NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5.

- Practical Geometry Class 8 Ex 4.1
- Practical Geometry Class 8 Ex 4.2
- Practical Geometry Class 8 Ex 4.3
- Practical Geometry Class 8 Ex 4.4

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Practical Geometry |

Exercise |
Ex 4.5 |

Number of Questions Solved |
1 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5

**Question 1.**

**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. Is it unique?

**Solution.**

**1. Steps of Construction**

- Draw RE = 5.1 cm.
- At R, draw a ray RX such that ∠ERX

- From ray RX, cut RD = 5.1 cm.
- At E, draw a ray EY such that ∠REY = 90°.
- From ray EY, cut EA = 5.1 cm.
- Join AD.

Then, READ is the required square.

**2. Steps of Construction**

[We know that the diagonals of a rhombus bisect each other at right angles. So in rhombus ABCD, the diagonals AC and BD will bisect each other at right angles.]

- Draw AC = 5.2 cm.
- Construct its perpendicular bisector. Let it intersect AC at O.

- Cut off \(\frac { 6.4 }{ 2 } \)= 3.2 cm lengths on either side of the bisector drawn in step 2, we get B and D.
- Join AB, BC, CD, and DA.

Then, ABCD is the required rhombus.

**3. Steps of Construction**

[We know that each angle of a rectangle is 90°. So, in rectangle PQRS,

∠P=∠Q=∠R=∠S= 90°.

Also, opposite sides of a rectangle are parallel.

So, in rectangle PQRS,

PQ || SR and PS || QR]

- Draw PQ = 5 cm.
- At Q, draw a ray QX such that ∠PQX = 90°.

- From ray QX, cut QR = 4 cm.
- At P, draw a ray PY parallel to QR.
- At R, draw a ray RZ parallel to QP to meet the ray drawn in step 4 at S.

Then, PQRS is the required rectangle.

**4. Steps of Construction**

[We know that in a parallelogram, opposite sides are parallel and equal. So,

OK = YA and OK || YA;

KA = OY and KA || OY]

- Draw OK = 5.5 cm.

- At K, draw a ray KX at any suitable angle from OK.
- From ray KX, cut KA = 4.2 cm.
- A, draw a ray AT parallel to KO.
- At O, draw a ray OZ parallel to KA to cut the ray drawn in step 4 at Y.

Then, OKAY is the required parallelogram.

This is not unique.

Note: We can construct countless parallelograms with these dimensions by varying ∠OKA

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