Practical Geometry

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 = 60⁰

Step1-

Draw a line segment QR of length 5.5 cm.

 

 

Step 2-

At Q,draw QX making 60⁰ 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 = 100⁰ and mZXY = 30⁰

Ans-

Step 1-

Draw XY of length 6cm.

 

 

Step 2-

At point X, draw a ray XP making an angle of 30⁰ with line XY.

 

 

Step 3-

At point Y, draw a ray YQ making an angle of 100⁰ 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 105⁰ at point I.

 

 

Step 4-

Take 6.5 cm distance on IX and take point S on it. Draw an angle of 120⁰ at point S.

 

Step 5-

Draw an angle of 75⁰ at point M. Line will intersect SY at point T. Join the Point T and M.

MIST is required quadrilateral.

First draw 90⁰ angle at point M applying same method at point I. Now bisect the 30⁰ angle inside 90⁰. We get 75⁰. 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

NCERT Solutions Class 7

NCERT Solutions Class 8

Worksheet Class 7

Construction of 30° angle

Construction of 60° angle

Construction of 75° angle

Construction of 90° angle

Construction of 105° angle

Construction of 120° angle