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

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

Question 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 =

Hence, the altitude = 300m and the base = 900 m

Question 3:
Let a = 42 cm, b = 34 cm and c = 20 cm

(i) Area of triangle =

(ii) Let base = 42 cm and corresponding height = h cm
Then area of triangle =

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 =

(ii) Let base = 18 cm and altitude = x cm
Then, area of triangle =

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

(s – a) = 50 cm, (s – b) = 15 cm, and (s – c) = 10 cm

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

Then, (s – a) = 29 m, (s – b) = 100 m, and (s – c) = 150m

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

Hence, area of the triangle = 60 cm2

Question 8:
Let the sides containing the right angle be x cm and (x × 7) cm

One side = 15 cm and other = (15 × 7) cm = 8 cm

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

Question 9:
Let the sides containing the right angle be x and (x × 2) cm

One side = 8 cm, and other (8 × 2) cm = 6 cm
= 10 cm

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

(ii) Height of the triangle=  (3√4×a) Sq.unit

Hence, area = 27.71 cm2 and height = 6.93 cm

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

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

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

Height of equilateral triangle

Question 14:
Base of right angled triangle = 48 cm
Height of the right angled triangle =

Question 15:
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 16:
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 17:
Let each side a = 13 cm and the base b = 20 cm

Hence, area of the triangle = 83.1 cm2.

Question 18:
Let each equal side be a cm in length.
Then,

Hence, hypotenuse = 28.28 cm and perimeter = 68.28 cm

Question 19:
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 20:
Perimeter of an isosceles triangle = 42 cm
(i) Let each side be a cm, then base =  32a
perimeter = (2a + b) cm

Hence each side = 12 cm and Base = 3212 = 18cm
(ii) Area of triangle =

(iii) Height of the triangle =

Question 21:
Let the height be h cm, then a= (h + 2) cm and b = 12 cm

Squaring both sides,

Therefore, a = h + 2 = (8 + 2)cm = 10 cm

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

(ii) The base = 154 cm and let the perpendicular = h cm

Hence, required length of the perpendicular of the triangle is 36 m.

Question 23:

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

Second we find area of ∆DBC which is right angled

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.

Area of right isosceles triangle ABC

Hence, area = 50 cm2 and perimeter = 34.14 cm

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