Chapter 4. Practical Geometry

Explanation:

Steps of construction:

1. First we draw ∆ABC using the given measurements A 7cm 4.5cm B 5.5cm C

2. Now taking A as centre, we draw an arc of radius 6cm and construct AD

3. Taking C as the centre, we now draw an arc of radius 4 cm, cutting the previous arc at point D. and then we join D yo A and C.

Hence, ABCD is the required quadrilateral.

Explanation:

1) ∆ JUP can be constructed by using the given measurements as follows.

2. Taking P and U as centres, draw arcs of radii 5 cm and 4 cm, respectively. Let the point of intersection be M.

(3) Join M to P and U.

JUMP is the required quadrilateral.

Explanation:

Since, the opposite sides of a parallelogram is of same length, we can say that OR=ME=6cm and MO=ER=4.5cm

(1) First we construct ∆ EOR can by using the given measurements.

2. Vertex M is 4.5 cm away from vertex O and 6 cm away from vertex E. Therefore, taking O and E as centres, we draw arcs of 4.5 cm radius and 6 cm radius, respectively. These will intersect each other at point M.

(3) Join M to O and E.

MORE is the required parallelogram.

Explanation:

Since, all sides of a rhombus are of the same length. We can say that BE = ES = ST = TB

1. First we construct ∆ BET by using the given measurements as follows.

2. Vertex S is 4.5 cm away from vertex E and also from vertex T. Therefore, while taking E and T as centres, draw arcs of 4.5 cm radius, which will intersect each other at point S.

(3) Join S to E and T.

BEST is the required rhombus.

Explanation:

1. First we construct ∆ ITL by using the given measurements as follows.

(2)Vertex F is 4.5 cm from vertex L and 3 cm from vertex I. ∴ With L and I as centers, draw arc s of radius 4.5 cm and radius 3 cm respectively, intersecting at point f.

3) Join F to T and I

LIFT is the required quadrilateral.

Explanation:

1. First we draw ∆ GDL by using the given measurements as follows.
   

(2)Vertex O is 10 cm from vertex D and 7.5 cm from vertex L. Therefore, with D and L as cente rs, draw arcs with radii of 10 cm and 7.5 cm respectively. They will intersect at point O.

(3) Join O to G and L.

GOLD is the required quadrilateral.

Explanation:

We know that the diagonals of a rhombus always intersect at 90º.

Suppose they intersect at point O of this rhombus. So, EO = OD = 3.25 cm.

1. First we draw a line segment BN of 5.6 cm, and its perpendicular bisector. Let it intersect the line segment BN at point O.

2. Taking O as the centre, draw arcs of 3.25 cm radius to intersect the perpendicular bisector at points D and E.

(3) Join points D and E to points B and N.

BEND is the required quadrilateral.

Explanation:

1) At point O draw a line segment MO of 6 cm and an angle of 105º. Since vertex R is 4.5 cm fr om vertex O, cut a line segment OR at 4.5 cm from this radius.

(2) Again, draw an angle of 105º at point R.

(3) Draw a 60º angle at point M. Let this ray intersect the ray R previously drawn at point E.

MORE is the required quadrilateral.

Explanation:

The sum of the angles of a quadrilateral is 360°. In quadrilateral PLAN,

∠P+∠L+∠A+∠N=360°

90° + ∠L + 110° + 85° = 360°

285° + ∠L = 360°

∠L = 360° − 285° = 75°

1. Draw a line segment PL of 4 cm and an angle of 75º at point L. Since vertex A is 6.5 cm f rom vertex L, cut a line segment LA at 6.5 cm from this radius

(2) Again, draw an angle of 110º at point A.

3. Draw an angle of 90º at point P. This ray will meet the previously drawn ray from A at point N.
   

PLAN is the required quadrilateral.

Explanation:

1. Draw a line segment HE of 5 cm and an angle of 85º at point E. As vertex A is 6 cm away from vertex E, cut a line segment EA of 6 cm from this ray.

(2)Vertex R is 6 cm and 5 cm from vertices H and A respectively, draw arcs from point H an d point A with radii of 6 cm and 5 cm respectively. They will intersect at point R.

(3) Join R to H and A.

HEAR is the required quadrilateral.

Explanation:

1. At point K, draw a line segment OK of 7 cm and an angle of 90º. Since vertex A is 5 cm from vertex K, cut a line segment KA of 5 cm from this radius.

(2)Vertex Y is 5 cm and 7 cm from vertices O and A respectively, draw arcs of O and A with ra dii of 5 cm and 7 cm respectively. They will intersect at point Y.

(3) Join Y to A and O.

OKAY is the required quadrilateral.

Explanation:

1. At point E draw a line segment DE of 4 cm and an angle of 60º. Since vertex A is 5 cm from vertex E, cut a 5 cm segment EA from this radius

(2)Again draw a 90º angle at point A. Since vertex R is 4.5 cm from vertex A, cut a 4.5 cm seg-ment RA from this radius.

3) Join D to R

DEAR is the required quadrilateral.

Explanation:

(1)At point U draw a line segment RU of 3 cm and an angle of 120º. Since vertex E is 4 cm from vertex U, cut a segment UE of 4 cm from this radius.

2. Then draw an angle of 75 degrees at point R, since vertex T is 3.5 cm from vertex R, cut a line segment whose RT is 3.5 cm from this beam.

(3) Join T to E.

TRUE is the required quadrilateral.

Explanation:

All the sides of a square are of the same length, and also, all the interior angles of a square are 90 degree measure. Therefore, the given square READ can be drawn as follows.

(1) Draw a line segment RE of 5.1 cm and an angle of 90º at points R and E.

(2)Since vertices A and D are 5.1 cm from vertices E and R, respectively, they intersect line seg ments EA and RD, each 5.1 cm from these radii.

(3) Join D to A.

READ is the required square.

Explanation:

In a rhombus, diagonals bisect each other at 90º.

• The given rhombus ABCD can be drawn as follows.

1. Draw a line segment AC of 5.2 cm and draw its perpendicular bisector. Let it intersect the line segment AC at point O.

2. Draw an arc of 6.4/2 = 3.2 on either side of this center line. Let the arc intersect the center line at points B and D.

(3) Join points B and D with points A and C.

ABCD is the required rhombus.

Explanation:

Opposite sides of a rectangle are of same lengths, and also, all the interior angles of a rectangle are of 90 degree. The given rectangle ABCD can be drawn as follows.

(1) Draw a line segment AB of 5 cm and an angle of 90º at points A and B.

2. As vertex C and D are 4 cm away from vertex B and A, respectively, cut line segments AD and BC, each of 4 cm, from these rays.

(3) Join D to C.

ABCD is the required rectangle.

Explanation:

We know, opposite sides of a parallelogram are equal and parallel to each other. The given parallelogram OKAY can be drawn as follows.

(1) Draw a line segment OK of 5.5 cm and a ray at point K at a convenient angle.

2. Draw a ray at point O parallel to the ray at K. As the vertices A and Y are 4.2 cm away from the vertices K and O, respectively, cut line segments KA and OY, each of 4.2 cm, from these rays.

(3) Join Y to A.

OKAY is the required rectangle.