Perimeter and Area: Important Questions and Answers on Measuring Space for Class 9

Important questions on Measuring Space: Perimeter and Area for Class 9 is provided in this Maths article. These important questions help students revise key concepts, formulas, and problem-solving techniques related to perimeter and area with ease. Exploring Measuring Space: Perimeter and Area covers important topics such as the perimeter and area of rectangles, squares, triangles, circles, and other plane figures, along with their practical applications aligned with the CBSE syllabus and the NCERT textbook for Class 9. Whether you follow the standard NCERT textbook or the Ganita Manjari, this material supports quick revision and better performance in examinations.

Table of Contents

• Key Concepts on Important Questions on Measuring Space: Perimeter and Area for Class 9
• Important Questions on Triangles (Heron's Formula) for Class 9
• Important Questions on Quadrilaterals for Class 9
• Important Questions on Circles & Sectors for Class 9
• Important Questions on Combined / Composite Figures for Class 9
• Case Study Questions on Perimeter and Area for Class 9

Key Concepts on Important Questions on Measuring Space: Perimeter and Area for Class 9

Make sure these formulas of perimeter and area for Class 9 are at your fingertips. The new Ganita Manjari textbook places special emphasis on Heron's formula, sectors, and composite figures.

Important Questions on Triangles (Heron's Formula) for Class 9

Q1: A triangular plot has sides of 40 m, 24 m, and 32 m. Find the cost of turfing it at ₹5 per m².

Solution: Sides: a = 40 m, b = 24 m, c = 32 m. 

Check: 24² + 32² = 576 + 1024 = 1600 = 40². So this is a right triangle.

Area (right triangle) = ½ × base × height 

= ½ × 24 × 32 = 384 m² 

Cost of turfing = ₹5 per m²

Cost = 384 × ₹5 = ₹1,920

Cost of turfing = ₹1,920

Q 2: An isosceles triangle has a perimeter of 30 cm and the two equal sides are 12 cm each. Find its area using Heron's formula.

Solution: Let the third (unequal) side = b. 

Perimeter = 12 + 12 + b = 30 

⇒ b = 6 cm

s = 30/2 = 15 cm 

Area = √[15 × (15−12) × (15−12) × (15−6)] 

= √[15 × 3 × 3 × 9] = √[1215] 

= 9√15 ≈ 34.85 cm²

Area = 9√15 cm² ≈ 34.85 cm²

Q 3: The perimeter of an equilateral triangle is 60 m. Find its area.

Solution: Side a = 60/3 = 20 m 

⇒ Area = (√3/4) × a² = (√3/4) × 400 = 100√3 m² ≈ 173.2 m²

Therefore, Area = 100√3 m²

Important Questions on Quadrilaterals for Class 9

Q 4: The diagonal of a square field is 50 m. Find its area and perimeter.

Solution: Given, d = 50 m

Area of square = d²/2 = (50)²/2 

= 2500/2 = 1250 m² 

Side a = d/√2 = 50/√2 = 25√2 m 

Perimeter = 4a = 4 × 25√2 

= 100√2 ≈ 141.4 m

Therefore, area = 1250 m² and perimeter = 100√2 m

Q 5: A rhombus-shaped field has diagonals of 18 m and 24 m. Find the cost of fencing its boundary at ₹30 per metre.

Solution: Diagonals of a rhombus bisect each other at right angles. 

So each side = √[(d₁/2)² + (d₂/2)²]

Each half-diagonal = 9 m and 12 m 

Side = √(9² + 12²) = √(81 + 144) = √225 = 15 m 

Perimeter = 4 × 15 = 60 m 

Cost = 60 × ₹30 = ₹1,800

Cost of fencing = ₹1,800

Q 6: Find the area of a trapezium with parallel sides 25 cm and 13 cm, and non-parallel sides 10 cm and 10 cm.

Solution: Drop a perpendicular from one end of the shorter parallel side. 

The base of this right triangle = (25 − 13)/2 = 6 cm, hypotenuse = 10 cm.

Height h = √(10² − 6²) 

= √(100 − 36) = √64 = 8 cm 

Area = ½ × h × (a + b) 

= ½ × 8 × (25 + 13) = ½ × 8 × 38 = 152 cm²

Area of trapezium = 152 cm²

Important Questions on Circles & Sectors for Class 9

Q 7: A circular track runs around a circular park. If the difference between the circumference of the track and the park is 66 m, find the width of the track.

Solution: Let R = radius of track, r = radius of park, and width = R − r.

2πR − 2πr = 66 

⇒ 2π(R − r) = 66 

⇒ R − r = 66 / (2 × 22/7) 

⇒ R − r = 66 × 7 / 44 = 10.5 m

Width of the track = 10.5 m

Q 8: Find the area of a sector of a circle with radius 14 cm and angle 90°.

