Practical Geometry

 Mind Maps

Class VII - maths: Practical Geometry
Q) Definition of Dimension?

Q) Definition of Points?

Q) Definition of Plane or Plane Surface?

Q) Definition of Solid Geometry?

Q) Definition of Line?

Q) Definition of Practical Geometry?

Q) How many sides a rectangle has?

Q) Name the straight line which joins two opposite corners of a square, rectangle or other straight sided shape?

Q) A four sided polygon is called?

Q) A simple closed curve made up of only line segments is called?

Q) What is a six sided quadrilateral?

Q) Draw a triangle ABC such that BC = 5 cm, AB = 6.5 cm, AC = 7.5 cm?

Q) (a)Draw a line segment of length 3.9 cm using a ruler?
(b)Construct a copy of a given segment?

Q) Construct a ?ABC, in which ?B = 700, AB = 4.8 cm and BC = 5.2 cm.

Q) Construct an isosceles triangle PQR such that PQ = PR = 7cm and P = 1400?

Q) Construct a triangle ABC in which BC = 6 cm, ∠B = 35° and ∠C = 100°?

Q) The angle sum of a convex polygon with number of sides n is?

Q) What is the number of sides of a quadrilateral?

Q) The angle sum of a convex polygon with no of sides 10 is?

Q) How many diagonals does a regular hexagon have?

Q) How many diagonals does a regular hexagon have?

Q) Draw a triangle ABC such that BC = 6.5 cm, AB = 5.5 cm, AC = 7 cm.

Q) Theorem on parallel lines and plane are explained step-by-step along with the converse of the theorem?

Q) If two straight line are both perpendicular to a plane then they are parallel?

Q) (a) Construct a right triangle PQR in which ?Q = 105° , PR = 8 cm and QR = 5 cm. (b)Construct a triangle ABC in which BC = 5 cm, ∠B = 65° and ∠C = 90° ?

Q) Given a line l and a point not on it, We used of some ruler and compasses construction. transversal diagram to draw a line parallel to l. We could also have used the idea of 'equal corresponding angles' to do the construction?

Q) (a) Construct a right-angle triangle whose hypotenuse is 8 cm long and one of the legs is 5 cm long. (b) Construct an isosceles right-angled triangle ABC, where m∠ACB = 9° and AC = 5 cm.

Q) The angle sum of a convex polygon with number of sides n is?

Q) What is the number of sides of a quadrilateral?

Q) The angle sum of a convex polygon with no of sides 10 is?

Q) How many diagonals does a regular hexagon have?

Q) Construct quadrilaterals PQRS in which PQ=6,QR=5,RS=4,QS=7,PS=5?

Q) If the straight line joining two points on a surface lies wholly on the surface then the surface is called a?

Q) If length of a side of a rhombus is 6cm, then the perimeter of the rhombus is ?

• Construction of parallel lines
• Theorem on Parallel Lines and Plane
• Converse of the theorem on parallel lines and plane
• Solid Geometry
• Definitions on solid geometry terms
• Angle Between Two Skew Lines
• Orthogonal Projection
• Angle Between a Straight Line and a Plane
• Dihedral Angle
• Construction of Triangles
• Construction of Line Segment
• SSS Construction
• SSS Construction of Δ
• SAS Construction
• SAS Triangle Construction
• ASA Triangle Construction
• RHS Construction
• RHS Triangle Construction
• In this Chapter, we looked into the methods of some ruler and compasses constructions.
1. Given a line l and a point not on it, we used the idea of ‘equal alternate angles’ in a transversal diagram to draw a line parallel to l. We could also have used the idea of ‘equal corresponding angles’ to do the construction.
• We studied the method of drawing a triangle, using indirectly the concept of congruence of triangles.
The following cases were discussed:
(i) SSS: Given the three side lengths of a triangle.
(ii) SAS: Given the lengths of any two sides and the measure of the angle between these sides.
(iii) ASA: Given the measures of two angles and the length of side included between them.
(iv) RHS: Given the length of hypotenuse of a right-angled triangle and the length of one of its legs.
Construction of parallel lines

Theorem on Parallel Lines and Plane

Theorem on parallel lines and plane are explained step-by-step along with the converse of the theorem.

