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

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.

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.