Quadrilaterals — Karnataka (SSLC) Class 9 Mathematics Solutions (Free)
Free step-by-step Karnataka (SSLC) Class 9 Mathematics solutions for "Quadrilaterals" — important questions with detailed answers, download PDF for board exam preparation.
TL;DR: Free step-by-step Karnataka (SSLC) Class 9 Mathematics solutions for "Quadrilaterals" — important questions with detailed answers, download PDF for bo…
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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: N…
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)
…
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…
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 …
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:…
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