ML Aggarwal Class 6 Solutions for ICSE Maths Model Question Paper 5

ML Aggarwal Class 6 Solutions for ICSE Maths Model Question Paper 5

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
Solution:
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
Solution:
Perimeter of pentagon = 5 × Side
60 = 5 × Side
∴ Side = \(\frac{60}{5}\) = 12 (b)

Question 3.
If the perimeter of a square is 42 cm, then find its area.
Solution:
Perimeter of square = 42 cm
= 4 × Side
∴ 4 × Side = 42 cm
Side = \(\frac{42}{4}=\frac{21}{2}\) cm
Area of square = (side)²
= \(\left(\frac{21}{2}\right)^{2}=\frac{21}{2} \times \frac{21}{2}\)
= \(\frac{441}{4}\) = 110.25 cm²

Question 4.
Using a ruler and compass, construct an angle of 90°.
Solution:
Steps of construction:

1. Draw a line AB.
2. Place the point of the compass at A and draw an at C.
3. The point where arc meets AB, name it as P.
4. Place compass at P and make an arc of radius PA
   to cut the previous arc and name that point as Q.
5. With the compass point at Q.
   Draw the arc with the same radius and mark that point R.
6. With the same radius, draw two arcs, intersecting at point T.
7. Join AT and external that line to C.
8. Hence, ∠CAB = 90°.
   

Question 5.
On a squared paper, sketch a quadrilateral with exactly two lines of symmetry. Also, sketch the lines of symmetry.
Solution:

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.
Solution:
Area of plot 396 sq. m
Length (l) = ?
Breadth (b) = 18m
Area = l × b
⇒ 396 = l × 18
⇒ l = \(\frac{396}{18}\)
⇒ 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).
Solution:
Steps of construction :

1. Draw a line AB = 6.4 cm.
2. Mark point P, such that AP = 4.5 cm.
3. With a certain radius, draw an arc at point P.
4. With the same radius,
   mark an arc on the previous arc draw from point P, name that point
5. Again with same radius,
   mark an arc from point Q, name that point = R.
6. From point Q, R draw an arc with same radius and
   the point where it intersect, name it as S.
7. Join PS and extend it to T.
8. ∠TPB = 90°.
   

Question 8.
Copy the given figure. How many lines of symmetry it has? Draw its all lines of symmetry.

Solution:

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.

Solution:
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) cm²
= 48 cm²

Question 10.
Copy the given figure on a squared paper and complete the figure such that the resultant figure is symmetrical about both the dotted lines.

Solution:

ML Aggarwal Class 6 Solutions for ICSE Maths