## RS Aggarwal Class 7 Solutions Chapter 20 Mensuration CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper.

**Other Exercises**

- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G
- RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper

**Question 1.**

**Solution:**

We know that all the angles of a rectangle are 90° and the diagonal divides the rectangle into two right angled triangles.

So, one side of the triangle will be 48 m and the diagonal, which is 50 m, will be the hypotenuse.

According to Pythagoras theorem :

(Hypotenuse)² = (Base)² + (Perpendicular)²

Perpendicular

Other side of the rectangular plot = 14m

Area of the rectangular plot = 48 m x 14 m = 672 m²

Hence, the area of a rectangular plot is 672 m².

**Question 2.**

**Solution:**

Length = 9 m; Breadth = 8 m

Height = 6.5 m

Area of the four walls = {2 (l + b) x h} sq. units

= {2 (9 + 8) x 6.5} m² = {34 x 6.5) m² = 221 m²

Area of one door = (2 x 1 .5) m² = 3m²

Area of one window = (1.5 x 1) m² = 1.5 m²

Area of four windows = (4 x 1.5) m² = 6 m²

Total area of one door and four windows = (3 + 6) m² = 9 m²

Area to be painted = (221 – 9) m² = 212 m²

Rate of painting = ₹ 50 per m²

Total cost of painting = ₹ (212 x 50) = ₹ 10,600

**Question 3.**

**Solution:**

Given that the diagonal of a square is 64 cm

**Question 4.**

**Solution:**

Let ABCD be the square lawn

and PQRS be the outer boundary of the square path

Let one side of the lawn (AB) be x m

Area of the square lawn = x²

Length PQ = (x m + 2 m + 2 m) =(x + 4) m

Area of PQRS = (x + 4)² = (x² + 8x + 16) m²

Now, Area of the path = Area of PQRS – Area of the square lawn

⇒ 136 = x² + 8x + 16x – x²

⇒ 136 = 8x + 16

⇒ 136 – 16 = 8x

⇒ 120 = 8x

⇒ x = 15

Side of the laws = 15 m

Area of the lawn = (Side)² = (15 m)² = 225 m²

**Question 5.**

**Solution:**

Let ABCD be the rectangular park

EFGH and IJKL are the two rectangular roads with width 2 m.

Length of the rectangular park AD = 30 cm

Breadth of the rectangular park CD = 20 cm

Area of the road EFGH = 30 m x 2 m = 60 m²

Area of the road IJKL = 20 m x 2m = 40 m²

Clearly, area of MNOP is common to the two roads.

Area of MNOP = 2m x 2m = 4m²

Area of the roads = Area(EFGH) + Area (IJKL) – Area (MNOP)

= (60 + 40) m² – 4 m² = 96 m²

**Question 6.**

**Solution:**

Let ABCD be the rhombus whose diagonals intersect at O.

Then, AB = 13 cm

AC = 24 cm

The diagonals of a rhombus bisect each other at right angles.

Therefore, ∆AOB is a right-angled triangle, right angled at O, such that:

OA = \(\frac { 1 }{ 2 }\) AC = 12 cm

AB = 13 cm

By Pythagoras theorem :

(AB)² = (OA)² + (OB)²

⇒ (13)² = (12)² + (OB)²

⇒ (OB)² = (13)² – (12)²

⇒ (OB)2 = 169 – 144 = 25

⇒ (OB)² = (5)²

⇒ OB = 5 cm

BD = 2 x OB = 2 x 5 cm = 10 cm

Area of the rhombus ABCD = \(\frac { 1 }{ 2 }\) x AC x BD cm²

= \(\frac { 1 }{ 2 }\) x 24 x 10

= 120 cm²

**Question 7.**

**Solution:**

Let the base of the parallelogram be x m.

The, the altitude of the parallelogram will be 2x m.

It is given that the area of the parallelogram is 338 m².

