IMO Practice Test — Mid-point Theorem
6 Questions • 15 min • Olympiad level
15:00
Question 1 of 6
hard
In a quadrilateral ABCD, the mid-points of all four sides are connected in sequence to form an inner four-sided shape PQRS. This new inner shape PQRS is guaranteed to be a:
Rectangle
Rhombus
Parallelogram
Square
Explanation: The diagonals of ABCD act as base lines for pairs of triangles. By the Mid-point Theorem, opposite sides of PQRS are parallel to the diagonals, making it a parallelogram.
Question 2 of 6
hard
In a triangle ABC, AD is the median drawn to the base side BC. A line segment EF is drawn connecting the mid-points of sides AB and AC. If EF intersects the median line AD at a point G, determine the ratio of segment AG to GD.
1:2
1:1
2:1
3:1
Explanation: EF is parallel to BC, so inside triangle ABD, EG is parallel to BD and starts at mid-point E, making G the mid-point of median line AD.
Question 3 of 6
hard
In a right-angled triangle ABC, where the right angle is located at vertex B, the points P, Q, and R represent the mid-points of sides AB, BC, and CA respectively. If AB = 6 cm and BC = 8 cm, find the absolute length of segment PR.
5 cm
4 cm
3 cm
10 cm
Explanation: By the Mid-point Theorem, PR = 1/2 × BC = 1/2 × 8 = 4 cm. Note that the measurement of side AB is extra information not needed to solve the problem.
Question 4 of 6
hard
In a trapezium ABCD, side AB runs parallel to side DC. Points E and F are chosen as the exact mid-points of the diagonals AC and BD respectively. If AB = 10 cm and DC = 6 cm, find the length of segment EF.
4 cm
8 cm
2 cm
3 cm
Explanation: The length of the segment connecting the mid-points of the diagonals of a trapezium is equal to half the difference of the parallel sides: 1/2 × (10 - 6) = 2 cm.
Question 5 of 6
hard
Three horizontal lines run parallel to one another. A transversal line crosses them at a 30-degree tilt, creating equal intercepts of length 6 cm. A second transversal crosses them at a right angle (90 degrees). Find the length of the intercepts formed on this second transversal.
6 cm
3 cm
$3\sqrt{3}$ cm
4 cm
Explanation: The perpendicular distance between the lines is found using a 30-60-90 right triangle: Height = 6 × sin(30°) = 6 × 0.5 = 3 cm.
Question 6 of 6
hard
Inside a large triangle ABC, a line segment DE is drawn joining the mid-points of AB and AC. A point P is chosen anywhere along the base line BC. If the area of the large triangle ABC is 32 square centimeters, find the area of the smaller triangle PDE.
16 cm²
8 cm²
4 cm²
12 cm²
Explanation: Triangle PDE shares the same base line DE (which is half of BC) and its height is exactly half of the height of triangle ABC. Therefore, Area = 1/4 × 32 = 8 cm².