Construct the following quadrilaterals.
(i) Quadrilateral LIFT
LI = 4 cm
IF = 3 cm
TL = 2.5 cm
LF = 4.5 cm
IT = 4 cm
(ii) Quadrilateral GOLD
OL = 7.5 cm
GL = 6 cm
GD = 6 cm
LD = 5 cm
OD = 10 cm
(iii) Rhombus BEND
BN = 5.6 cm
DE = 6.5 cm
(i) A rough sketch of this quadrilateral:
(a) Δ ITL can be constructed as follows:
(b) Vertex F is 4.5 cm away from vertex L and 3 cm away from vertex I.
Hence, taking L and I as centers, draw arcs of 4.5 cm radius and 3 cm radius respectively, which will intersect each other at point F
(c) Join F to T and F to I
LIFT is the required quadrilateral
(ii) A rough sketch of this quadrilateral:
(a) Δ GDL can be constructed as follows:
(b) Vertex O is 10 cm away from vertex D and 7.5 cm away from vertex L.
Hence, taking D and L as centers, draw arcs of 10 cm radius and 7.5 cm radius respectively These will intersect each other at point O
(c) Join O to G and L
GOLD is the required quadrilateral
(iii) We know that the diagonals of a rhombus always bisect each other at 90°
So in this case let us assume these are intersecting each other at point O
Therefore,
EO = OD = 3.25 cm
A rough sketch of this rhombus:
(a) Draw a line segment BN of 5.6 cm and its perpendicular bisector. Let it will be intersecting the line segment BN at point O
(b) Taking O as centre, draw arcs of 3.25 cm radius to intersect the perpendicular bisector at point D and E
(c) Join points D and E to points B and N
BEND is the required quadrilateral