Quadrilaterals — Karnataka (SSLC) Class 9 Mathematics Solutions (Free)

Given: quadrilateral PQRS with L, M, N, O as the midpoints of PQ, QR, RS, SP respectively. To prove: LMNO is a parallelogram. Proof: Draw diagonal PR. In triangle PQR: L and M are the midpoints of PQ and QR. By the Midpoint Theorem for triangles, the segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half its length. So LM ∥ PR and LM = ½PR. In triangle PRS: O and N are the midpoints of SP and RS. By the same theorem, ON ∥ PR and ON = ½PR. Since both LM and ON are parallel to PR and equal to half of PR, LM ∥ ON and LM = ON. A quadrilateral with one pair of opposite sides both parallel AND equal is a parallelogram. Hence, LMNO is a parallelogram.

A parallelogram is a quadrilateral in which BOTH pairs of opposite sides are parallel (and consequently equal in length). A trapezium is a quadrilateral in which only ONE pair of opposite sides is parallel; the other pair need not be parallel or equal. Key difference: a parallelogram requires two pairs of parallel sides, while a trapezium requires only one — every parallelogram technically satisfies the "at least one pair parallel" condition too, but a trapezium is specifically the case where the second pair is NOT parallel. Area calculation: For a parallelogram, Area = base × height (the perpendicular distance between the parallel base and its opposite side). Given: base AB = 10 cm, height = 6 cm. Area = 10 × 6 = 60 cm²

In a rectangle, each angle is 90°, so any diagonal, together with two adjacent sides, forms a right-angled triangle. Given: PQ = 9 cm, QR = 12 cm. Using the Pythagorean theorem in triangle PQR (right-angled at Q): PR² = PQ² + QR² PR² = 9² + 12² = 81 + 144 = 225 PR = √225 = 15 cm Verifying O bisects both diagonals: in a rectangle, the two diagonals are always equal in length AND bisect each other — this follows from the fact that a rectangle is a parallelogram (whose diagonals always bisect each other) with the additional property that its diagonals are equal. Since PR = QS = 15 cm and both pass through the same intersection point O splitting each into two equal 7.5 cm segments, this property is confirmed.

A key property of a rhombus is that its diagonals bisect each other AT RIGHT ANGLES. Given: AC = 10 cm, BD = 24 cm. Since the diagonals bisect each other: AO = AC/2 = 10/2 = 5 cm BO = BD/2 = 24/2 = 12 cm Since the diagonals intersect at 90°, triangle AOB is a right triangle with legs AO and BO, and the rhombus's side AB as the hypotenuse. Using the Pythagorean theorem: AB² = AO² + BO² = 5² + 12² = 25 + 144 = 169 AB = √169 = 13 cm Since all four sides of a rhombus are equal, the side length is 13 cm.

For a trapezium, the area is found by averaging the lengths of the two parallel sides and multiplying by the perpendicular distance (height) between them: Area = ½ × (sum of parallel sides) × height Given: AB = 14 cm, CD = 8 cm, height = 5 cm. Area = ½ × (14 + 8) × 5 Area = ½ × 22 × 5 Area = 55 cm²

A diagonal of a square splits it into two right-angled triangles, with the diagonal as the hypotenuse and two adjacent sides of the square as the legs. Given: side length = 7 cm. Using the Pythagorean theorem: diagonal² = side² + side² = 7² + 7² = 49 + 49 = 98 diagonal = √98 = 7√2 ≈ 9.9 cm This confirms the well-known general result: the diagonal of any square is always equal to (side length) × √2.

Given: a quadrilateral ABCD. To prove: ∠A + ∠B + ∠C + ∠D = 360° Proof: Step 1: Draw diagonal AC, dividing the quadrilateral into two triangles: △ABC and △ACD. Step 2: By the angle sum property, the angles of △ABC total 180°: ∠BAC + ∠B + ∠BCA = 180°. Step 3: Similarly, the angles of △ACD total 180°: ∠DAC + ∠D + ∠DCA = 180°. Step 4: Adding these two equations: (∠BAC + ∠DAC) + ∠B + (∠BCA + ∠DCA) + ∠D = 360° Step 5: Since ∠BAC + ∠DAC together make up the full angle ∠A, and ∠BCA + ∠DCA together make up the full angle ∠C: ∠A + ∠B + ∠C + ∠D = 360° Hence proved.

Two key properties of a parallelogram's angles: opposite angles are EQUAL, and adjacent angles are SUPPLEMENTARY (add up to 180°) — because adjacent angles lie along a pair of parallel sides cut by a transversal. Given: ∠A = 72°. Opposite angle: ∠C = ∠A = 72° (opposite angles are equal) Adjacent angles: ∠B = 180° − ∠A = 180° − 72° = 108° (co-interior angles on parallel sides are supplementary) ∠D = ∠B = 108° (∠D is opposite ∠B, so it equals ∠B by the same equal-opposite-angles rule) Verification: ∠A + ∠B + ∠C + ∠D = 72° + 108° + 72° + 108° = 360° ✓