The document NCERT Solutions(Part- 3)- Practical Geometry Class 8 Notes | EduRev is a part of the Class 8 Course Class 8 Mathematics by VP Classes.

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**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 âˆ 90Â° 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 âˆ REX = 90Â°.

**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 âˆ QPX = 90Â°

**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.

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