# RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C

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.

### 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

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