Quadrilaterals — Telangana (SSC) Class 9 Mathematics Solutions (Free)

Free step-by-step Telangana (SSC) Class 9 Mathematics solutions for "Quadrilaterals" — important questions with detailed answers, download PDF for board exam preparation.

Quadrilaterals — Telangana (SSC) Class 9 Mathematics Solutions (Free)

Free step-by-step Telangana (SSC) Class 9 Mathematics solutions for "Quadrilaterals" — important questions with detailed answers, download PDF for board exam preparation.

✓ 100% Free ✓ No Login Needed ✓ NCERT / CBSE Aligned ✓ Download as PDF

TL;DR: Free step-by-step Telangana (SSC) Class 9 Mathematics solutions for "Quadrilaterals" — important questions with detailed answers, download PDF for boa…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated

🤖 Stuck on any question? Ask Syllab's free AI Tutor for a step-by-step explanation — instant, unlimited, no login.

Quadrilaterals — Textbook Questions Answered (8)

Q1: State the midpoint theorem for quadrilaterals: In quadrilateral PQRS, if L, M, N, O are the midpoints of sides PQ, QR, RS, SP respectively, prove that LMNO is a parallelogram.

Step 1: Midpoint Theorem for Quadrilaterals: The figure formed by joining the midpoints of the sides of any quadrilateral is a parallelogram. Step 2: Let PQRS be a quadrilateral with: L = midpoint of PQ M = midpoint of QR N = midpoint of RS O = midpoint of SP Step 3: In triangle PQR, L and M are midpoints of PQ and QR respectively. By midpoint theorem in triangle: LM ∥ PR and LM = PR/2 Step 4: In triangle RPS, N and O are midpoints of RS and SP respectively. By midpoint theorem in triangle: NO ∥ PR and NO = PR/2 Step 5: From steps 3 and 4: LM ∥ NO and LM = NO Step 6: Similarly, in triangle PQS, L and O are midpoints of PQ and PS respectively. By midpoint theorem: LO ∥ QS and LO = QS/2 Step 7: In triangle QRS, M and N are midpoints of QR and RS respectively. By midpoint theorem: MN ∥ QS and MN = QS/2 Step 8: From steps 6 and 7: LO ∥ MN and LO = MN Step 9: Since LM ∥ NO, LM = NO and LO ∥ MN, LO = MN, the quadrilateral LMNO has both pairs of opposite sides equal and parallel. Answer: LMNO is a parallelogram.

Q2: Define and differentiate between a parallelogram and a trapezium. If ABCD is a parallelogram with AB = 8 cm and perpendicular distance between AB and CD is 5 cm, find the area.

Step 1: Parallelogram: A quadrilateral with both pairs of opposite sides parallel and equal. Properties: Opposite sides equal, opposite angles equal, diagonals bisect each other. Step 2: Trapezium: A quadrilateral with only ONE pair of parallel sides. Properties: One pair of parallel sides, other sides (legs) are not necessarily equal. Step 3: Key Difference: Parallelogram has TWO pairs of parallel sides; trapezium has only ONE pair. Step 4: For the given parallelogram ABCD: AB = 8 cm (base) Perpendicular distance between AB and CD = 5 cm (height) Step 5: Area of parallelogram = base × height Area = AB × h Area = 8 × 5 Area = 40 cm² Step 6: Note: Since ABCD is a parallelogram, CD = AB = 8 cm (opposite sides equal) and the perpendicular distance is the same throughout. Answer: Area of parallelogram ABCD = 40 cm².

Q3: In rectangle PQRS, the diagonals PR and QS intersect at O. If PQ = 6 cm and QR = 8 cm, find the length of diagonal PR and verify that O divides both diagonals into equal parts.

Step 1: In rectangle PQRS: PQ = 6 cm (length) QR = 8 cm (breadth) All angles are 90° Step 2: Triangle PQR is a right triangle with right angle at Q. Step 3: By Pythagoras theorem: PR² = PQ² + QR² PR² = 6² + 8² PR² = 36 + 64 PR² = 100 PR = 10 cm Step 4: In a rectangle, both diagonals are equal and bisect each other. Therefore, QS = PR = 10 cm Step 5: Since diagonals bisect each other at O: PO = PR/2 = 10/2 = 5 cm OR = PR/2 = 10/2 = 5 cm So PO = OR ✓ Step 6: Similarly: QO = QS/2 = 10/2 = 5 cm OS = QS/2 = 10/2 = 5 cm So QO = OS ✓ Step 7: Verification: O divides PR into PO = 5 cm and OR = 5 cm (equal parts). O divides QS into QO = 5 cm and OS = 5 cm (equal parts). Answer: Length of diagonal PR = 10 cm; O divides both diagonals into equal parts of 5 cm each.

Q4: In rhombus ABCD, the diagonals AC and BD intersect at O. If AC = 12 cm and BD = 16 cm, find the side length of the rhombus.

