**Notes of chapter: Practical Geometry are presented below. Indepth notes along with worksheets and NCERT Solutions.**

**Practical Geometry-**

A branch of mathematics which is related to different shapes, angles, lines and their properties is known as geometry. The use of properties of geometry is called** practical geometry.**

**(1)**Construction of a parallel line using ruler and compasses only

**Step 1-**

Draw a line ‘l’ and took a point ‘A’ outside ‘l’.

**Step 2-**

Take any point B on l and join B to A.

**Step 3-**

With B as centre and a convenient radius, draw an arc cutting l at C and BA at D.

**Step 4-**

Now with A as centre and the same radius as in Step 3, draw an arc EF cutting AB at G.

**Step 5-**

Place the pointed tip of the compasses at C and adjust the opening so that the pencil tip is at D.

**Step 6-**

With the same opening as in Step 5 and with G as centre, draw an arc cutting the arc EF at H.

**Step7-**

Join AH and draw a line m.

**(2)Construction of Triangles**

**(i) Constructing a triangle when the lengths of its three sides are known (SSS Criterion)**

**Eg.:- **Construct a triangle ABC, given that AB = 5cm, BC = 6 cm and AC = 7 cm.

**Ans-**

**Step 1-**

Draw a line segment BC of length 6 cm.

**Step 2 –**

With B as center, draw an arc of radius 5 cm.

**Step 3-**

With C as center, draw an arc of radius 7 cm.

**Step 4-**

Mark the point of intersection of arcs as A. Join AB and AC.ABC is required triangle.

**(ii)Constructing a triangle when the lengths of two sides and the measure of the angle between them are known (SAS Criterion) **

**Eg.-**Construct a triangle PQR, given that PQ = 3 cm, QR = 5.5 cm and PQR = 60^{0}

**Step1-**

Draw a line segment QR of length 5.5 cm.

**Step 2-**

At Q,draw QX making 60^{0} with QR.

**Step 3-**

With Q as center, draw an arc of radius 3 cm. It cuts QX at the point P.

** Step 4-**

Join PR, ΔPQR is mow obtained.

**(iii)Constructing a triangle when the measures of two of its angles sand the length of the side included between them is given.(ASA criterion)**

**Eg- **Construct XYZ if XY = 6cm, mXYZ = 100^{0} and mZXY = 30^{0}

**Ans-**

**Step 1-**

Draw XY of length 6cm.

**Step 2-**

At point X, draw a ray XP making an angle of 30^{0} with line XY.

**Step 3-**

At point Y, draw a ray YQ making an angle of 100^{0} with line YX.

**Step 4-**

Z has to lie on both the rays XP and YQ. So, the point of intersection of the two rays is Z.

ΔXYZ is required triangle.

**(iv)Constructing a right- angled triangle when the length of one leg and its hypotenuse are given (RHS Criterion)**

**Eg- **Construct LMN, right-angled at M, given that LN = 5cm and MN = 3cm.

**Ans-**

**Step1-**

Draw MN of length 3 cm.

**Step 3-**

With N as center, draw an arc of radius 5 cm.

**Step 4-**

L has to be on the perpendicular line MX as well as on the arc drawn with centre N.

Therefore,

L is the meeting point of these two.

ΔLMN is now obtained.

**(3) Construction of quadrilateral**

**(i) When the lengths of four sides and a diagonal are given**

**Eg :- **Construct a quadrilateral PQRS where PQ = 4 cm, QR = 6cm, RS = 5cm, PS = 5.5cm and PR = 7cm.

**Step 1-**

Draw rough sketch of required quadrilateral PQRS.

**Step 2-**

Draw diagonal PR = 7cm.

**Step 3-**

Draw an arc of 4 cm with P as a centre.

**Step 4-**

Draw an arc of 6 cm with R as a centre.

**Step 5-**

Mark point of intersection as Q and join point Q to point P and R.

**Step 6-**

Draw an arc of 5.5cm with P as a centre in other side of the diagonal.

**Step 7-**

Draw an arc of 5 cm with R as centre.

**Step 8-**

Mark point of intersection as S. Join point S to point P and R.

We get required quadrilateral PQRS.

**(ii) When two diagonals and three sides are given**

**Eg :-** Construct a quadrilateral ABCD, given that BC = 4.5 cm, AD = 5.5 cm, CD = 5 cm the diagonal AC = 5.5 cm and diagonal BD = 7cm.

**Step 1-**

Draw rough sketch of required quadrilateral ABCD.

**Step 2-**

Draw a diagonal AC = 5.5cm.

**Step 3-**

Draw an arc of 5.5 cm with A as centre.

**Step 4 –**

Draw an arc of 5 cm with C as a centre.

**Step 5 –**

Mark point of intersection as D. Join the point D with point A and point C.

**Step 6- **

Draw an arc of 7 cm with D as a centre.

**Step 7 –**

Draw an arc of 4.5 cm with C as centre.

**Step 8-**

Mark point of intersection as B. Join point B to point A, C and D.

ABCD is required quadrilateral.

**(iii)When two adjacent sides and three angles are given**

**Eg:- **Construct a quadrilateral MIST where MI = 3.5 cm, IS = 6.5 cm, M = 75º ,

∠I = 105º and ∠S = 120º

**Step 1-**

Draw rough sketch of required quadrilateral MIST.

**Step 2 –**

Draw a line MI = 3.5cm.

**Step 3-**

Draw an angle of 105^{0} at point I.

**Step 4-**

Take 6.5 cm distance on IX and take point S on it. Draw an angle of 120^{0} at point S.

**Step 5-**

Draw an angle of 75^{0} at point M. Line will intersect SY at point T. Join the Point T and M.

MIST is required quadrilateral.

First draw 90^{0} angle at point M applying same method at point I. Now bisect the 30^{0} angle inside 90^{0}. We get 75^{0}. Join the line and make quadrilateral MIST.

**(iv) When three sides and two included angles are given**

**Eg:-** Construct a quadrilateral ABCD, where AB = 4cm,BC = 5cm, CD = 6.5 cm and ∠B = 105º and ∠C = 80º .

**Step 1-**

Draw rough sketch of required quadrilateral ABCD.

**Step 2-**

Draw a line BC of 5 cm length. Draw an angle of 105º at point B.Join point B to A at the length of 4 cm.

**Step 3-**

Draw an angle of 80º at point C. Draw an arc of 6.5 cm from point C. Mark this point as D.

**Step 4-**

Join point D to point A. Quadrilateral ABCD is required quadrilateral.

**Helping Topics**