MATHS 3 Marks Question

Arranging the two pieces in such a way that they form a new shape with the desired perimeter:

Total length of boundary = AB + BC + CD + DE + EF + FG + GH + HA
= 2 + 1 + 2 + 6 + 2 + 1 + 2 + 6
= 22 cm

Smallest Perimeter (12 units): Only the 3 × 3 square achieves this.
Largest Perimeter (20 units): Only the straight line achieves this or any pattern made by folding two straight lines.

Perimeter of 18 units: Multiple shapes can achieve this. For example, a T-shaped figure or other L-shaped configurations.

The reasoning is based on the arrangement of the unit squares and the number of exposed edges. The more compact the shape, the smaller the perimeter; the more elongated, the larger the perimeter. For intermediate perimeters like 18 units, various configurations can be created by adjusting the arrangement of the squares.

Inner Track (100 m per side)
Perimeter Calculation: The perimeter of the inner track is (4 times 100 = 400) meters.
Distance to Run: The runner on the inner track needs to run 350 meters.
Starting Position (A): Since the perimeter is 400 meters, the runner will start 50 meters before the common finish line (400 – 350 = 50 meters).
Outer Track (150 m per side)
Perimeter Calculation: The perimeter of the outer track is (4 times 150 = 600) meters.
Distance to Run: The runner on the outer track also needs to run 350 meters.
Starting Position (B): Since the perimeter is 600 meters, the runner will start 250 meters before the common finish line (600 – 350 = 250 meters).

As given, the Area of the rectangle of size 12 units × 8 units = 96 sq units
And the area of an inner rectangle that occupies exactly half the area = 48 sq units

For shape A, can be arrange it as 9 units by 2 units, giving a perimeter of 22 units.

For shape B, can be arrange it as 5 units by 4 units, giving a perimeter of 18 units.

Dimensions of rectangle 1: 5 m × 10m
Dimensions of rectangle 2:2m × 7m
Area of rectangle 1 = 50 sq m
Area of rectangle 2 = 14 sq m
Now, area of rectangle = sum of areas of rectangle 1 and 2 = 50 sq m + 14 sq m = 64 sq m
So possible dimensions of a rectangle with area 64 sq m are 1 m × 64 m; 2 m × 32 m; 4 m × 16 m; 8 m × 8 m, etc.

When arranging the 7 pieces to form a rectangle, the area of the rectangle will be the same as that of area of square.

Area of rectangle = 16 × area of shape C.

Let’s say the area of C = x
Area of D = Area of 2C = 2x
Area of E = Area of C = x
Area of F = Area of 2C = 2x
Area of G = Area of 2C = 2x
Area of A = Area of 2F = 2 × 2x = 4x
Area of B = Area of A = 4x
Hence total area of big shape = Area of A + B + C + D + E + F + G
= 4x + 4x + x + 2x + x + 2x + 2x
= 16x
= 16C
That means the area of a big square is 16 times the area of shape C.

Shape D is twice as big as shape C. This means that if you place two shape C pieces together. Then, they exactly cover shape D.
The relationship between these shapes
Shape D can be completely filled by combining shape C and shape E. So, area of shape D is equal to the sum of the area of shape C and E.
Each of shapes C and E has half the area of shape D.

Perimeter of a rectangular field = 2 (length + breadth)
= 2 (230 m + 160 m)
= 780 m
The farmer wants 3 rounds of rope to fence.
Total length of rope needed = 780 m × 3
= 2340 m