Areas of Parallelograms and Triangles — Karnataka (SSLC) Class 9 Mathematics Solutions (Free)
Free step-by-step Karnataka (SSLC) Class 9 Mathematics solutions for "Areas of Parallelograms and Triangles" — important questions with detailed answers, download PDF for board exam preparation.
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Areas of Parallelograms and Triangles — Textbook Questions Answered (8)
Q1: Two parallelograms ABCD and PQRS have the same base AB = PQ = 8 cm and are on the same side of the base. If the perpendicular distance from C to AB is 5 cm and from S to PQ is 6 cm, find the ratio of their areas.
Step 1: Area of parallelogram = base × height (perpendicular distance from opposite side).
Step 2: Area of parallelogram ABCD = AB × perpendicular distance from C to AB = 8 × 5 = 40 cm².
Step 3: Area of parallelogram PQRS = PQ × perpendicular distance from S to PQ = 8 × 6 = 48 cm².
Step 4: Ratio of areas = Area of ABCD : Area of PQRS = 40 : 48 = 5 : 6.
Final Answer: The ratio of areas of parallelograms ABCD and PQRS is 5:6.
Q2: Triangle ABC and triangle PQR have the same base BC = QR = 10 cm. The perpendicular distance from A to BC is 8 cm, and from P to QR is 6 cm. What is the ratio of their areas?
Step 1: Area of triangle = (1/2) × base × height.
Step 2: Area of triangle ABC = (1/2) × BC × perpendicular distance from A to BC = (1/2) × 10 × 8 = 40 cm².
Step 3: Area of triangle PQR = (1/2) × QR × perpendicular distance from P to QR = (1/2) × 10 × 6 = 30 cm².
Step 4: Ratio of areas = Area of ABC : Area of PQR = 40 : 30 = 4 : 3.
Final Answer: The ratio of areas of triangles ABC and PQR is 4:3.
Q3: A parallelogram and a rectangle have equal bases 12 cm and equal heights 5 cm. What is the ratio of their areas?
Step 1: Area of parallelogram = base × height = 12 × 5 = 60 cm².
Step 2: Area of rectangle = length × breadth. Since height = 5 cm and base = 12 cm, Area = 12 × 5 = 60 cm².
Step 3: Ratio of areas = 60 : 60 = 1 : 1.
Final Answer: A parallelogram and rectangle with equal bases and heights have equal areas. Ratio = 1:1.
Q4: Triangle ABC has base BC = 16 cm. Triangle DEF has base EF = 20 cm. Both triangles have the same area of 80 cm². Find the heights (perpendicular distances) of both triangles.
Step 1: For triangle ABC, Area = (1/2) × base × height.
80 = (1/2) × 16 × h₁
80 = 8 × h₁
h₁ = 10 cm.
Step 2: For triangle DEF, Area = (1/2) × base × height.
80 = (1/2) × 20 × h₂
80 = 10 × h₂
h₂ = 8 cm.
Step 3: Height of triangle ABC = 10 cm, Height of triangle DEF = 8 cm.
Final Answer: Height of triangle ABC is 10 cm and height of triangle DEF is 8 cm.
Q5: Triangles ABC and PQR are on the same base AB = PQ = 15 cm and on the same side of the base. If their vertices C and R lie on a line parallel to AB, prove that triangles ABC and PQR have equal area.
Step 1: Let the common base be AB = PQ = 15 cm.
Step 2: Vertices C and R both lie on a line parallel to AB.
Step 3: Since AB and line CR are parallel, the perpendicular distance from any point on CR to line AB is the same.
Step 4: Let the perpendicular distance from CR to AB be h cm.
Step 5: Area of triangle ABC = (1/2) × AB × h = (1/2) × 15 × h = 7.5h cm².
Step 6: Area of triangle PQR = (1/2) × PQ × h = (1/2) × 15 × h = 7.5h cm².
Step 7: Since both areas equal 7.5h cm², Area of ABC = Area of PQR.
Final Answer: Triangles ABC and PQR have equal areas (both = 7.5h cm²) because they have the same base and their vertices lie on parallel lines.
Q6: In triangle ABC, if a line parallel to side BC divides the triangle into two parts of equal area, in what ratio does this line divide the sides AB and AC?
Step 1: Let the line parallel to BC intersect AB at D and AC at E.
Step 2: Triangle ADE is similar to triangle ABC (by AA similarity, since angle A is common and corresponding angles are equal).
Step 3: Let AD/AB = AE/AC = k (ratio of corresponding sides).
Step 4: Ratio of areas = (ratio of corresponding sides)² = k².
Step 5: Given that triangle ADE has area equal to area of trapezium DBCE.
Area of ADE + Area of trapezium = Area of ABC.
Area of ADE = (1/2) × Area of ABC.
Step 6: Area of ADE / Area of ABC = 1/2.
k² = 1/2
k = 1/√2 = √2/2.
Step 7: AD/AB = √2/2, so AD : DB = √2 : (2 - √2) or simply AD : DB = 1 : (√2 - 1).
Alternatively, AD/DB = 1 : (√2 - 1) ≈ 1 : 0.414.
Step 8: Line DE divides AB and AC in ratio √2 : (2 - √2) from A, or DB : AD = (2 - √2) : √2.
Final Answer: The line parallel to BC divides AB and AC in the ratio AD : DB = 1 : (√2 - 1), making ADE have half the area of ABC.
Q7: A parallelogram ABCD has area 120 cm². If the perpendicular distance between parallel sides AB and DC is 8 cm, what is the length of side AB?
Step 1: Area of parallelogram = base × height.
120 = AB × 8.
AB = 120 / 8 = 15 cm.
Step 2: The perpendicular distance between AB and DC is the height of the parallelogram.
Final Answer: Length of side AB is 15 cm.
Q8: Triangles ABC and ABD share the same base AB = 10 cm. C and D lie on the same side of AB on a line parallel to AB at distance 6 cm from AB. Find the ratio of their areas.
Step 1: Both triangles have the same base AB = 10 cm.
Step 2: Both C and D lie on a line parallel to AB, so the perpendicular distance from both C and D to line AB is the same = 6 cm.
Step 3: Area of triangle ABC = (1/2) × AB × perpendicular distance from C = (1/2) × 10 × 6 = 30 cm².
Step 4: Area of triangle ABD = (1/2) × AB × perpendicular distance from D = (1/2) × 10 × 6 = 30 cm².
Step 5: Ratio of areas = 30 : 30 = 1 : 1.
Final Answer: Triangles ABC and ABD have equal areas. Ratio = 1:1.
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