# Model Question Paper-5 Class-6 ML Aggarwal ICSE Maths Solutions

Model Question **Paper-5 Class-6 ML Aggarwal** ICSE Maths Solutions . APC Understanding Mathematics for ICSE Class-6 Model Question Paper-5 Solutions based on Chapter-12 to 14. Visit official Website **CISCE** for detail information about ICSE Board **Class-6** Mathematics.

## Model Question **Paper-5 Class-6 ML Aggarwal** ICSE Maths Solutions

( based on chapter-12 to 14)

Time Allowed-1 hour

max mark-25

**Note :**

(i) All Questions are compulsory

(ii) Questions 1-2 carry 1 mark each, Questions 3-5 carry 2 marks each, Questions 6-8 carry 3 marks each and Questions 9-10 carry 4 marks each.

### Paper-5 Class-6 ML Aggarwal

**Choose the correct answer from the given four options (1-2):**

Question 1.

The number of lines of symmetry of a protractor is

(a) 0

(b) 1

(c) 2

(d) unlimited

**Answer**

1 (b)

Question 2.

If the perimeter of a regular pentagon is 60 cm, then its each side is

(a) 10 cm

(b) 12 cm

(c) 15 cm

(d) 20 cm

**Answer**

Perimeter of pentagon = 5 × Side

60 = 5 × Side

∴ Side = = 12 (b)

Question 3.

If the perimeter of a square is 42 cm, then find its area.

**Answer**

Perimeter of square = 42 cm

= 4 × Side

∴ 4 × Side = 42 cm

Side = cm

Area of square = (side)^{2}

=

= = 110.25 cm^{2}

Question 4.

Using a ruler and compass, construct an angle of 90°.

**Answer**

Steps of construction:

- Draw a line AB.
- Place the point of the compass at A and draw an at C.
- The point where arc meets AB, name it as P.
- Place compass at P and make an arc of radius PA

to cut the previous arc and name that point as Q. - With the compass point at Q.

Draw the arc with the same radius and mark that point R. - With the same radius, draw two arcs, intersecting at point T.
- Join AT and external that line to C.
- Hence, ∠CAB = 90°.

Question 5.

On a squared paper, sketch a quadrilateral with exactly two lines of symmetry. Also, sketch the lines of symmetry.

**Answer**

Question 6.

If the area of a rectangular plot is 396 sq. m and its breadth is 18 m. Find the length of the plot and the cost of fencing it at the rate of ₹7.50 per metre.

**Answer**

Area of plot 396 sq. m

Length (l) = ?

Breadth (b) = 18m

Area = l × b

⇒ 396 = l × 18

⇒ l =

⇒ l = 22 m

Cost of fencing is ₹7.50 per metre

Perimeter of field = 2(l + b) = 2(22 + 18) m = 80 m

∴ Cost of fencing = 80 × ₹7.50 = ₹600

Question 7.

Draw a line segment AB of length 6.4 cm. Take a point P on AB such that AP = 4.5 cm. Draw a perpendicular to AB at P. (use ruler and compass).

**Answer**

Steps of construction :

- Draw a line AB = 6.4 cm.
- Mark point P, such that AP = 4.5 cm.
- With a certain radius, draw an arc at point P.
- With the same radius,

mark an arc on the previous arc draw from point P, name that point - Again with same radius,

mark an arc from point Q, name that point = R. - From point Q, R draw an arc with same radius and

the point where it intersect, name it as S. - Join PS and extend it to T.
- ∠TPB = 90°.

Question 8.

Copy the given figure. How many lines of symmetry it has? Draw its all lines of symmetry.

**Answer**

Four lines of symmetry.

Question 9.

In the given figure, all adjacent sides are at right angles. Find:

(i) the perimeter of the figure.

(ii) the area enclosed by the figure.

**Answer**

Perimeter of given figure,

= AB + BC + CD + DE + EF + FA

= 7 + 4 + CD + DE + 4 + 9

= 11 +CD + DE+ 13

= 11 + 3 + (9.4) + 13

[As GB = CD and GB = (7 – 4) = 3 cm]
Also, DE = GE – DG = 9 – 4 = 5 cm = 14 + 5 + 13 = 32 cm

Area of figure = Area of AGEF + Area of GBCD

= (EF × AF) + (CD × BC)

= (9 × 4) + (3 × 4)

= (36 + 12) cm^{2}

= 48 cm^{2}

Copy the given figure on a squared paper and complete the figure such that the resultant figure is symmetrical about both the dotted lines.

**Answer**

— End of Model Question **Paper-5 Class-6 ML Aggarwal** Solutions :–

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