In this chapter, we provide RS Aggarwal Solutions for Class 7 Chapter 20 Mensuration Ex 20C for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 7 Chapter 20 Mensuration Ex 20C Maths pdf, free RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 7 |

Subject | Maths |

Chapter | Chapter 20 |

Chapter Name | Mensuration |

Exercise | 20C |

**RS Aggarwal Solutions for Class 7 Chapter 20 Mensuration Ex 20C Download PDF**

**Question 1.****Solution:**

In parallelogram ABCD,

Base AB = 32cm

Height DL = 16.5cm.

Area = Base x height = 32 x 16.5 cm² = 528 cm²

**Question 2.****Solution:**

Base of parallelogram = 1 m 60m = 160 cm

and height = 75 cm

Area = Base x height = 160 x 75 = 12000 cm²

= 1200010000 m² = 1.2m²

**Question 3.****Solution:**

Base of parallelogram = 14dm = 140cm

and height = 6.5 dm = 65cm

Area (in cm²) = Base x height = 140 x 65 = 9100 cm²

Area (in m²) = 140100 x 65100 = 910010000

= 0.91 m²

**Question 4.****Solution:**

Area of parallelogram = 54 cm²

Base = 15 cm

**Question 5.****Solution:**

Area of parallelogram ABCD = 153 cm²

**Question 6.****Solution:**

In parallelogram ABCD

AB || DC and AD || BC and AB = DC, AD = BC

AB = DC = 18cm, BC = 12cm

Area of parallelogram ABCD = Base x altitude = DC x AL = 18 x 6.4cm2 = 115.2 cm²

and area of parallelogram ABCD = BC x AM

⇒ 115.2 = 12 x AM

⇒ AM = 9.6 cm

**Question 7.****Solution:**

In parallelogram ABCD

AB = DC = 15 cm

BC = AD = 8 cm.

Distance between longer sides AB and DC is 4cm

i.e. perpendicular DL = 4cm.

DM ⊥ BC.

Area of parallelogram = Base x altitude = AB x DL = 15 x 4 = 60 cm²

Again let DM = x cm

area ABCD = BC x DM = 8 x x = 8x cm²

8x cm² = 60 cm²

⇒ x = 7.5 cm

Distance between shorter lines = 7.5 cm

**Question 8.****Solution:**

Let Base of the parallelogram = x

⇒ x² = 108 x 3 = 324 = (18)²

⇒ x = 18

Base = 18 cm

and altitude = 13 x 18 = 6 cm

**Question 9.****Solution:**

Area of parallelogram = 512 cm²

Let height of the parallelogram = x

Then base = 2x

Area = Base x height

⇒ 512 = 2x x x

⇒ 2x² = 512

⇒ x² = 256 = (16)²

⇒ x = 16

Base = 2x = 2 x 16 = 32 cm

and height = x = 16 cm

**Question 10.****Solution:**

(i) Each side of rhombus = 12 cm

height = 7.5 cm

Area = Base x height = 12 x 7.5 = 90 cm²

(ii) Each side = 2 cm = 20 cm

Height = 12.6 cm

Area = Base x height = 20 x 12.6 = 252 cm²

**Question 11.****Solution:**

(i) Diagonals of rhombus ABCD are 16 cm and 28 cm

**Question 12.****Solution:**

In rhombus ABCD, diagonals AC and BD intersect each other at right angles at O.

AO = OC and BO = OD

AO = 12 x AC = 12 x 24 cm = 12cm

Let OB = x

Each side of rhombus = 20cm

In right ∆AOB

AO² + OB² = AB² (Pythagoras Theorem)

⇒ (12)² + OB² = (20)²

⇒ 144 + OB² = 400

⇒ OB² = 400 – 144 = 256 = (16)²

⇒ OB = 16

But BD = 2 BO = 2 x 16 = 32cm

Now, area of rhombus = Productofdiagonals2

= 24×322 cm2 = 384 cm²

**Question 13.****Solution:**

Area of rhombus = 148.8 cm²

one diagonal = 19.2 cm

Let second diagonal = x

**Question 14.****Solution:**

Area of rhombus = 119 cm²

Perimeter = 56 cm

Its side = 564 = 14 cm

**Question 15.****Solution:**

Area of rhombus = 441 cm²

Height = 17.5 cm

**Question 16.****Solution:**

Base of a triangle = 24.8 cm

Corresponding height = 16.5 cm

**All Chapter RS Aggarwal Solutions For Class 7 Maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class** **7**

**All Subject NCERT Solutions For Class 7**

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