Test: Building With Bricks - 3 - Class 4

The triangles formed are.

A, B, C, D, E, F., G, H, BG, BE, CE and CG, i.e., 12 in number

GH symmetrically divides the figure.

All are closed figures with different number of sides and a closed figure is called a polygon.

Perimeter of the given figure = 44 cm.

Only one square must be shaded to divide the figure into two equal parts.

                    
The squares formed are 12, 34, 56, 78, 2367 and 12345678 i.e., 6 in number.

We have,
AB + GH + CD = FE
⇒ 3 cm + GH + 6 cm = 12 cm
⇒ GH = 12 cm - 9 cm = 3 cm

Perimeter of a figure is the sum of all of its sides.
Perimeter of I=5+2+5+2= 14cm Perimeter of II = 2+2+2=6cm
Perimeter of III = 4+4+4+4=16cm Hence III has highest perimeter,

GH symmetrically divides the figure.

In rectangle IJKL, we have
IJ = LM + MK (opposite sides are equal in a rectangle.) ⇒8=5+MK
∴ MK = 3 cm
In rectangle MNOP, we have
MP = NO ⇒ MK + KP = NO
⇒ 3 + KP = 12
⇒ KP = 9 cm.