Theorem: If two straight lines are parallel and if one of them is perpendicular to a plane, then the other is also perpendicular to the same plane.

Let PQ and RS be two parallel straight lines of which PQ is perpendicular to the plane XY. We are to prove that the straight line RS is also perpendicular to the plane XY.

Construction: Let us assume straight line PQ and RS intersect the plane XY at Q and S respectively.
Join QS. Evidently, QS lies in the XY plane.
Now, through S draw ST perpendicular to QS in the XY plane. Then, join QT, PT and PS.

Proof: By construction, ST is perpendicular to QS. Therefore, from the right-angled triangle QST we get,

QT2 = QS2 + ST2 ..............(1)

• Since PQ is perpendicular to the plane XY at Q and the straight lines QS and QT lie in the same plane, therefore PQ is perpendicular to both the lines QS and QT.Therefore, from the right-angle PQS We get,
PS2 = PQ2 + QS2 ..............(2)
And from the right-angle PQT we get,
PT2= PQ2 + QT2= PQ2+ QS2 + ST2 [using (1)]
or, PT2 = PS2 + ST2 [using (2)]
• Therefore, ∠PST = 1 right angle. i.e., ST is perpendicular to PS. But by construction, ST is perpendicular to QT.

Thus, ST is perpendicular to both PS and QS at S.Therefore, ST is perpendicular to the plane PQS, containing the lines PS and QS.

Now, S lies in the plane PQS and RS is parallel to PQ; hence, RS lies in the plane of PQ and PS i.e., in the plane PQS. Since ST is perpendicular to the plane PQS at S and RS lies in this plane, hence ST is perpendicular to RS i.e., RS is perpendicular to ST.
Again, PQ and RS are parallel and ∠PQS = 1 right angle.
Therefore, ∠RSQ = 1 right angle i.e., RS is perpendicular to QS. Therefore, RS is perpendicular to both QS and ST at S; hence, RS is perpendicular to the plane containing QS and ST i.e., perpendicular to the XY.

Converse of the theorem on parallel lines and plane

If two straight lines are both perpendicular to a plane then they are parallel.

Let two straight lines PQ and RS be both perpendicular to the plane XY. We are to prove that the lines PQ and RS are parallel.

Following the same construction as in theorem on parallel lines and plane

• it can be proved that ST is perpendicular to PS.
• Since, RS is perpendicular to the plane XY, hence RS is perpendicular to TS, a line through S in the plane XY i.e.,
• TS is perpendicular to RS. Again, by construction, TS is perpendicular QS.
• Therefore, TS is perpendicular to each of the straight lines QS, PS and RS at S.
• Hence, QS, PS and RS are co-planar (by theorem on co-planar).
• Again, PQ, QS and PS are co-planar (Since they lie in the plane of the triangle PQS).
• Thus, PQ and RS both lie in the plane of PS and QS i.e.,
• PQ and RS are co-planar.

Again, by hypothesis,
∠PQS = 1 right angle and ∠RSQ = 1 right angle.
Therefore, ∠PQS + ∠RSQ = 1 right angle + 1 right angle = 2 right angles.

Solid Geometry

Definitions on solid geometry terms

(i) Dimension: Each of length, breadth and thickness of any body is called a dimension of the body.

(ii) Point: A point has no dimension, that is, it has neither length nor breadth nor thickness ; it has position only.

(iii) Line: A line has length only but no breadth and thickness. Therefore, a line has one dimension, that is, it is one dimensional.

(iv) Surface: A surface has length and breadth but no thickness. Therefore, a surface has two dimensions, that is, it is two dimensional.

(v) Solid: A solid has length, breadth and thickness. Therefore, a solid has three dimensions, that is, it is three dimensional.

The book is a solid, each of its six faces is a surface, each of its edges is a line and each of its corners is a point.

A line is bounded by points, a surface is bounded by lines and a solid is bounded by surfaces. In other words, a line is generated by the motion of a point, a surface is generated by the motion of a line and a solid is generated by the motion of a surface.

