Ch-Test of Area of Plane Figures Class 9 OP Malhotra ICSE Maths Solutions 2026-27. We Provide Step by Step Solutions / Answer of Area of Plane Figures OP Malhotra Maths. Visit official Website CISCE  for detail information about ICSE Board Class-9 Mathematics.

Area of Plane Figures Class 9 OP Malhotra Ch-Test ICSE Maths Solutions

Board	ICSE
Publications	S Chand
Subject	Maths
Class	9th
Chapter-16	Area of Plane Figures
Writer	OP Malhotra
Ch-Test	Extra Practice Questions
Edition	2026-2027

Ch-Test on Area of Plane Figures

Area of Plane Figures Class 9 OP Malhotra ICSE Maths Solutions 2026-27

Que-1: What is the area of an equilateral triangle having altitude equal to 2√3 cm?
(a) √3cm²   (b) 2√3cm²   (c) 3√3cm²   (d) 4√3cm²

Sol:  (d)
Altitude = √3/2 sides
⇒ √3/2 = 2√3 cm
∴ Side = (2√3×2)/√3 = 4 cm
Area = (√3/4) side² = (√3/4) × 4 × 4
= 4√3 cm²

Que-2: The altitude of an equilateral triangle is √3 cm. What is its perimeter?
(a) 3 cm   (b) 3√3 cm   (c) 6 cm   (d) 6√3 cm

Sol: Altitude of an equilateral triangle = √3
∴ √3/2 side = √3 ⇒ side = 2 cm
and perimeter = 3 × Side
= 3 × 2 = 6 cm

Que-3: The sides of a triangle are in the ratio 2 : 3 : 4. The perimeter of the triangle is 18 cm. The area (in cm²) of the triangle is
(a) 9   (b) 36   (c) √42   (d) 6√15

Sol: Sides of a triangle are in the ratio = 2 : 3 : 4
and perieter = 18 cm
Sum of ratios = 2 + 3 + 4 = 9
∴ First side = (18/9) × 2 = 4 cm
Second side = (18/9) × 3 = 6 cm
Third side = (18/9) × 4 = 8 cm
∴ s = (a+b+c)/2
= 18/2 = 9
Area = √{s(s-a)(s-b)(s-c)}
= √{9(9-4)(9-5)(9-6)}
= √(9×5×4×3)
= √540
= √(36×15)
= 6√15 cm²

Que-4: A parallelogram has sides 15 cm and 7 cm long. The length of one of the diagonals is 20 cm. The area of the parallelogram is
(a) 42 cm²   (b) 60 cm²   (c) 84 cm²   (d) 96 cm²

Sol: Sides of a parallelogram = 15 cm and 7 cm
and length of one diagonal = 20 cm
Area (s) = (a+b+c)/2
s = (15+7+20)/2
= 42/2 = 21
Area = √{s(s-a)(s-b)(s-c)}
= √{21(21-15)(21-7)(21-20)}
= √(21×6×14×1)
= √(21×84)
= √(21×21×2×2)
= 21×2 = 42 cm²
Area of ||gm = 2 × area of ΔABC
= 2 × 42 = 84 cm²

Que-5: The perimeter of two squares are 68 cm and 60 cm. The perimeter of a third square whose area is equal to the difference of the areas of the two squares.
(a) 64 cm   (b) 60 cm   (c) 32 cm   (d) 8 cm

Sol: (c)
Perimeter of one square = 68 cm
∴ Side = 68/4 = 17 cm
Perimeter of second square = 60 cm
∴ Side = 60/4 = 15 cm
Area of first square = (Side)²
= (17)² = 289 cm²
Area of first square = (Side)²
=(17)² = 289 cm²
Area second square = (15) = 225 cm²
∴ Area of third square = 289 – 225 = 64 cm²
∴ Side = √Area = √64 = 8 cm
and perimeter = 4 × side
= 4 × 8 = 32 cm

Que-6: A square is of area 200 m². A new square is formed in such a way that the length of its diagonal is √2 times of the diagonal of the given square. Then the area of the new square formed is

(a) 200√2 sq m
(b) 400√2 sq m
(c) 400 sq m
(d) 800 sq m

Sol: (c)
Area of a square = 200 m²
∴ Its diagonal = √(Area × 2)
= √(200×2) = √400 = 20 m
∴ Diagonal of second square = √2 × 20 m
Area of second square = (diagonal)²/2
= (√2×20)²/2 = (2×400)/2 = 400 m²

Que-7: The sides of a parallelogram are in the ratio 5 : 4. Its area is 1000 sq. units. Altitude on the greater side is 20 units. Altitude on the smaller side is
(a) 30 units   (b) 25 units   (c) 10 units   (d) 15 units

Sol: Ratio in the sides of a parallelogram = 5 : 4
Area = 1000 sq. units
Altitude on the greater side = 20 units
Let greater side = 5x
and smaller side = 4x
∴ Area = Greater side × Altitude on it
1000 = 5x × 20 ⇒ x = 1000/(5×20) = 10
∴ Greater side = 5 × 10 = 50 units
and altitude on smaller side = Area / Smaller side
= 1000/(4×10) = 25 units

Que-8: One diagonal and perimeter of a rhombus are 24 cm and 52 cm respectively. The other diagonal is
(a) 13 cm   (b) 12 cm   (c) 10 cm   (d) 15 cm

Sol: Perimeter of a rhombus = 52 cm
∴ Side = Perimeter / 4 = 52/4 = 13 cm
One diagonal = 24 cm
By joining the diagonals
They bisect each other at O at right angle
∴ AO = OC = 24/2 = 12 cm and BO = OD
In right △AOB
AB² = AO² + BO²
⇒ (13)² = 12² + B²
⇒ 169 = 144 + B²
⇒ BO² = 169 – 144 = 25 = (5)²
BO = 5 cm
∴ Diagonal BD = 2 × BO
= 2 × 5 = 10 cm

Que-9: The area of a field in the shape of a trapezium measures 1440 m². The perpendicular distance between its parallel sides is 24 m. If the ratio of the parallel sides is 5 : 3, the length of the longer parallel sides is:
(a) 45 m   (b) 60 m   (c) 75 cm   (d) 120 m

Sol: Area of a trapezium shaped field = 1440 m2
Perpendicular distance = 24 m
∴ Sum of parallel sides = (Area × 2) / Height
= (1440×2)/24 = 120 cm
Ratio in side = 5 : 3
Sum of ratio = 5 + 3 = 8
∴ First longer side = (120/8) × 5 = 75 m

Que-10: A lawn 30 m long and 16 m wide is surrounded by a path 2 m wide. What is the area of the path?

(a) 200 m²
(b) 280 m²
(c) 300 m²
(d) 320 m²

Sol: (a)
Length of a lawn = 30 m
Breadth of a lawn = 16 m
Width of path around it = 2 m
Outer length = 30 + 2 × 2
= 30 + 4 = 34 m
and outer breadth = 16 + 2 × 2
=16 + 4 = 20 m
∴ Area of path = Outer Area – Inner area
= 34 × 20 – 30 × 16 m²
& = 680 – 480 = 200 m²

— : End of Area of Plane Figures Class-9 OP Malhotra Ch-Test ICSE Maths Solutions :–

Return to :–  OP Malhotra S Chand Solutions for ICSE Class-9 Maths

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