Quadrilaterals — Chapter Test | Class 9 IMO

Q1

Find the length of the line segment joining the mid-points of the non-parallel sides of a trapezium whose parallel sides are 12 cm and 18 cm long.

The length of the line segment joining the mid-points of the non-parallel sides of a trapezium is equal to half the sum of the parallel sides. Length = (12 + 18) / 2 = 30 / 2 = 15 cm.

Q2

In a square ABCD, the diagonal AC and BD intersect at O. The triangle AOB is always a/an:

In a square, diagonals are equal, bisect each other at 90°, so OA = OB and angle AOB = 90°. Thus, triangle AOB is an isosceles right-angled triangle.

Q3

ABCD is a parallelogram. The bisectors of angle A and angle B intersect at a point P. Find the measure of angle APB.

In a parallelogram, adjacent angles are supplementary, so angle A + angle B = 180°. The bisectors mean we take half of each angle: angle PAB + angle PBA = (angle A)/2 + (angle B)/2 = (angle A + angle B)/2 = 180°/2 = 90°. In triangle APB, angle APB = 180° − (angle PAB + angle PBA) = 180° − 90° = 90°.

Q4

Let ABCD be a square. An equilateral triangle EBC is constructed on side BC externally. Find the measure of angle AED.

Since EBC is an equilateral triangle, EB = BC = EC and angle EBC = angle ECB = 60°. Since ABCD is a square, AB = BC = CD = DA and angle ABC = angle BCD = 90°. Now, angle ABE = angle ABC + angle EBC = 90° + 60° = 150°. Since AB = EB, triangle ABE is isosceles with angle BAE = angle BEA = (180° − 150°)/2 = 15°. By symmetry, angle CDE = 150° and angle CDE = angle CED = 15°. Now, angle AED = angle BEC − (angle BEA + angle CED) = 60° − (15° + 15°) = 30°.

Q5

The diagonals of a rhombus are 16 cm and 12 cm long. Find the length of each side of the rhombus.

The diagonals of a rhombus bisect each other at right angles (90°). Half lengths of the diagonals are 8 cm and 6 cm. Using Pythagoras theorem for the right-angled triangle formed by the half-diagonals: side² = 8² + 6² = 64 + 36 = 100. Side = √100 = 10 cm.

Q6

An angle tracking device notes that in a trapezium ABCD with AB parallel to CD, angle A = 50° and angle B = 70°. What are the values of angle D and angle C respectively?

Since AB || CD, adjacent angles on the same transverse side are supplementary. Angle A + angle D = 180° → 50° + angle D = 180° → angle D = 130°. Angle B + angle C = 180° → 70° + angle C = 180° → angle C = 110°.

Q7

Complete the logic sequence: In triangle ABC, D and E are mid-points of AB and AC respectively. If DE = 4.5 cm, then the parallel side BC must logically be equal to:

By the Mid-point theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of it. So, DE = BC/2 → BC = 2 × DE = 2 × 4.5 = 9 cm.

Q8

If the mid-points of the sides of any arbitrary quadrilateral are joined in order, the resulting inner figure logically always forms a:

By using the mid-point theorem on the diagonals of the quadrilateral, the sides of the inner quadrilateral are parallel to and half of the diagonals, meaning opposite sides are equal and parallel. Thus, it is always a parallelogram.

Q9

ABCD is a trapezium with AB || CD. If a line segment EF is drawn joining the mid-points of AD and BC respectively, and AB = 10 cm, CD = 16 cm, find the length of EF.

The line joining the mid-points of non-parallel sides of a trapezium is parallel to the bases and its length is the average of the two bases: EF = (AB + CD)/2 = (10 + 16)/2 = 26/2 = 13 cm.

Q10

Harish is measuring a piece of land that is a parallelogram ABCD. He knows that the fence lengths are AB = 40 m and BC = 30 m. If he walks from corner A directly to corner C along a straight path of length 50 m, what is the angle ABC between the two fences?

In triangle ABC, the sides are 40 m, 30 m, and 50 m. Since 30² + 40² = 900 + 1600 = 2500 = 50², triangle ABC is a right-angled triangle with the right angle opposite to the side of 50 m (AC). Thus, angle ABC = 90°.

Q11

ABCD is a rectangle whose diagonals AC and BD intersect at O. If angle OAB = 28°, find angle OBC.

In a rectangle, each corner angle is 90°. Therefore, angle ABC = 90°. Also, diagonals are equal and bisect each other, so OA = OB, which means angle OBA = angle OAB = 28°. Therefore, angle OBC = angle ABC − angle OBA = 90° − 28° = 62°.

Q12

In a parallelogram ABCD, diagonal AC and BD intersect at O. If AO = 3 cm and BD = 8 cm, find the length of OC and OD respectively.

Diagonals of a parallelogram bisect each other. So, OC = AO = 3 cm. OD = BD / 2 = 8 / 2 = 4 cm.

Q13

An agricultural land patch is shaped like a rhombus with diagonals of lengths 30 m and 40 m. A fence is to be put around its boundaries. What is the total length of the fence required?

Half diagonals are 15 m and 20 m. Side of rhombus = √(15² + 20²) = √(225 + 400) = √625 = 25 m. Total fence length (perimeter) = 4 × 25 = 100 m.

Q14

ABCD is a rhombus with side length 5 cm. If diagonal AC = 6 cm, find the area of the rhombus.

The diagonals of a rhombus bisect each other at right angles. Half of AC is 3 cm. Let half of BD be x. Using Pythagoras theorem: 3² + x² = 5² → 9 + x² = 25 → x² = 16 → x = 4 cm. So, the full diagonal BD = 2 × 4 = 8 cm. Area of rhombus = ½ × d1 × d2 = ½ × 6 × 8 = 24 cm².

Q15

In a logical diagram, if a rectangle has its diagonals intersecting at point O such that angle BOC = 60°, what is the value of angle OAD?

In a rectangle, diagonals are equal and bisect each other, so OA = OB = OC = OD. Angle AOD = angle BOC = 60° (vertically opposite). In triangle AOD, OA = OD, which implies angle OAD = angle ODA. Since angle AOD = 60°, angle OAD + angle ODA = 180° − 60° = 120° → angle OAD = 60°.