(vi) Solid Geometry: The branch of geometry which deals with the properties of points, lines, surfaces and solids in three dimensional space is called solid geometry.

(vii) Plane or Plane Surface: If the straight line joining two points on a surface lies wholly on the surface then the surface is called a plane surface or a plane.

A straight line may be extended indefinitely in either direction, that is, straight lines are supposed to be of infinite length. Similarly, planes are also assumed to be of infinite extent, unless otherwise stated. The statement that a straight line lies wholly on a surface signifies that every point on the line (however produced in both directions) lies on the surface.

A surface is called curved surface when it is not a plane surface.

(i) Lines or points are said to be co-planar if they lie on the same plane; in other words, lines or points are co-planar if a plane can be made to pass through them.

(ii) Two co-planar straight lines are either parallel or they intersect at a point. Two straight lines are said to be parallel when they are co-planar and they do not meet however indefinitely they are produced in both directions.

(iii) Two straight lines are said to be skew (or non-coplanar) if a plane cannot be made to pass through them. In other words, two straight lines are said to be skew when they do not meet at a point and they are not parallel.

(iv) Two planes are said to be parallel if they do not meet when extended infinitely in all directions.

(v) A straight line is said to be parallel to a plane if they do not meet when both are produced infinitely.

In the given picture we observe that, the lines LM, MN, NO and OL lie in the plane LMNO, that is, they are co-planar. The lines LM and LO meet at L and the lines LM and ON are parallel. LP and MN are skew lines and the line QR is parallel to the plane LPSO. The planes ABFE and DCGH are parallel.

A straight line is said to be perpendicular to a plane if it is perpendicular to every straight lines drawn in the plane through the point where the line meets the plane.

A straight line perpendicular to a plane is called a normal to the plane. In the given figure, the straight line OP meets the plane XY at O ; if OP is perpendicular to every straight lines OI, OJ, OK, OD etc. drawn through O in the XY plane then OP is perpendicular (or a normal) to the plane XY.

A straight line parallel to the direction of a plumb-line hanging freely at rest is called a vertical line. A plane which is perpendicular to a vertical line is called the horizontal plane. A straight line drawn in a horizontal plane is called a horizontal line.

Angle Between Two Skew Lines:

The angle between two skew lines (i.e., two non-co-planar straight lines) is measured by the angle between one of them and a straight line drawn parallel to the other through a point on the first line. In the given figure, let MN and QR be two skew straight lines. Take any point O on the line MN and draw the straight line OP parallel to QR through O. Then ∠NOP gives the measure of the angle between the skew straight lines MN and QR.

A triangle is a plane figure since all its three sides lie in one plane. Similarly a parallelogram is also a plane figure. But a quadrilateral may or may not be plane figures since all its four sides always do not lie in one plane. A quadrilateral whose two adjacent sides lie in one plane and other two adjacent sides lie in a different plane is called a skew quadrilateral.

Orthogonal Projection:

(a) If a perpendicular be drawn from an external point on a given line then the foot of the perpendicular is called the orthogonal projection (or simply the projection) of the external point on the given line.

In the above left side figure, Pp is perpendicular from the external point P on the straight line AB. Since the foot of the perpendicular is p hence, p is the projection of P on the on the line AB. Again, we can observe that in the above right side figure, 5 the point P lies on the line AB ; hence, in this case the projection of P on AB is the point P itself.

(b) The locus of the feet of the perpendiculars drawn from all points of a line (straight or curved) on a given straight line is called the projection of the line on the given straight line.

In the above left side figure, Pp and Qq are perpendiculars from P and Q respectively on the straight line AB; p and q are the respective feet of perpendiculars. Then, pq is the projection of the straight line PQ on the straight line AB. Again, previous left hand side figure, the projection of the straight line PQ on the straight line AB is Pq. Again, similarly we can observe that in the above right side figure, pq is the projection of the curved line PQ on the straight line AB. Again, suppose the straight line PQ intersects the straight line AB at R; in this case, the projections of QR and RP on AB are qR and Rp respectively.

