Complete, line-by-line step-by-step solutions for DAV Class 7 Secondary Mathematics Chapter 11: Perimeter and Area (Brain Teasers, HOTS & Value Based Questions). Features full geometric proofs, circle circumference formulas, cross-path area computations, Higher Order Thinking Skills (HOTS), and Value-Based Questions (VBQs) aligned strictly with the DAV Board curriculum.
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Q.1. (A) Tick (✓) the correct option
(i) The diameter of a circle whose circumference is 22 cm is:
- (a) 3.5 cm
- (b) 7 cm
- (c) 14 cm
- (d) 12 cm
Circumference C = πd = 22 cm
⇒ (22/7) × d = 22
⇒ d = (22 × 7)/22 = 7 cm.
Correct Option: (b) 7 cm
(ii) Area of a triangle whose base is 3 cm and height is 7 cm is:
- (a) 10.5 cm2
- (b) 21 cm2
- (c) 10.5 cm
- (d) 105 cm2
Area = ½ × Base × Height
= ½ × 3 × 7 = 21/2 = 10.5 cm2.
Correct Option: (a) 10.5 cm2
(iii) The area of a field is 5,00,000 m2. Its area in hectares is:
- (a) 500 hectares
- (b) 50 hectares
- (c) 5 hectares
- (d) 5000 hectares
1 hectare = 10,000 m2
Area in hectares = 5,00,000/10,000 = 50 hectares.
Correct Option: (b) 50 hectares
(iv) Which of the following represents the circumference of a circle in terms of its diameter d?
- (a) 2πr2
- (b) πd
- (c) πr2
- (d) 2πd
Circumference = 2πr = π(2r) = πd (since diameter d = 2r).
Correct Option: (b) πd
(v) Area of a right-angled triangle whose sides containing the right angle are 12 dm and 5 dm is:
- (a) 60 dm2
- (b) 30 dm
- (c) 60 dm
- (d) 30 dm2
In a right triangle, the two perpendicular sides serve as Base and Height.
Area = ½ × 12 × 5 = 6 × 5 = 30 dm2.
Correct Option: (d) 30 dm2
Q.1. (B) Short Answer Questions
(a) What is the constant ratio of the circumference of any circle to its diameter?
Ans: π (pi), which is approximately equal to 22/7 or 3.14159.
(b) Find the area of a square park whose perimeter is 320 metres.
Area = Side2 = 80 × 80 = 6400 m2.
Ans: 6400 m2
(c) If the radius of a circle is doubled, by how many times does its area increase?
New radius R = 2r ⇒ New Area A2 = π(2r)2 = 4πr2 = 4A1.
Ans: The area increases by 4 times.
(d) Write the formula for the area of a circular ring (shaded path between two concentric circles).
Ans: Area = π(R2 − r2), where R is outer radius and r is inner radius.
Q.2. A wire is in the shape of a square of side 10 cm. If the wire is rebent into a rectangle of length 12 cm, find its breadth. Which figure encloses more area and by how much?
Side of square s = 10 cm
Perimeter = 4 × s = 4 × 10 = 40 cm.
Step 2: Find the breadth of the rectangle:
Length of wire remains constant ⇒ Perimeter of rectangle = 40 cm.
Formula: 2(l + b) = 40
⇒ 2(12 + b) = 40
⇒ 12 + b = 40/2 = 20
⇒ b = 20 − 12 = 8 cm.
Step 3: Compare enclosed areas:
• Area of square = s2 = 10 × 10 = 100 cm2.
• Area of rectangle = l × b = 12 × 8 = 96 cm2.
Conclusion:
The Square encloses more area than the rectangle.
Difference = 100 cm2 − 96 cm2 = 4 cm2.
Ans: Breadth = 8 cm | Square encloses more area by 4 cm2
Q.3. A circular grass lawn of radius 14 metres has a path of uniform width 2 metres around it. Find the area of the path. (Take π = 22/7)
• Inner radius of lawn (r) = 14 m.
• Width of path = 2 m.
• Outer radius including path (R) = 14 + 2 = 16 m.
Step 2: Calculate Area of Circular Path:
Area of path = Outer Area − Inner Area
= πR2 − πr2 = π(R2 − r2)
= 22/7 × (162 − 142)
= 22/7 × (256 − 196)
= 22/7 × 60
= 1320/7 ≈ 188.57 m2 (or 1884/7 m2).
Ans: 188.57 m2
🔥 Higher Order Thinking Skills (HOTS)
HOTS Q1: A copper wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is rebent into the form of a circle, find the area enclosed by the circle.
Area of square s2 = 121 cm2 ⇒ Side s = √121 = 11 cm.
Total length of wire (Perimeter) = 4 × 11 = 44 cm.
Step 2: Find radius of circle formed by same wire:
Circumference = 2πr = 44 cm
⇒ 2 × (22/7) × r = 44
⇒ r = (44 × 7)/44 = 7 cm.
Step 3: Area of Circle:
Area = πr2 = (22/7) × 7 × 7 = 22 × 7 = 154 cm2.
Ans: 154 cm2
HOTS Q2: In a rectangular park of dimensions 60 m × 40 m, two crossroads of width 2 m each run through the center parallel to the sides. Find the area of the crossroads.
• Area of road parallel to breadth = 40 m × 2 m = 80 m2.
• Common center intersection area (counted twice) = 2 m × 2 m = 4 m2.
Total Area of crossroads = 120 + 80 − 4 = 196 m2.
Ans: 196 m2
🌱 Value Based Questions (VBQ)
VBQ: A Resident Welfare Association (RWA) constructs a circular jogging track of outer radius 21 m and inner radius 14 m in a society park to promote senior citizen and youth fitness.
(a) Find the area of the jogging track.
Area = π(R2 − r2) = 22/7 × (212 − 142)= 22/7 × (441 − 196) = 22/7 × 245 = 22 × 35 = 770 m2.
(b) What social value does this initiative exhibit?
Core Values Depicted: Community health awareness, encouragement of active lifestyles, care for the well-being of elderly citizens, and communal harmony.Ans: (a) 770 m2 | (b) Health Consciousness & Community Care
📚 DAV Class 7 Maths Chapter 11: Perimeter and Area – Complete Solutions Index
| Worksheet / Section | Shapes & Concepts | Solution Link |
|---|---|---|
| Worksheet 1 | Perimeter & Area of Squares & Rectangles | View Worksheet 1 → |
| Worksheet 2 | Area of Parallelograms & Triangles | View Worksheet 2 → |
| Worksheet 3 | Circumference of Circle (2πr, πd) | View Worksheet 3 → |
| Worksheet 4 | Area of Circle (πr2) | View Worksheet 4 → |
| Worksheet 5 | Area of Circular Ring & Track Boundaries | View Worksheet 5 → |
| Worksheet 6 | Crossroads, Pathways & Verandahs | View Worksheet 6 → |
| Brain Teasers, HOTS & VBQs | Rebent Wires, Circular Tracks & Advanced Problems | Current Page |