Question 1. How will you construct a rectangle PQRS if you know only the lengths PQ and QR?
Solution: A rectangle can be constructed by taking PQ as the length.
Making an ∠90° at Q and cutting off QR, the breadth the ray Remaining two points R and S can be located by taking P and R as centres and radii as QR and PQ respectively draw arcs to intersect at S. Thus, PQRS is the required parallelogram.
Question 2. Construct the kite EASY if AY = 8 cm, EY = 4 cm and SY = 6 cm. Which properties of the kite did you use in the process?
Solution: Following properties have been used in constructing the KITE:
(i) Diagonals are at right angles.
(ii) One of the diagonal bisects the other.
(iii) Pairs of consecutive sides are equal.
Steps of construction:
I. Draw a line segment AY = 8 cm.
II. Draw , the perpendicular bisector of AY such that it meets AY at O.
III. We cannot locate a point E on PQ at 4 cm from Y and A, i.e. EY = 4 cm = EA is not possible. It is possible only when E and O coincide. In that case the kite does not exist.
EXERCISE 4.5
Question: Draw the following.
1. The square READ with RE = 5.1 cm.
2. A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
3. A rectangle with adjacent sides of lengths 5 cm and 4 cm.
4. A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm.
Solution:1. Steps of construction:
I. Draw a line segment RE = 5.1 cm.
II. At E, draw , such that ∠REX = 90°.
III. From , cut-off = 5.1 cm.
IV. With centre at A, draw an arc above RE of radius = 5.1 cm.
V. With centre at R, and radius = 5.1 cm, draw another arc to intersect the previous arc at D.
VI. Join DA and DR.
Thus, READ is the required square.
2. Steps of construction:
Note: The diagonals of a rhombus bisect each other at right angles.
I. Draw a line segment AC = 5.2 cm.
II. Draw , the perpendicular bisector of AC.
III. From XY, cut-off OD
IV. Similarly, cut-off OB
V. Join AD, DC, CB and BA.
Thus, ABCD is the required rhombus.
3. Steps of construction:
I. Draw a line segment PQ = 5 cm.
II. At P, draw , such that ∠QPX = 90°
III. From , cut-off PS = 4 cm.
IV. With centre at 5 and radius = 5 cm, mark an arc towards Q.
V. With centre Q and radius = 4 cm, mark an arc to intersect the previous arc at R.
VI. Join RQ and RS
Thus, PQRS is the required rectangle.
4. Steps of construction:
I. Draw a line segment OK = 5.5 cm.
II. At K, draw a ray .
III. From , cut-off KA = 4.2 cm.
IV. With centre at A and radius = 5.5 cm, draw an arc above OK.
V. With centre O and radius = 4.2 cm, draw another arc to intersect the previous arc at Y.
VI. Join YO and YA.
Thus, OKAY is the required parallelogram.
1. What are the different types of triangles based on their sides? |
2. What are the different types of triangles based on their angles? |
3. How can we construct an equilateral triangle using a compass and a ruler? |
4. How can we construct a perpendicular bisector of a line segment? |
5. How can we construct an angle bisector of an angle? |
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