RBSE Solutions for Class 8 Maths Chapter 7 Construction of Quadrilaterals Ex 7.4 is part of RBSE Solutions for Class 8 Maths. Here we have given Rajasthan Board RBSE Class 8 Maths Chapter 7 Construction of Quadrilaterals Exercise 7.4.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 7 |

Chapter Name |
Construction of Quadrilaterals |

Exercise |
Exercise 7.4 |

Number of Questions |
5 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 8 Maths Chapter 7 Construction of Quadrilaterals Ex 7.4

Question 1.

Construct a quadrilateral KLMN, in which KL = 4.8 cm., LM = 5.0 cm., MN = 3.8 cm., ∠L = 72°, ∠M = 105°.

Solution

First of all we draw a(RBSESolutions.com)rough sketch of the given measurements :

Steps of Construction

- First of all, draw a line segment KL = 4.8 cm.
- At point, L construct ∠KLX = 72°. We cut an arc of 5.0 cm on LX from point L. Mark the point as M.
- AT point M(RBSESolutions.com)construct ∠LMN = 105° and cut arc MN = 3.8 cm on MY. Mark the point as N.
- Join KN.

Thus, KLMN is a required quadrilateral.

Question 2.

Construct a quadrilateral ABCD in which AB = BC = 3 cm., AD = 5 cm., ∠A = 90° and ∠B = 105°.

Solution

First of all(RBSESolutions.com)we draw a rough sketch of the given measurements :

Steps of Construction

- First of all we draw a line segment AB = 3 cm.
- At point(RBSESolutions.com)A we construct ∠BAX = 90° with AB and cut AD = 5.0 cm.
- Similarly, we construct at point B, ∠ABY = 105° with AB and an arc of 3 cm. on BY is cut. Mark this point as C.
- Join CD.

Thus, we obtained the required quadrilateral ABCD.

Question 3.

Construct a quadrilateral PQRS in which QR = 3.6 cm., RS = 4.5 cm., PS = 5.0 cm., ∠R = 75° and ∠S = 120°.

Solution

First of all we draw a rough sketch of given measurements :

Steps of Construction

- First of all we draw a line segment of line RS = 4.5 cm.
- At point R we construct ∠SRX = 75° with SR with the help of pencil and compass and

cut an arc 3.6 cm from R on RX mark this point as Q. - Similarly, we(RBSESolutions.com)construct an ∠RSY = 120° with RS with the help of pencil and compass from point SR cut an arc of 5 cm. is cut mark this point as P.
- Join PQ.

Thus, PQRS is a required quadrilateral.

Question 4.

Construct a quadrilateral DEAR in which DE = 4.7 cm., EA = 5.0 cm., AR = 4.5 cm., ∠E = 60° and ∠A = 90°.

Solution

First of all we draw a rough sketch of given measurements :

Steps of Construction

- Draw a line of segment DE = 4.7 cm.
- From point E, a ray EX in drawn making an angle of 60° with DE from point E an arc of 5.0 cm is cut on EX and marked the point as A.
- Similarly, at A, a ray AY is drawn making an angle of 90°. Cut an arc of 4.5 cm on AY and marked the point as R.
- Join DR.
- Thus, we obtained(RBSESolutions.com)a required quadrilateral DEAR.

Question 5.

Construct a quadrilateral ABCD in which ∠B = 135°, ∠C = 90°, BC = 5.0 cm., AB = 9.0 cm. and CD = 7.0 cm.

Solution

First of all we draw a rough sketch of given measurements :

Steps of Construction

- First of all a line segment, AB = 9.0 cm indrawn.
- Construct ∠ABC = 135° at B with BX cut an arc of 5.0 cm from BX. Mark this points as C.
- Also at(RBSESolutions.com)point C, make an angle of 90° by a ray CY and cut an arc of 7.0 cm mark this point at D.
- Join AD.

- Thus, we obtained a required quadrilateral ABCD.

We hope the given RBSE Solutions for Class 8 Maths Chapter 7 Construction of Quadrilaterals Ex 7.4 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 7 Construction of Quadrilaterals Exercise 7.4, drop a comment below and we will get back to you at the earliest.