## RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.4

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.4

**Other Exercises**

- RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1
- RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.2
- RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.3
- RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.4
- RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5

**Question 1.**

**Construct a quadrilateral ABCD, in which AB = 6 cm, BC = 4 cm, CD = 4 cm, ZB = 95° and ∠C = 90°.**

**Solution:**

**Steps of construction :**

(i) Draw a line segment BC = 4 cm.

(ii) At B, draw a ray BX making an angle of 95° and cut off BA = 6 cm.

(iii) At C, draw a ray CY making an angle of 90° and cut off CD = 4 cm.

(iv) Join AD.

Then ABCD is the required quadrilateral.

**Question 2.**

**Construct a quadrilateral ABCD, where AB = 4.2 cm, BC = 3.6 cm, CD = 4.8 cm, ∠B = 30° and ∠C = 150°.**

**Solution:**

**Steps of construction :**

(i) Draw a line segment BC = 3.6 cm.

(ii) At B, draw a ray BX making an angle of 30° and cut of BA = 4.2 cm.

(iii) At C, draw another ray CY making an angle of 150° and cut off CD = 4.8 cm.

(iv) Join AD.

Then ABCD is the required quadrilateral.

**Question 3.**

**Construct a quadrilateral PQRS, in which PQ = 3.5 cm, QR = 2.5 cm, RS = 4.1 cm, ∠Q = 75° and ∠R = 120°.**

**Solution:**

**Steps of construction :**

(i) Draw a line segment QR = 2.5 cm.

(ii) At Q, draw a ray QX making an angle of 75° and cut off QP = 3.5 cm.

(iii) At R, draw another ray RY making an angle of 120° and cut off RS = 4.1 cm.

(iv) Join PS.

Then PQRS is the required quadrilateral.

**Question 4.**

**Construct a quadrilateral ABCD given BC = 6.6 cm, CD = 4.4 cm, AD = 5.6 cm and ∠D = 100° and ∠C = 95°.**

**Solution:**

**Steps of construction :**

(i) Draw a line segment CD = 4.4 cm.

(ii) At C, draw a ray CX making an angle of 95° and cut off CB = 6.6 cm

(iii) At D, draw another ray DY making an angle of 100° and cut off DA = 5.6 cm.

(iv) Join AB.

Then ABCD is the required quadrilateral.

**Question 5.**

**Construct a quadrilateral ABCD in which AD = 3.5 cm, AB = 4.4 cm, BC = 4.7 cm, ∠A = 125° and ∠B = 120°.**

**Solution:**

**Steps of construction :**

(i) Draw a line segment AB 4.4 cm.

(ii) At A, draw a ray AX making an angle of 125° and cut off AD = 3.5 cm.

(iii) At B, draw another ray BY making an angle of 120° and cut off BC = 4.7 cm.

(iv) Join CD.

Then ABCD is the required quadrilateral.

**Question 6.**

**Construct a quadrilateral PQRS in which ∠Q = 45°, ∠R = 90°, QR = 5 cm, PQ = 9 cm and RS = 7 cm.**

**Solution:**

**Steps of construction :**

This quadrilateral is not possible to construct as shown in the figure.

**Question 7.**

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

**Solution:**

**Steps of construction :**

(i) Draw a line segment AB = 3 cm.

(ii) At A, draw a ray AX making an angle of 90° and cut off AD = 5 cm.

(iii) At B, draw another ray BY making an angle of 105° and cut off BC = 3 cm.

(iv) Join CD.

Then ABCD is the required quadrilateral.

**Question 8.**

**Construct a quadrilateral BDEF where DE = 4.5 cm, EF = 3.5 cm, FB = 6.5 cm and ∠F = 50° and ∠E = 100°**

**Solution:**

**Steps of construction :**

(i) Draw a line segment EF = 3.5 cm.

(ii) At E, draw a ray EX making an angle of 100° and cut off ED = 4.5 cm.

(iii) At F, draw another ray FY making an angle of 45° and cut off FB = 6.5 cm.

(iv) Join DB.

Then BDEF is the required quadrilateral.

Hope given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.4 are helpful to complete your math homework.

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