/3 of the corresponding sides of ΔABC, we need to draw a ray BX such that CBX is an acute angle and X is on the opposite side of A with respect to BC. In order to locate points at equal distances on ray BX, we need at least three points. This is because with two points, we can only determine a line segment, but with three points, we can determine a ray. Therefore, the minimum number of points to be located at equal distances on ray BX is 3.

15) The exterior angle of a triangle is equal to___:

Sum of all the angles in a triangle

Sum of vertically opposite angles

Sum of opposite interior angles in a triangle

None of these

The exterior angle of a triangle is equal to the sum of the two opposite interior angles. Vertically opposite angles are formed when two lines intersect, and they are equal in measure. Therefore, the sum of the vertically opposite angles is equal to the exterior angle of a triangle.

Explanation

The exterior angle of a triangle is equal to the sum of the two opposite interior angles. Vertically opposite angles are formed when two lines intersect, and they are equal in measure. Therefore, the sum of the vertically opposite angles is equal to the exterior angle of a triangle.