In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17 A Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17 A Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 17 Perimeter and Areas of Plane Figures Ex 17 A Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 17 |

Chapter Name | Perimeter and Areas of Plane Figures |

Exercise | 17 A |

Category | RS Aggarwal Solutions |

**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 m^{2}

If the cost of sowing is Re. 1, area = 1000058 m^{2}

If the cost of sowing is Rs. 783, area = (1000058×783) m^{2}

Area of the field = 135000 m^{2}

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 m^{2}

**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 cm^{2}

**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 cm^{2}

**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 cm^{2}

**Question 17:**

Let each side a = 13 cm and the base b = 20 cm

Hence, area of the triangle = 83.1 cm^{2}.

**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 cm^{2}.

**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 cm^{2} and perimeter = 34.14 cm

**All Chapter RS Aggarwal Solutions For Class 10 Maths**

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**All Subject NCERT Solutions For Class 10**

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