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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
Asked by Harsh Last Modified
Answer (i) A rough sketch of this quadrilateral can be drawn as follows.
(1) Δ ITL can be constructed by using the given measurements as follows.
(2) Vertex F is 4.5 cm away from vertex L and 3 cm away from vertex I. Therefore, while taking L and I as centres, draw arcs of 4.5 cm radius and 3 cm radius respectively, which will be intersecting each other at point F.
(3) Join F to T and F to I.
|
|
| LIFT is the required quadrilateral. |
(ii)A rough sketch of this quadrilateral can be drawn as follows.
(1) Δ GDL can be constructed by using the given measurements as follows.
(2) Vertex O is 10 cm away from vertex D and 7.5 cm away from vertex L. Therefore, while taking D and L as centres, draw arcs of 10 cm radius and 7.5 cm radius respectively. These will intersect each other at point O.
(3) 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 90º. Let us assume that these are intersecting each other at point O in this rhombus.
Hence, EO = OD = 3.25 cm
A rough sketch of this rhombus can be drawn as follows.
(1) Draw a line segment BN of 5.6 cm and also draw its perpendicular bisector. Let it intersect the line segment BN at point O.
(2) Taking O as centre, draw arcs of 3.25 cm radius to intersect the perpendicular bisector at point D and E.
(3) Join points D and E to points B and N.
|
|
| BEND is the required quadrilateral. |
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