Area of a parallelogram = Base x Altitude

⇒ 338 = x x 2x

⇒ 338 = 2x²

⇒ x² = 169 m²

⇒ x = 13 m

Base = x m = 13 m

Altitude = 2x m = (2 x 13) m = 26 m

**Question 8.**

**Solution:**

Consider ∆ABC Here, ∠B = 90°

AB = 24 cm

AC = 25 cm

Now, AB² + BC² = AC²

BC² = AC² – AB² = (25² – 24²) =(625 – 576) = 49

BC = (√49) cm = 7 cm

Area of ∆ABC = \(\frac { 1 }{ 2 }\) x BC x AB Sq.units

= \(\frac { 1 }{ 2 }\) x 7 x 24 cm² = 84 cm²

Hence, area of the right angled triangle is 84 cm².

**Question 9.**

**Solution:**

Radius of the wheel = 35 cm

Circumference of the wheel = 2πr

**Question 10.**

**Solution:**

Let the radius of the circle be r cm

Area = (πr²) cm²

πr² = 616

⇒ \(\frac { 22 }{ 7 }\) x r x r = 616

Hence, the radius of (he given circle is 14 cm.

**Mark (✓) against the correct answer in each of the following:**

**Question 11.**

**Solution:**

(a) 14 cm

Let the radius of the circle be r cm

Then, its area will be (πr²) cm²

πr² = 154

**Question 12.**

**Solution:**

(b) 154 cm²

Let the radius of the circle be r cm.

Circumference = (2πr) cm

(2πr) = 44

**Question 13.**

**Solution:**

(c) 98 cm²

Given that the diagonal of a square is 14 cm

Area of a square

**Question 14.**

**Solution:**

(b) 10 cm

Given that the area of the square is 50 cm²

We know:

Area of a square

**Question 15.**

**Solution:**

(a) 192 m²

Let the length of the rectangular park be 4x.

Breadth = 3x

Perimeter of the park = 2 (l + b) = 56 m (given)

⇒ 56 = 2 (4x + 3x)

⇒ 56 = 14x

⇒ x = 4

Length = 4x = (4 x 4) = 16 m

Breadth = 3x = (3 x 4) = 12 m

Area of the rectangular park = 16 m x 12 m= 192 m²

**Question 16.**

**Solution:**

(a) 84 cm²

Let a = 13 cm, b = 14 cm and c = 15 cm

**Question 17.**

**Solution:**

(a) 16√3 cm²

Given that each side of an equilateral triangle is 8 cm

Area of the equilateral triangle

**Question 18.**

**Solution:**

(b) 91 cm²

Base = 14 cm

Height = 6.5 cm

Area of the parallelogram = Base x Height

= (14 x 6.5) cm² = 91 cm²

**Question 19.**

**Solution:**

(b) 135 cm²

Area of the rhombus = \(\frac { 1 }{ 2 }\) x (Product of the diagonals)

= \(\frac { 1 }{ 2 }\) x 18 x 15 = 135 cm²

Hence, the area of the rhombus is 135 cm².

**Question 20.**

**Solution:**

(i) If d_{1}, and d_{2} be the diagonals of a rhombus, then its area is \(\frac { 1 }{ 2 }\) d_{1}d_{2} sq. units.

(ii) If l, b and h be the length, breadth and height respectively of a room, then area of its 4 walls = [2h (l + b)] sq. units.

(iii) 1 hectare = (1000) m². (since 1 hecta metre = 100 m)

1 hectare = (100 x 100) m²

(iv) 1 acre = 100 m².

(v) If each side of a triangle is a cm, then its area = \(\frac {\surd 3 }{ 4 }\) a² cm².

**Question 21.**

**Solution:**

(i) False

Area of a triangle = \(\frac { 1 }{ 2 }\) x Base x Height

(ii) True

(iii) False

Area of a circle = πr²

(iv) True

Hope given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper are helpful to complete your math homework.

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