Solution: Given r = 14 cm

Area of sector = (θ/360°) × πr² 

= (90/360) × (22/7) × 14² 

= (1/4) × (22/7) × 196 

= (1/4) × 616 = 154 cm²

Area of sector = 154 cm²

Q 9: A wire of length 176 cm is bent to form a circle. Find its area. (Use π = 22/7)

Solution: Length of wire = Circumference = 176 cm 

⇒ 2πr = 176 r = 176 / (2 × 22/7) 

= 176 × 7 / 44 = 28 cm 

Area = πr² = (22/7) × 28² 

= (22/7) × 784 = 2464 cm²

Area = 2464 cm²

Important Questions on Combined / Composite Figures for Class 9

Q10: A figure is made up of a rectangle 20 cm × 14 cm with a semicircle on its shorter side. Find the total area and perimeter of the figure.

Solution: The semicircle sits on the 14 cm side, so its diameter = 14 cm 

⇒ radius = 7 cm.

Area of rectangle = 20 × 14 = 280 cm² 

Area of semicircle = πr²/2 = (22/7) × 49/2 = 77 cm² 

⇒ Total area = 280 + 77 = 357 cm² 

Perimeter: Two long sides of rectangle = 2 × 20 = 40 cm 

One short side (opposite semicircle) = 14 cm 

The curved arc of semicircle = πr = (22/7) × 7 = 22 cm 

⇒ Total perimeter = 40 + 14 + 22 = 76 cm

Q11: From a square sheet of side 14 cm, four equal circles of maximum possible size are cut out. Find the remaining area.

Solution: Four equal circles fit in a 2×2 arrangement. 

Each circle has diameter = 14/2 = 7 cm 

⇒ radius = 3.5 cm.

Area of square = 14² = 196 cm² 

Area of 4 circles = 4 × π × (3.5)² 

= 4 × (22/7) × 12.25 = 4 × 38.5 = 154 cm² 

Remaining area = 196 − 154 = 42 cm²

Remaining area = 42 cm²

Q12: A plot of land is in the shape of a quadrilateral ABCD where AB = 9 m, BC = 40 m, CD = 28 m, DA = 15 m, and ∠ABC = 90°. Find the area of the plot.

Solution: Draw diagonal AC to divide the quadrilateral into triangle ABC and triangle ACD.

Triangle ABC (right-angled at B): 

AC = √(AB² + BC²) = √(9² + 40²) = √(81 + 1600) = √1681 = 41 m 

⇒ Area of △ABC = ½ × 9 × 40 = 180 m² 

Triangle ACD with sides 41 m, 28 m, 15 m: 

s = (41 + 28 + 15)/2 = 84/2 = 42 m 

⇒Area = √[42 × (42−41) × (42−28) × (42−15)] 

= √[42 × 1 × 14 × 27] = √15876 = 126 m² 

Total area: Area of ABCD = 180 + 126 = 306 m²

Area of quadrilateral plot ABCD = 306 m²

Case Study Questions on Perimeter and Area for Class 9

C1: The School Garden Project Case Study

A school has a rectangular ground of dimensions 50 m × 30 m. The school plans to create a garden in one corner shaped like a right triangle with legs 10 m and 12 m. In the centre, a circular fountain of radius 3.5 m is built. The remaining area will be used for a play zone. Fencing is to be put around the entire rectangular ground.

i) What is the area of the rectangular ground?

Area = 50 × 30 = 1500 m²

ii) What is the area of the triangular garden?

Area = ½ × 10 × 12 = 60 m²

iii) What is the area of the circular fountain?

Area = πr² = (22/7) × (3.5)² = (22/7) × 12.25 = 38.5 m²

iv) What is the area of the play zone?

Play zone = 1500 − 60 − 38.5 = 1401.5 m²

v) What length of fencing is needed around the entire ground?

Perimeter of rectangle = 2(50 + 30) = 160 m

 C2: Flooring a Heritage Hall Case Study

A heritage hall has a floor shaped like a trapezium with parallel sides of 28 m and 16 m and a height of 10 m. There are two square pillars of side 1.5 m each, whose area must be excluded. Marble tiles of size 60 cm × 60 cm cost ₹250 each.

i) Area of the trapezoidal floor (before pillars)?

Area = ½ × 10 × (28 + 16) = 5 × 44 = 220 m²

ii) Area of both square pillars?

Area = 2 × (1.5)² = 2 × 2.25 = 4.5 m²

iii) Net area to be tiled?

Net area = 220 − 4.5 = 215.5 m²

iv) Area of one tile?

Area = 0.6 × 0.6 = 0.36 m²

v) Number of tiles needed and total cost?

Tiles needed = 215.5 / 0.36 ≈ 599 (round up to 599) 

Total cost = 599 × ₹250 = ₹1,49,750

Tiles needed ≈ 599 and Total cost ≈ ₹1,49,750