(c) The locus of the feet of the perpendiculars drawn from all points of a line (straight or curved) on a given plane is called the projection of the line on the plane. In this figure, the locus of the feet of the perpendiculars drawn from points of the line MN on the plane XY is the line mn; hence, the projection of the line MN on the plane XY is the line mn.

#### Note:

1. The projection of a straight line on a plane is a straight line ; but the projection of a curved line on a plane may be a straight line as well as a curved line. If the curved line MN lies in a plane which is perpendicular to the plane XY then the projection of MN on the plane XY is a straight line.
2. A straight line and its projection on a plane are co-planar.

Angle Between a Straight Line and a Plane

The angle between a straight line and a plane is measured by the angle between the given straight line and its projection on the given plane. Let mn be the projection of the straight line MN on the plane XY. Suppose, in the given figure, straight lines MN and mn (when produced) meet at the point R in the plane XY. Then the angle between the straight line MN and the plane XY is measured by ∠MRm.

Dihedral Angle

The plane angle between two intersecting planes is called a dihedral angle between the planes. The angle between two intersecting planes (i.e., a dihedral angle) is measured as follows: Take any point on the line of intersection of the two planes. From this point draw two straight lines, one in each plane, at right angles to the line of intersection. Then the plane angle between the drawn two lines gives the measure of the dihedral angle between the planes.

Let XY and LM be two intersecting planes and XL be their line of intersection. From any point A on XL draw the straight line AB perpendicular to XL in the XY plane and the straight line AC perpendicular to XL in the LM plane. Then the plane ∠BAC is the measure of the dihedral angle between the two intersecting planes XY and LM.

If the dihedral angle between two intersecting planes is a right angle then one plane is said to be perpendicular to the other.

Construction of Triangles

#### To Construct a Triangle whose Three Sides are given:

To construct a triangle whose three sides are given; let us follow the examples.

1. Draw a triangle ABC such that BC = 4 cm, AB = 3.5 cm, AC = 4.5 cm.
[Let us draw a rough sketch of the triangle ABC. Name the vertices and show the given measures.]

Steps of Construction:

Now let’s follow these steps one by one
(i) Draw a line segment BC = 4 cm.
(ii) With B as Centre and radius 3.5 cm, draw an arc.
(iii) With C as centre and radius 4.5 cm, draw another arc cutting the previous arc at A.
(iv) Join AB and AC.

Therefore, Δ ABC is the required triangle.

2. Draw a triangle ABC such that BC = 4.5 cm, AB = 3.5 cm, AC = 5 cm.
[Let us draw a rough sketch of the triangle ABC. Name the vertices and show the given measures.]

Steps of Construction:

Now let’s follow these steps one by one
(i) Draw a line segment BC = 4.5 cm.
(ii) With B as Centre and radius 3.5 cm, draw an arc.
(iii) With C as centre and radius 5 cm, draw another arc cutting the previous arc at A.
(iv) Join AB and AC.

Therefore, Δ ABC is the required triangle.

Construction of Line Segment

1). Draw a line segment of length 3.9 cm using a ruler.

Solution :

Step 1: Draw a line l. Mark a point on the line l.

Step 2: Place the compasses pointer on the zero mark of the ruler. Open it to place the pencil point up to 3.9 cm.

Step 3: Taking caution that the opening of the compasses has not changed, place the pointer on A and swing an arc to cut l at B.

Step 4: segment AB is a line segment of required length.

2). Construct a copy of a given segment.

Step 1: Draw line segment CD of any length.

Step 2: Fix the compasses pointer on C and the pencil end on D. Now this opening of the compasses gives the length of segment CD.

Step 3: Draw a line l. Mark a point A on the line l. Without changing the compasses setting, place the pointer on A.

Step 4: Cut an arc that cuts l at a point B. Now, AB is a copy of CD.

SSS Construction

Construction Of Triangle

Note that it is not sufficient to know only three angles.

