RBSE Solutions for Class 8 Maths Chapter 7 Construction of Quadrilaterals Ex 7.5 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 7.5.

Board |
RBSE |

Textbook |
SIERT, Rajasthan |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 7 |

Chapter Name |
Construction of Quadrilaterals |

Exercise |
Exercise 7.5 |

Number of Questions |
4 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 8 Maths Chapter 7 Construction of Quadrilaterals Ex 7.5

Question 1.

Construct a quadrilateral MORE, in which MO = 6 cm., OR = 4.5 cm., ∠M = 60°, ∠O = 105° and ∠R = 105°.

Solution

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

Steps of Construction

(i) First of all we draw a line segment MO = 6 cm.

(ii) At point M and O construct ∠MOY 105° and ∠OMX = 60° with MO respectively.

(iii) Cut an(RBSESolutions.com)arc of 4.5 cm from point O on ray OY and mark the point as R.

(iv) At point M construct ∠OME = 60° with OM. Intersecting MX at E.

(v) Thus we. obtained required quadrilateral MORE.

Question 2.

Construct a quadrilateral ABCD, in which ∠B = 70°, ∠A = 54°, ∠D = 105°, AB = 6.2 cm. and AD = 5.7 cm.

Solution

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

Steps of Construction

(i) First of all draw a line segment AB = 6.2 cm.

(ii) From A, a ray AX is drawn making an angle of 54° with line segment AB, cut an arc of 5.7 cm on AX and marked the point as D.

(iii) From(RBSESolutions.com)point D, making an angle of 105° a ray DZ is drawn.

(iv) From point B, a ray BY is drawn, making an angle of 70° with cut ray DZ at point C.

(v) Thus, ABCD is a required quadrilateral.

Question 3.

Construct a quadrilateral PQRS in which ∠P = 75°, ∠Q = 85°, ∠R = 110°, PQ = 4.1 cm. and QR = 3.9 cm.

Solution

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

Steps of Construction

(i) First we draw a line segment PQ = 4.1 cm.

(ii) At point P, construct ∠QPX = 75° with PQ.

(iii) At(RBSESolutions.com)point Q, construct ∠PQY = 85° with PQ and cut QR = 3.9 cm from QY.

(iv) At point R, construct ∠QRZ = 110° with QR intersecting PX at point S.

(v) Thus, we obtained a quadrilateral PQRS.

Question 4.

Construct a quadrilateral DNSI, in which DN = 2.5 cm., NS = 3.7 cm. and ∠I = 60°, ∠N = 120°, ∠S = 90°.

Solution

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

We know that sum of all angles of a quadrilateral = 360°

∴ ∠D + ∠N + ∠S + ∠I= 360°

∴ ∠D = 360° – (120° + 90° + 60°)

= 360° – 270° = 90*

∴ ∠D = 90°

Steps of Construction

(i) First of all draw a line segment NS = 3.7 cm.

(ii) At(RBSESolutions.com)point N and S construct ∠SNX = 120° and ∠NSY = 90° with N.S respectively.

(iii) Cut DN = 2.5 cm from NX.

(iv) At point D, construct ∠NDZ = 90° with DN and intersecting SY at I.

Thus, we obtained required quadrilateral DNSI

We hope the given RBSE Solutions for Class 8 Maths Chapter 7 Construction of Quadrilaterals Ex 7.5 will help you. If you have any query regarding Rajasthan Board RBSE Class 8 Maths Chapter 7 Construction of Quadrilaterals Exercise 7.5, drop a comment below and we will get back to you at the earliest.

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