Class 9 Maths Ganita Manjari Chapter 12 MCQ – Quadrilaterals

A quadrilateral has four sides joined to form a closed boundary in a plane. Its four vertices are distinct, no three are collinear, and its sides do not cross. Equal side lengths are not required.

A diagonal joins two opposite vertices of a polygon. In ABCD, A and C are opposite vertices, as are B and D. Therefore, AC and BD are the diagonals. The other listed segments are sides of the quadrilateral.

In a convex quadrilateral, every interior angle is less than 180°. Here, the angle of 220° is a reflex angle, so the quadrilateral is non-convex. The presence of this angle rules out a rectangle or a parallelogram.

The sum of the four interior angles of a quadrilateral is 360°. The three given angles add to 75° + 95° + 110° = 280°. Therefore, the remaining angle is 360° − 280° = 80°.

A parallelogram is defined by having both pairs of opposite sides parallel. It need not have all four sides equal or equal diagonals. Having just one pair of equal opposite sides does not guarantee a parallelogram.

pposite sides of a parallelogram are equal, so AB = CD. Thus, 3x + 2 = 5x − 8. Rearranging gives 2x = 10, so x = 5. Substitution gives AB = CD = 17 cm.

Adjacent angles of a parallelogram are supplementary, so their sum is 180°. Let the angles be 2x and 3x. Then 5x = 180°, giving x = 36°. The smaller angle is 2 × 36° = 72°.

The diagonals of a parallelogram bisect each other. Therefore, AO = OC = 7 cm and BO = OD = 5 cm. Adding the two halves gives AC = 14 cm and BD = 10 cm. The diagonals need not be equal.

A quadrilateral is a parallelogram if one pair of opposite sides is both equal and parallel. Both conditions must apply to the same pair. Equality alone, parallelism alone, or equal diagonals alone does not establish a parallelogram.

Both pairs of opposite angles are equal, which is a sufficient condition for a quadrilateral to be a parallelogram. Since the angle sum is 360°, ∠A + ∠B = 180°. The supplementary adjacent angles establish the required parallel sides.

For parallelogram ABCD with AC = BD, triangles ABC and BAD are congruent by SSS. Hence, the adjacent angles ∠ABC and ∠BAD are equal. They also add to 180°, so each is 90°. Therefore, the parallelogram is a rectangle.

Diagonals that bisect each other establish a parallelogram. Their perpendicularity makes adjacent sides equal, using congruent right triangles around the intersection. Thus, all four sides are equal and the figure is a rhombus. Equal diagonals are not guaranteed.

Since AC bisects ∠DAB, ∠DAC = ∠CAB. Also, AD ∥ BC gives ∠DAC = ∠ACB. Thus, ∠CAB = ∠ACB, so AB = BC. Opposite sides of a parallelogram are equal, making all four sides equal.

Two pairs of corresponding sides are given equal: AB = CD and BC = DA. The third pair is the common side AC = CA. Therefore, all three corresponding sides match, so SSS directly proves the triangles congruent.

The Midpoint Theorem states that the segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length. Hence, PQ ∥ BC and PQ = 18/2 = 9 cm.

By the converse of the Midpoint Theorem, a line through the midpoint of one side and parallel to another side bisects the third side. Therefore, Q is the midpoint of AC, giving AQ = QC = 6 cm and AC = 12 cm.

Each side of the triangle formed by the three midpoints is half the corresponding side of the original triangle. Its side lengths are therefore 5 cm, 6 cm and 7 cm. Adding them gives a perimeter of 18 cm.

Joining the three side midpoints divides the original triangle into four congruent triangles. This follows from the Midpoint Theorem and SSS congruence. All four have equal areas, so the triangle formed by the midpoints has area 64/4 = 16 cm².

A median joins a vertex of a triangle to the midpoint of the opposite side. Thus, joining A to the midpoint of BC gives a median. Perpendicularity or angle bisection alone does not make a segment a median.

The centroid G divides each median in the ratio 2 : 1, with the longer part next to the vertex. Therefore, AG = (2/3) × 15 = 10 cm and GD = (1/3) × 15 = 5 cm.

Applying the Midpoint Theorem in triangles ABC and ADC gives PQ ∥ AC and SR ∥ AC. Similarly, QR and PS are parallel to BD. Hence, both pairs of opposite sides of PQRS are parallel, making it a parallelogram.

Each pair of opposite sides of the Varignon parallelogram is half the length of one original diagonal. Its adjacent sides are therefore 8 cm and 5 cm. Its perimeter is 2 × (8 + 5) = 26 cm.

The sides of a Varignon parallelogram are half the lengths of the original quadrilateral’s diagonals. If those diagonals are equal, all four sides of the new parallelogram are equal. It is therefore a rhombus, but need not be a square.

Adjacent sides of the Varignon parallelogram are parallel to the two diagonals of the original quadrilateral. If those diagonals are perpendicular, these adjacent sides are perpendicular too. Thus, the Varignon parallelogram has right angles and is a rectangle.

The segment joining the midpoints of the non-parallel sides of a trapezium is parallel to its bases. Its length is half the sum of the base lengths. Therefore, EF = (AB + CD)/2 = (8 + 14)/2 = 11 cm.

The point O bisects diagonal AC. Since AB ∥ CD, triangles AOP and COQ have equal corresponding angles and AO = CO, so they are congruent. Hence, OP = OQ = 4 cm and PQ = 8 cm.

Tiling means covering the plane without gaps or overlaps. Identical copies of any non-self-intersecting quadrilateral can be arranged to do this. The construction works for both convex and non-convex quadrilaterals; the tile need not have equal sides or right angles.

A half-turn is a rotation through 180°. Rotating about the midpoint of AB sends A to B and B to A, so the two copies share the same side. Repeating such placements gives the quadrilateral tiling construction.

The longest side must be shorter than the sum of the other three sides. We need x < 2 + 5 + 11 and 11 < 2 + 5 + x, giving 4 < x < 18. Only 10 cm satisfies these conditions. Equality would give a degenerate figure.

From each vertex of a hexagon, diagonals can be drawn to three vertices, excluding itself and its two neighbours. This gives 6 × 3 = 18 counts. Each diagonal is counted from both ends, so there are 18/2 = 9 diagonals.