**Practical Geometry**- It covers all the fundamentals of

**drawing parallel lines**and some types of

**triangles**. The chapter first talks about the

**construction of a line parallel to a given**

**line.**There are two methods of doing the same:

1. Through

**paper folding**when the

**point is not on the line**

2. Using

**ruler and compasses**only

The topic is explained stepwise with a well labelled diagrammatic description.

Exercise 10.1 is based on the same concept.

Later the focus is given on the concepts involved in the

**construction of triangles.**

Explanation of the following topics with the help of useful figures are covered in a comprehensive manner:

**Constructing a triangle when the lengths of its three sides are known (SSS criterion).****Constructing a triangle when the lengths of two sides and the measure of the angle between them are known(SAS criterion).****Constructing a triangle when the measures of two of its angles and the length of the side included between them is given(ASA criterion).****Constructing a Right-Angled Triangle when the length of one leg and its hypotenuse are given(RHS criterion).**

A short overview of the chapter is given at the end for quick revision.

#### Page No 196:

#### Question 1:

Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.

#### Answer:

The steps of construction are as follows.

(i)Draw a line AB. Take a point P on it. Take a point C outside this line. Join C to P.

(ii)Taking P as centre and with a convenient radius, draw an arc intersecting line AB at point D and PC at point E.

(iii) Taking C as centre and with the same radius as before, draw an arc FG intersecting PC at H.

(iv) Adjust the compasses up to the length of DE. Without changing the opening of compasses and taking H as the centre, draw an arc to intersect the previously drawn arc FG at point I.

(v) Join the points C and I to draw a line ‘*l*’.

This is the required line which is parallel to line AB.

##### Video Solution for integers (Page: 196 , Q.No.: 1)

NCERT Solution for Class 7 math - integers 196 , Question 1

#### Page No 196:

#### Question 2:

Draw a
line *l*. Draw a perpendicular to *l* at any point on *l*.
On this perpendicular choose a point X, 4 cm away from* l*.
Through X, draw a line *m* parallel to *l*.

#### Answer:

The steps of construction are as follows.

(i) Draw a line *l* and take a point P on line *l*. Then,
draw a perpendicular at point P.

(ii) Adjusting the compasses up to the length of 4 cm, draw an arc to
intersect this perpendicular at point X. Choose any point Y on line
*l*. Join X to Y.

(iii) Taking Y as centre and with a convenient radius, draw an arc
intersecting *l* at A and XY at B.

(iv) Taking X as centre and with the same radius as before, draw an arc CD cutting XY at E.

(v)Adjust the compasses up to the length of AB. Without changing the opening of compasses and taking E as the centre, draw an arc to intersect the previously drawn arc CD at point F.

(vi) Join
the points X and F to draw a line *m*.

Line *m* is the required line which is parallel to line *l*.

#### Page No 196:

#### Question 3:

Let *l*
be a line and P be a point not on* l*. Through P, draw a line *m*
parallel to *l*. Now join P to any point Q on *l*. Choose
any other point R on *m*. Through R, draw a line parallel to PQ.
Let this meet *l* at S. What shape do the two sets of parallel
lines enclose?

#### Answer:

The steps of construction are as follows.

(i)Draw a line *l* and take a point A on it. Take a point P not on *l* and join A to P.

(ii) Taking A as centre and with a convenient radius, draw an arc cutting *l* at B and AP at C.

(iii)Taking P as centre and with the same radius as before, draw an arc DE to intersect AP at F.

(iv) Adjust the compasses up to the length of BC. Without changing the opening of compasses and taking F as the centre, draw an arc to intersect the previously drawn arc DE at point G.

(v)Join P to G to draw a line *m*. Line *m* will be parallel to line *l*.

(vi)Join P to any point Q on line *l*. Choose another point R on line *m*. Similarly, a line can be drawn through point R and parallel to PQ.

Let it meet line *l* at point S.

In quadrilateral PQSR, opposite lines are parallel to each other.