Step 1: Properties of rhombus: - All sides are equal - Diagonals bisect each other at right angles (90°) - Diagonals divide the rhombus into four congruent right triangles Step 2: Given: AC = 12 cm (first diagonal) BD = 16 cm (second diagonal) Step 3: Since diagonals bisect each other: AO = AC/2 = 12/2 = 6 cm BO = BD/2 = 16/2 = 8 cm Step 4: In right triangle AOB (with right angle at O): AB² = AO² + BO² AB² = 6² + 8² AB² = 36 + 64 AB² = 100 AB = 10 cm Step 5: Since all sides of a rhombus are equal: AB = BC = CD = DA = 10 cm Step 6: Verification using triangle BOC: BC² = BO² + OC² BC² = 8² + 6² BC² = 64 + 36 BC² = 100 BC = 10 cm ✓ Answer: Side length of rhombus ABCD = 10 cm.

Q5: In trapezium ABCD, AB ∥ CD, AB = 10 cm, CD = 6 cm, and the perpendicular distance between the parallel sides is 4 cm. Calculate the area of the trapezium.

Step 1: Trapezium ABCD has one pair of parallel sides: AB = 10 cm (one parallel side) CD = 6 cm (other parallel side) Perpendicular distance = 4 cm (height) Step 2: Formula for area of trapezium: Area = (1/2) × (sum of parallel sides) × height Area = (1/2) × (a + b) × h Where a and b are the parallel sides, and h is the perpendicular distance. Step 3: Substitute values: Area = (1/2) × (AB + CD) × height Area = (1/2) × (10 + 6) × 4 Area = (1/2) × 16 × 4 Area = (1/2) × 64 Area = 32 cm² Step 4: Verification: If we imagine the trapezium as two rectangles or use the average of parallel sides: Average of parallel sides = (10 + 6)/2 = 8 cm Area = 8 × 4 = 32 cm² ✓ Answer: Area of trapezium ABCD = 32 cm².

Q6: In square PQRS with side length 5 cm, find the length of the diagonal and verify using the Pythagoras theorem.

Step 1: Square PQRS with side length = 5 cm All sides are equal: PQ = QR = RS = SP = 5 cm All angles are 90° Step 2: Consider triangle PQR formed by two adjacent sides and a diagonal: angle Q = 90° PQ = 5 cm QR = 5 cm Step 3: By Pythagoras theorem: PR² = PQ² + QR² PR² = 5² + 5² PR² = 25 + 25 PR² = 50 PR = √50 PR = √(25 × 2) PR = 5√2 cm Step 4: Alternative formula for diagonal of square: For a square with side a, diagonal = a√2 Diagonal = 5√2 cm ✓ Step 5: Numerical value: 5√2 ≈ 5 × 1.414 5√2 ≈ 7.07 cm Step 6: Verification: In square, both diagonals are equal. Second diagonal QS = 5√2 cm ✓ Answer: Length of diagonal = 5√2 cm ≈ 7.07 cm.

Q7: Prove that the sum of the interior angles of a quadrilateral is 360°.

Step 1: Consider quadrilateral ABCD. Step 2: Draw diagonal AC, which divides the quadrilateral into two triangles: triangle ABC and triangle ACD. Step 3: In triangle ABC: angle BAC + angle ABC + angle BCA = 180° ... (by angle sum property of triangles) Step 4: In triangle ACD: angle CAD + angle ACD + angle ADC = 180° ... (by angle sum property of triangles) Step 5: Add the equations from steps 3 and 4: (angle BAC + angle ABC + angle BCA) + (angle CAD + angle ACD + angle ADC) = 180° + 180° Step 6: Rearrange the terms: (angle BAC + angle CAD) + angle ABC + (angle BCA + angle ACD) + angle ADC = 360° Step 7: Observe that: angle BAC + angle CAD = angle BAD = angle A of the quadrilateral angle BCA + angle ACD = angle BCD = angle C of the quadrilateral angle ABC = angle B of the quadrilateral angle ADC = angle D of the quadrilateral Step 8: Therefore: angle A + angle B + angle C + angle D = 360° Answer: The sum of all interior angles of a quadrilateral = 360°.

Q8: In parallelogram ABCD, angle A = 65°. Find the measures of angles B, C, and D, and verify that the sum of all angles is 360°.

Step 1: Properties of parallelogram: - Opposite angles are equal - Adjacent angles are supplementary (sum to 180°) Step 2: Given: angle A = 65° Step 3: Find angle C (opposite to angle A): angle C = angle A = 65° (opposite angles in parallelogram are equal) Step 4: Find angle B (adjacent to angle A): angle A + angle B = 180° (adjacent angles are supplementary) 65° + angle B = 180° angle B = 180° - 65° angle B = 115° Step 5: Find angle D (opposite to angle B): angle D = angle B = 115° (opposite angles in parallelogram are equal) Step 6: Alternatively, angle D is also adjacent to angle A: angle A + angle D = 180° 65° + angle D = 180° angle D = 115° ✓ Step 7: Verify the sum of all angles: angle A + angle B + angle C + angle D = 65° + 115° + 65° + 115° = 130° + 230° = 360° ✓ Step 8: Summary: angle A = 65°, angle B = 115°, angle C = 65°, angle D = 115° Answer: angle B = 115°, angle C = 65°, angle D = 115°; Sum = 360°.

Related

More free resources for this chapter