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RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17A
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17B
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17A
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17B
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Correct option: (d)

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17A
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17B
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

Question 1.
Solution:
Let the length of plot be x meters
Its perimeter = 2 [length + breadth]
= 2(x + 16) = (2x + 32) meters

Length of the rectangle is 21. 5 meter
Area of the rectangular plot = length × breadth
= (16 × 21.5)m²
= 344 m²
The length = 21.5 m and the area = 344 m²

Question 2.
Solution:
The length of a rectangular park is twice its breadth
and its perimeter is 840 m.
Find the area of the park.
Let the breadth of a rectangular park = x m
Then, length of a rectangular park = 2x m
Perimeter of a rectangular park = 840 m

∴ Area of a rectangular park = length x breadth = 140x 280 = 39200 m2

Question 3.
Solution:
Let ABCD be the rectangle in which AB = 12 cm and AC = 37 m

Thus, length = 35 cm and breadth = 12 cm
Area of rectangle = (12 × 35) = 420
Hence, the other side = 35 cm and the area = 420 cm²

Question 4.
Solution:
Let the breadth of the plot be x meter
Area = Length × Breadth = (28 × x) meter
= 28x m²
Breadth of plot is = 16. 5 m
Perimeter of the plot is = 2(length + breadth)

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:
Area of verandah = (36 × 15)m² = 540 m²
Area of stone = (0.6 × 0.5)m² [10 dm = 1 m]
Number of stones required = \(\frac { Area\quad of\quad verandah }{ Area\quad of\quad stone } =\frac { 540 }{ 0.3 } =1800\)
Hence, 1800 stones are required to pave the verandah

Question 9.
Solution:

Question 10.
Solution:
Length of the park = 35 m
Breadth of the park = 18 m

Area of the park = (35 18)m² = 630 m²
Length of the park with grass =(35 – 5) = 30 m
Breadth of the park with grass = (18 – 5) m = 13 m
Area of park with grass = (30 × 13) m² = 390 m²
Area of path without grass = Area of the whole park area of park with grass
= 630 m² – 390 m² = 240 m²
Hence, area of the park to be laid with grass = 240 m²

Question 11.
Solution:
Length of the plot = 125 m
Breadth of the plot = 78 m

Area of plot ABCD = (125 × 78)m² = 9750 m²
Length of the plot including the path= (125 + 3 + 3) m = 131 m
Breadth of the plot including the path = (78 + 3 + 3) m = 84 m
Area of plot PQRS including the path
= (131 × 84) m² = 11004 m²
Area of path = Area of plot PQRS Area of plot ABCD
= (11004 – 9750)m² = 1254 m²
Cost of gravelling = Rs 75 per m²
Cost of gravelling the whole path = Rs. (1254 × 75) = Rs. 94050
Hence, cost of gravelling the path = Rs. 94050.

Question 12.
Solution:
Area of rectangular field including the foot path = (54 × 35)m²
Let the width of the path be x m
Then, area of rectangle plot excluding the path = (54 × 2x) (35 × 2x)
Area of path = (54 × 35) (54 × 2x) (35 × 2x) = 54 ×35) (54 × 2x) (35 × 2x) = 420

Question 13.
Solution:
Let the length and breadth of a rectangular garden be 9x and 5x.

Then, area of garden = (9x 5x)m = 45x² m²
Length of park excluding the path = (9x 7) m
Breadth of the park excluding the path = (5x 7) m
Area of the park excluding the path = (9x 7)(5x 7) m²

Question 14.
Solution:

Question 15.
Solution:
Let the width of the carpet = x meter
Area of floor ABCD = (8 ×5) m²
Area of floor PQRS without border
= (8 × 2x)(5 × 2x)
= 40 × 16 x 10x + 4x²
= 40 × 26x + 4x²
Area of border = Area of floor ABCD Area of floor PQRS

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Question 26.
Solution:

Question 27.
Solution:

Question 28.
Solution:
Area of the ||gm = (base height) sq. unit
= (25 × 16.8) cm² = 420 cm²

Question 29.
Solution:

Question 30.
Solution:

Question 31.
Solution:
Area of parallelogram = 2 area of DABC
Opposite sides of parallelogram are equal
AD = BC = 20 cm
And AB = DC = 34 cm
In ∆ABC we have
a = AC = 42 cm
b = AB = 34 cm
c = BC = 20 cm

Question 32.
Solution:

We know that the diagonals of a rhombus, bisect each other at right angles
OA = OC = 15 cm,
And OB = OD = 8 cm
And ∠AOB = 90⁰
∴By Pythagoras theorem, we have

Question 33.
Solution:

Question 34.
Solution:

Question 35.
Solution:

Question 36.
Solution:

Question 37.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17b are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17a

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17a.

Other Exercises

• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17A
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17B
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures MCQ
• RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Test Yourself

Question 1.
Solution:

Question 2.
Solution:
Let a = 42 cm, b = 34 cm and c = 20 cm

(ii)Let base = 42 cm and corresponding height = h cm

Hence, the height corresponding to the longest side = 16 cm

Question 3.
Solution:
Let a = 18 cm, b = 24 cm, c = 30 cm
Then,2s = (18 + 24 + 30) cm = 72 cm
s = 36 cm
(s a) = 18cm, (s b) = 12 cm and (s c) = 6 cm

(ii)Let base = 18 cm and altitude = x cm

Hence, altitude corresponding to the smallest side = 24 cm

Question 4.
Solution:
On dividing 150 m in the ratio 5 : 12 : 13, we get

Question 5.
Solution:
On dividing 540 m in ratio 25 : 17 : 12, we get

Question 6.
Solution:
Let the length of one side be x cm
Then the length of other side = {40 (17 + x)} cm = (23 – x) cm
Hypotenuse = 17 cm
Applying Pythagoras theorem, we get

Question 7.
Solution:
Let the sides containing the right – angle be x cm and (x – 7) cm

∴ perimeter of triangle (15 + 8 + 17) cm = 40 cm

Question 8.
Solution:
Let the sides containing the right angle be x and (x 2) cm

Therefore, perimeter of the triangle = 8 + 6 + 10 = 24 cm

Question 9.
Solution:
Side of an equilateral triangle = a = 10 cm

Question 10.
Solution:
Let each side of the equilateral triangle be a cm

Question 11.
Solution:
Let each side of the equilateral triangle be a cm

Perimeter of equilateral triangle = 3a = (3 12) cm = 36 cm

Question 12.
Solution:
Let each side of the equilateral triangle be a cm

Question 13.
Solution:
Base of right angled triangle = 48 cm

Question 14.
Solution:
Let the hypotenuse of right – angle triangle = 6.5 m
Base = 6 cm

Hence, perpendicular = 2.5 cm and area of the triangle =7.5 cm²

Question 15.
Solution:
The circumcentre of a right – triangle is the midpoint of the hypotenuse

Hypotenuse = 2 × (radius of circumcircle)
= (2 × 8) cm = 16 cm
Base = 16 cm, height = 6 cm
Area of right angled triangle=

Hence, area of the triangle= 48 cm²

Question 16.
Solution:
Let each equal side be a cm in length.
Then,

Hence, hypotenuse = 28.28 cm and perimeter = 68.28 cm

Question 17.
Solution:
Let each equal side be a cm and base = 80 cm

perimeter of triangle = (2a + b) cm
= (2 41 + 80) cm
= (82 + 80) cm = 162 cm
Hence, perimeter of the triangle = 162 cm

Question 18.
Solution:
Let the height be h cm, then a= (h + 2) cm and b = 12 cm

Hence, area of the triangle = 48 cm².

Question 19.
Solution:
Let ∆ABC is a isosceles triangle. Let AC, BC be the equal sides
Then AC = BC = 10cm. Let AB be the base of ∆ABC right angle at C.

Hence, area = 50 cm² and perimeter = 34.14 cm

Question 20.
Solution:

Area of shaded region = Area of ABC – Area of DBC
First we find area of ABC

Area of shaded region = Area of ∆ABC – Area of ∆DBC
= (43.30 – 24) cm² = 19. 30 cm²
Area of shaded region = 19.3 cm²

Hope given RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17a are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.