PQ || RS and PR || QS

Thus, the quadrilateral PQSR is a parallelogram.

#### Page No 199:

#### Question 1:

Construct ΔXYZ in which XY = 4.5 cm, YZ = 5 cm and ZX = 6 cm.

#### Answer:

The rough figure of this triangle is as follows.

The required triangle is constructed as follows.

(i) Draw a line segment YZ of length 5 cm.

(ii) Point X is at a distance of 4.5 cm from point Y. Therefore, taking point Y as centre, draw an arc of 4.5 cm radius.

(iii) Point X is at a distance of 6 cm from point Z. Therefore, taking point Z as centre, draw an arc of 6 cm radius. Mark the point of intersection of the arcs as X. Join XY and XZ.

XYZ is the required triangle.

#### Page No 199:

#### Question 2:

Construct an equilateral triangle of side 5.5 cm.

#### Answer:

An equilateral triangle of side 5.5 cm has to be constructed. We know that all sides of an equilateral triangle are of equal length. Therefore, a triangle ABC has to be constructed with AB = BC = CA = 5.5 cm.

The steps of construction are as follows.

(i) Draw a line segment BC of length 5.5 cm.

(ii) Taking point B as centre, draw an arc of 5.5 cm radius.

(iii) Taking point C as centre, draw an arc of 5.5 cm radius to meet the previous arc at point A.

(iv) Join A to B and C.

ABC is the required equilateral triangle.

#### Page No 199:

#### Question 3:

Draw ΔPQR with PQ = 4 cm, QR = 3.5 cm and PR = 4 cm. What type of

triangle is this?

#### Answer:

The steps of construction are as follows.

(i) Draw a line segment QR of length 3.5 cm.

(ii) Taking point Q as centre, draw an arc of 4 cm radius.

(iii) Taking point R as centre, draw an arc of 4 cm radius to intersect the previous arc at point P.

(iv) Join P to Q and R.

ΔPQR is the required triangle. As the two sides of this triangle are of the same length (PQ = PR), therefore, ΔPQR is an isosceles triangle.

##### Video Solution for integers (Page: 199 , Q.No.: 3)

NCERT Solution for Class 7 math - integers 199 , Question 3

#### Page No 199:

#### Question 4:

Construct ΔABC such that AB = 2.5 cm, BC = 6 cm and AC = 6.5 cm. Measure ∠B.

#### Answer:

The steps of construction are as follows.

(i) Draw a line segment BC of length 6 cm.

(ii) Taking point C as centre, draw an arc of 6.5 cm radius.

(iii) Taking point B as centre, draw an arc of radius 2.5 cm to meet the previous arc at point A.

(iv) Join A to B and C.

ΔABC is the required triangle. ∠B can be measured with the help of protractor. It comes to 90º.

#### Page No 200:

#### Question 1:

Construct ΔDEF such that DE = 5 cm, DF = 3 cm and m∠EDF = 90°.

#### Answer:

The rough sketch of the required ΔDEF is as follows.

The steps of construction are as follows.

(i)Draw a line segment DE of length 5 cm.

(ii) At point D, draw a ray DX making an angle of 90° with DE.

(iii) Taking D as centre, draw an arc of 3 cm radius. It will intersect DX at point F.

(iv) Join F to E. ΔDEF is the required triangle.

#### Page No 200:

#### Question 2:

Construct an isosceles triangle in which the lengths of each of its equal sides

is 6.5 cm and the angle between them is 110°.

#### Answer:

An isosceles triangle PQR has to be constructed with PQ = QR = 6.5 cm. A rough sketch of the required triangle can be drawn as follows.

The steps of construction are as follows.

(i) Draw the line segment QR of length 6.5 cm.

(ii) At point Q, draw a ray QX making an angle 110° with QR.

(iii) Taking Q as centre, draw an arc of 6.5 cm radius. It intersects QX at point P.

(iv) Join P to R to obtain the required triangle PQR.

