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

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TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 17
Chapter NamePerimeter and Areas of Plane Figures
Exercise17 A
CategoryRS Aggarwal Solutions

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

Question 1:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 1.1

Question 2:
If the cost of sowing the field is Rs. 58, then area = 10000 m2
If the cost of sowing is Re. 1, area =  1000058 m2
If the cost of sowing is Rs. 783, area =  (1000058×783) m2
Area of the field = 135000 m2
Let the attitude of the field be x meters
Then, Base = 3x meter
Area of the field = RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17 Q2
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 9a 2.1
Hence, the altitude = 300m and the base = 900 m

Question 3:
Let a = 42 cm, b = 34 cm and c = 20 cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 3.1
(i) Area of triangle = RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 3.2
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 3.3
(ii) Let base = 42 cm and corresponding height = h cm
Then area of triangle = RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 3.4
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 3.5
Hence, the height corresponding to the longest side = 16 cm

Question 4:
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
(i) Area of triangle = RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 4.1
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 4.2
(ii) Let base = 18 cm and altitude = x cm
Then, area of triangle = RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 4.3
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 4.4
Hence, altitude corresponding to the smallest side = 24 cm

Question 5:
On dividing 150 m in the ratio 5 : 12 : 13, we get
Length of one side =  (150×530)m=25m
Length of the second side =  (150×1230)m=60m
Length of third side =  (150×1330)m=65m
Let a = 25 m, b = 60 m, c = 65 m
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 5.1
(s – a) = 50 cm, (s – b) = 15 cm, and (s – c) = 10 cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 9a 5.2
Hence, area of the triangle = 750 m2

Question 6:
On dividing 540 m in ratio 25 : 17 : 12, we get
Length of one side =  (540×2554)m=250m
Length of second side =  (540×1754)m=170m
Length of third side =  (540×1254)m=120m
Let a = 250m, b = 170 m and c = 120 m
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 6.1
Then, (s – a) = 29 m, (s – b) = 100 m, and (s – c) = 150m
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 9a 6.2
The cost of ploughing 100 area is = Rs. 18. 80
The cost of ploughing 1 is =  Rs. 18.80100
The cost of ploughing 9000 area = Rs. (18.80100×9000)
= Rs. 1692
Hence, cost of ploughing = Rs 1692.

Question 7:
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
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 9a 7.1
Hence, area of the triangle = 60 cm2

Question 8:
Let the sides containing the right angle be x cm and (x × 7) cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 8.1
One side = 15 cm and other = (15 × 7) cm = 8 cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 8.2
perimeter of triangle (15 + 8 + 17) cm = 40 cm

Question 9:
Let the sides containing the right angle be x and (x × 2) cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 9a 9.1
One side = 8 cm, and other (8 × 2) cm = 6 cm
= 10 cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 9.2
Therefore, perimeter of the triangle = 8 + 6 + 10 = 24 cm

Question 10:
(i) Here a = 8 cm
Area of the triangle =  (3√4×a2) Sq.unit
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 10.1
(ii) Height of the triangle=  (3√4×a) Sq.unit
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 10.2
Hence, area = 27.71 cm2 and height = 6.93 cm

Question 11:
Let each side of the equilateral triangle be a cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 11.1

Question 12:
Let each side of the equilateral triangle be a cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 12.1
Perimeter of equilateral triangle = 3a = (3 × 12) cm = 36 cm

Question 13:
Let each side of the equilateral triangle be a cm
area of equilateral triangle =  3√4a2
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 13.1
Height of equilateral triangle
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 13.2

Question 14:
Base of right angled triangle = 48 cm
Height of the right angled triangle = RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Q14
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 14.1

Question 15:
Let the hypotenuse of right angle triangle = 6.5 m
Base = 6 cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 9a 15.1
Hence, perpendicular = 2.5 cm and area of the triangle =7.5 cm2

Question 16:
The circumcentre of a right triangle is the midpoint of the hypotenuse
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 16.1
Hypotenuse = 2 ×(radius of circumcircle)
= (2 × 8) cm = 16 cm
Base = 16 cm, height = 6 cm
Area of right angled triangle
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 16.2
Hence, area of the triangle= 48 cm2

Question 17:
Let each side a = 13 cm and the base b = 20 cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 17.1
Hence, area of the triangle = 83.1 cm2.

Question 18:
Let each equal side be a cm in length.
Then,
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 9a 18.1
Hence, hypotenuse = 28.28 cm and perimeter = 68.28 cm

Question 19:
Let each equal side be a cm and base = 80 cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 9a 19.1
perimeter of triangle = (2a + b) cm
= (2 ×41 + 80) cm
= (82 + 80) cm = 162 cm
Hence, perimeter of the triangle = 162 cm

Question 20:
Perimeter of an isosceles triangle = 42 cm
(i) Let each side be a cm, then base =  32a
perimeter = (2a + b) cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 20.1
Hence each side = 12 cm and Base = 3212 = 18cm
(ii) Area of triangle = RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 20.2

RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 9a 20.3
(iii) Height of the triangle = RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 20.4

RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 9a 20.5

Question 21:
Let the height be h cm, then a= (h + 2) cm and b = 12 cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 21.1
Squaring both sides,
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 21.2
Therefore, a = h + 2 = (8 + 2)cm = 10 cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 9a 21.3
Hence, area of the triangle = 48 cm2.

Question 22:
Perimeter of triangle = 324 cm
(i) Length of third side = (324 – 85 – 154) m = 85 m
Let a = 85 m, b = 154 m, c = 85 m
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 22.1
(ii) The base = 154 cm and let the perpendicular = h cm
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 22.2
Hence, required length of the perpendicular of the triangle is 36 m.

Question 23:
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 9a 23.1
Area of shaded region = Area of ∆ABC – Area of ∆DBC
First we find area of ∆ABC
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 23.2
Second we find area of ∆DBC which is right angled
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 9a 23.3
Area of shaded region = Area of ∆ABC – Area of ∆DBC
= (43.30 – 24) = 19. 30
Area of shaded region = 19.3

Question 24:
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.
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 17a 24.1
Area of right isosceles triangle ABC
RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures 9a 24.2
Hence, area = 50 cm2 and perimeter = 34.14 cm

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