In order to be able to construct a Δ from three segments, one of the segments must be shorter than the sum of the other two:
a < b+c, or b < a+c, or c < a+b

SSS Construction of Δ

Construct ΔXYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6cm.
Step 1. Draw a line YZ of length 5 cm.
Step 2. From Y, point X is at a distance of 4.5 cm. So, with Y as center, draw an arc of radius 4.5 cm.
Step 3. From Z, point X is at a distance of 6 cm. So, with Z as center, draw an arc of radius 6 cm.
Step 4. X has to be on both the arcs drawn. So, it is the point of intersection of arcs. Mark the point of interaction of arcs as X. Join XY and XZ. ΔXYZ is the required triangle as shown in the figure.

SAS Construction

Triangle given two sides and included angle (SAS)
Multiple triangles possible
It is possible to draw more than one triangle has the side lengths and angle measure as given. Depending on which line you start with, which end of the line you draw the angles, and whether they are above or below the line, the four triangles below are possible. All four are correct in that they satisfy the requirements, and are congruent to each other.

#### Printable step-by-step instructions

The above animation is available as a printable step-by-step instruction sheet, which can be used for making handouts or when a computer is not available.

#### Proof

The image below is the final drawing above with the red items added.

Argument Reason
1 Line Segment MN Is Congruent To AB. Drawn With The Same Compass Width.For Proof See Copying A Line Segment
2 Line Segment ML Is Congruent To AC. Drawn With The Same Compass Width.
3 The Angle LMN Is Congruent To The Angle A Copied Using The Procedure Shown In Copying An
Angle.See That Page For The Proof
4 Triangle LNM satisfies the side lengths and angle measure given.

SAS Triangle Construction

#### Δ Construction

1.Construct a ΔABC, in which ∠B = 70°, AB = 4.8 cm and BC = 5.2 cm.

Before doing construction, draw a rough sketch, so that you will get an idea which side is taken as a base.Here after drawing a rough sketch, we are sure that the base is AB = 4.8 cm.

Step 1 : Draw AB = 4.8 cm.
Step 2 : Using protractor, draw ∠ABK = 70° .
Step 3 : On the line segment BK, cut off BC = 5.2 cm.
Step 4 : Join A and C.

ΔABC is the required triangle.

2.Construct an isosceles triangle PQR such that PQ = PR = 5cm and P = 110°.

Before doing construction, draw a rough sketch, so that you will get an idea which side is taken as a base.Here after drawing a rough sketch, we are sure that the base is AB = 4.8 cm.

Step 1 : Draw PR = 5 cm.

Step 2 : Using protractor, draw ∠QPM = 110°.

Step 3 : On the line segment PM, cut off PQ = 5 cm.

Step 4 : Join Q and R.

ΔPQR is the required triangle.

ASA Triangle Construction

To construct a triangle when two of its angles,say B and C, and the included side BC are given,we proceed as follows :

Step 1 : Draw a line segment BC.

Step 2 : Draw ∠CBX of measure equal to that of ∠B.

Step 3 : Draw ∠BCY with Y on the same side of BC as X, such that its measure is equal to that of ∠C. Let BX and CY intersect at A. Then, ΔABC is the required triangle.

ΔPQR is the required triangle.

#### Example

Construct a ΔABC in which BC = 6 cm, ∠B = 35° and ∠C = 100°.

Step 1 : Draw a line segment BC = 6 cm.

Step 2 :Draw ∠CBX, such that ∠CBX = 35°

Step 3 : Draw ∠BCY with Y on the same side of BC as X such that ∠BCY = 100°

Step 4 : Let BX and CY intersect at A.

ΔABC is the required triangle.

RHS Construction

#### HL Triangle Construction

To construct a right triangle ABC right angled at C when its hypotenuse is AC and one side is BC is given, we follow the following steps :

Step 1 : Draw a line segment BC of given length.
Step 2: Draw ∠BCX of measure 90°.
Step 3 : With center B and radius equal to the hypotenuse AB, draw an arc of the circle to intersect ray CX at A.
Step 4 : Join BA to obtain the required triangle ABC

#### Example

RHS Triangle Construction

Construct a right triangle PQR in which ∠Q = 90° , PR = 6 cm and QR = 4 cm.

Step 1: Draw QR = 4 cm.

Step 2 : Using compass, at Q draw ∠RQK = 90° .

Step 3 : Cut off RP = 6 cm.

Step 4 : Join P and R.

Now we get the required ΔPQR.

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