Practical Geometry

Given: (i) a line AB

(ii) a point C outside it

To construct: a line parallel to AB passing through C

Steps for construction

Step 1: We have to draw a line segment AB and take a point c outside AB.

Step 2: We have to take any point D on AB and join C to D.

Step 3 : Taking D as a center , we have to draw an arc cutting AB at E and CD at F.

Step 4: Taking C as a center and same radius as in step 3, we have to draw an arc GH cutting CD at I.

Step 5 : We have to draw an arc cutting GH at J and taking I as center, keeping the length of the arc same as EF

Step 6: Join JC to draw a line L

Line l is parallel to AB

Given: (i) X is 4cm away from point I

To construct: a line m parallel to I

Steps for construction

Step 1: Draw a line L, take a point P on it.

Step 2: We will draw a semicircle of any radius, taking P as a center and mark the points A and B .

Step 3 : We have to draw an arc , taking A as a center and radius more than AP. Now with the same radius we will draw an arc taking B as center.

Step 4 : We have to mark the intersection points of the arc as Q, and join P and Q.

Step 5 : We have to mark a point X on PQ at 4cm distance from P. For this we will make an arc , taking P as center and radius 4cm

Step 7 : We will draw an arc , taking P as a center and intersecting l at C and PX at D.

Step 8 : We will draw an arc , taking X as a center and same radius , intersecting PX at E.

Step 9 : We have to measure the length of CD using the compass and draw an arc taking E as center and cutting at F.

Step 10 : We have to draw line m passing through X and F.

Thus, m is the required line

Given: (i) P is parallel to line l

To construct: a parallel line through R

Steps for construction

Step 1 : We have to draw a line l with point P not on line l .

Step 2 : We have to take a point B on line l an join them and then we have to draw an arc taking b as center , cutting l at C and PB at D

Step 3 : Taking the same radius we have to draw an arc with P as center and intersecting PB at E.

Step 4: We have to measure the length of CD using the compass and draw an arc taking E as center and cutting at F.

Step 5 : We have to draw a line m passing through P and F. Line l and m are parallel.

Step 6 : We have taken any point R on m and any point Q on l and join PQ.

Step 7 : Taking P as center , we will draw an arc intersecting line PQ at G and line m at H

Step 8 : We have to consider the same radius as before and taking R as center , we will draw an arc intersecting line m at I.

Step 9: We have to measure the length of GH using the compass and draw an arc taking I as center and cutting at K. We will draw a line joining R and K and intersecting line l at S

Now PQ is parallel to RS andd line l is parallel to line m

Therefore we can say that it is a parallelogram.

Given: (i) XY = 4.5cm

(ii) YZ = 5 cm

(iii) ZX = 6 cm

To construct: a XYZ

Steps for construction

Step 1 : First we will draw a line segment XY of length 4.5cm

Step 2 : We have to draw an arc , taking X as center and radius 6cm

Step 3 : We have to another arc, with radius 5cm and center at Y and mark the intersection of arcs as Z.

Step 4 : We have to join XZ and XY

This is the required triangle

Given: length of side of equilateral triangle is 5.5cm

To construct: an equilateral triangle

Steps for construction

Step 1: We have to draw a line segment AB as 5.5cm

Step 2 : We have to draw an arc , taking A as center and radius 5.5cm

Step 3 : We have to another arc, with radius 5.5cm and center at B and mark the intersection of arcs as C.

Step 4 : We have to join AC and BC

This is the required triangle

Given: (i) PQ = 4 cm

(ii) QR = 3.5 cm

(iii) PR = 4 cm

To construct: a PQR

Steps for construction

Step 1 : We have to draw a line segment PQ of 4cm

Step 2 : We have to draw an arc , taking P as center and radius 4cm

Step 3 : We have to another arc, with radius 3.5cm and center at Q and mark the intersection of arcs as R.

Step 4 : We have to join PR and RQ

This is the required triangle. Here we can see that PQ=PR. Therefore it is an isosceles triangle.

Given: (i) AB = 2.5cm

(ii) BC = 6 cm

(iii)AC = 6.5 cm

To construct: a ABC

Steps for construction

Step 1 : We have to draw a line segment AB of 2.5cm

Step 2 : We have to draw an arc , taking A as center and radius 6.5cm

Step 3 : We have to another arc, with radius 6cm and center at B and mark the intersection of arcs as C.

Step 4 : We have to join AC and BC

Here we on measuring the angle B is 90°

Given: (i) DE = 5cm

(ii) DF = 3 cm

(iii)

To construct: a ABC

Steps for construction

Step 1 : We have to draw a line segment DF as 3cm

Step 2 : We will draw from point D

Step 3 : We will draw an arc of 5 cm , taking D as center and cutting at point E

Step 4 : We will join EF to get the required triangle

Here,

According to the question,

We have to draw figure using following steps of construction:

Step 1: Draw a line segment AB of 6.5 cm

Step 2: Now, from point A draw a ray AC making an angle of 110° from QR (with the help of protractor)

Step 3: Now, taking A as a centre and radius 6.5 cm mark an arc intersecting the line drawn in previous step. Mark the intersecting point as C. Final figure is

Hence ΔABC is the required triangle.

Given: (i) BC = 7.5 cm

(ii) AC = 5 cm

To construct: a triangle

Steps for construction

Step 1 : We will draw a line segment CB of 7.5cm

Step 2 : We will draw from point C

Step 3 : We will take C as center and draw an arc of radius 5cm intersecting at point A.

Step 4 : We will join AB and we will get the required triangle.

Given: (i)

(ii)

(iii) AB = 5.8 cm

To construct: a triangle

Steps for construction

Step 1 : We will draw a line segment AB of 5.8 cm

Step 2 : We will draw from point A

Step 3 : Now, we will draw from point B and mark the intersection points of both the ray as C.

Step 4: The required triangle will be

Given: (i)

(ii)

(iii) PQ = 5 cm

To construct: a triangle

In we have,

We know that sum of angle in a triangle is always equals to

Now , we can draw the required triangle

Steps for construction

Step 1 : We will draw a line segment PQ of length 5cm

Step 2 : We will draw 35° using protractor

Step 3 : Now we will draw 105° from Q using protractor .

Step 4 : Now we will mrk the intersection point as R snd we will get the required triangle

Here,

According to the question,

It is given that,

∠E = 110° and ∠F = 80°

That shows that ∠E + ∠F = 110 + 80

= 190°

This is greater than 180°

And, we know that,

The sum of interior angles of triangle is 180°

Hence,

The given measurements cannot form a triangle.

We have to draw figure using following steps of construction:

Step 1: Draw a line segment EF of 7.2 cm

Step 2: Now, from point E draw a ray EX making an angle of 110° from EF.

Step 3: From F, draw a ray FY from EF making 80° angle

Now, we can observe that EX and FY does not intersect.

Hence,

The ΔDEF is not possible.

Given: (i)

(ii) QR = 8 cm

(iii) PR = 10 cm

To construct: a Right – Angled Triangle

Steps for construction

Step 1: We will draw a line segment QR of 8 cm

Step 2 : We will draw from point Q

Step 3 : We will draw an arc taking R as center and radius as 10 cm and mark the intersection point as P.

Step 4 : Now, we will join PR and we will get the required triangle

Given: (i) Hypotenuse is 6 cm

(ii) One leg is 4 cm

To construct: a Right – Angled Triangle

Steps for construction

Step 1 : We will draw a line segment BC of 4 cm

Step 2 : We will draw from point B

Step 3 : We will draw an arc from C , taking 6 cm as radius and mark the intersection point as A

Step 4 : We will join AC and we will get the required triangle .

Given: (i)

(ii) AC is 6 cm

To construct: an Isosceles Right Angled Triangle

Steps for construction

Step 1 : We will draw a line segment CA of length 6 cm

Step 2 : We will draw from point C

Step 3 : We will draw an arc from C , taking 6 cm as radius and mark the intersection point as B

Step 4 : We will join BA and we will get the required triangle .