#### Page No 200:

#### Question 3:

Construct ΔABC with BC = 7.5 cm, AC = 5 cm and m∠C = 60°.

#### Answer:

A rough sketch of the required triangle is as follows.

The steps of construction are as follows.

(i) Draw a line segment BC of length 7.5 cm.

(ii) At point C, draw a ray CX making 60º with BC.

(iii) Taking C as centre, draw an arc of 5 cm radius. It intersects CX at point A.

(iv) Join A to B to obtain triangle ABC.

##### Video Solution for integers (Page: 200 , Q.No.: 3)

NCERT Solution for Class 7 math - integers 200 , Question 3

#### Page No 202:

#### Question 1:

Construct ΔABC, given m∠A = 60°, m∠B = 30° and AB = 5.8 cm.

#### Answer:

A rough sketch of the required ΔABC is as follows.

The steps of construction are as follows.

(i)Draw a line segment AB of length 5.8 cm.

(ii)At point A, draw a ray AX making 60º angle with AB.

(iii) At point B, draw a ray BY, making 30º angle with AB.

(iv) Point C has to lie on both the rays, AX and BY. Therefore, C is the point of intersection of these two rays.

This is the required triangle ABC.

#### Page No 202:

#### Question 2:

Construct ΔPQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°.

(Hint: Recall angle sum property of a triangle).

#### Answer:

A rough sketch of the required ΔPQR is as follows.

In order to construct ΔPQR, the measure of ∠RPQ has to be calculated.

According to the angle sum property of triangles,

∠PQR + ∠PRQ + ∠RPQ = 180º

105º + 40º + ∠RPQ = 180º

145º + ∠RPQ = 180º

∠RPQ = 180° − 145° = 35°

The steps of construction are as follows.

(i) Draw a line segment PQ of length 5 cm.

(ii) At P, draw a ray PX making an angle of 35º with PQ.

(iii) At point Q, draw a ray QY making an angle of 105º with PQ.

(iv)Point R has to lie on both the rays, PX and QY. Therefore, R is the point of intersection of these two rays.

This is the required triangle PQR.

##### Video Solution for integers (Page: 202 , Q.No.: 2)

NCERT Solution for Class 7 math - integers 202 , Question 2

#### Page No 202:

#### Question 3:

Examine whether you can construct ΔDEF such that EF = 7.2 cm, m∠E =

110° and m∠F = 80°. Justify your answer.

#### Answer:

Given that,

m∠E = 110° and m∠F = 80°

Therefore,

m∠E + m∠F = 110° + 80° = 190°

However, according to the angle sum property of triangles, we should obtain

m∠E + m∠F + m∠D = 180°

Therefore, the angle sum property is not followed by the given triangle. And thus, we cannot construct ΔDEF with the given measurements.

Also, it can be observed that point D should lie on both rays, EX and FY, for constructing the required triangle. However, both rays are not intersecting each other. Therefore, the required triangle cannot be formed.

#### Page No 203:

#### Question 1:

Construct the right angled ΔPQR, where m∠Q = 90°, QR = 8 cm and PR = 10 cm.

#### Answer:

A rough sketch of ΔPQR is as follows.

The steps of construction are as follows.

(i) Draw a line segment QR of length 8 cm.

(ii) At point Q, draw a ray QX making 90º with QR.

(iii) Taking R as centre, draw an arc of 10 cm radius to intersect ray QX at point P.

(iv) Join P to R. ΔPQR is the required right-angled triangle.

##### Video Solution for integers (Page: 203 , Q.No.: 1)

NCERT Solution for Class 7 math - integers 203 , Question 1

#### Page No 203:

#### Question 2:

Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4 cm long.

#### Answer:

A right-angled triangle ABC with hypotenuse 6 cm and one of the legs as 4 cm has to be constructed. A rough sketch of ΔABC is as follows.

The steps of construction are as follows.

(i) Draw a line segment BC of length 4 cm.

(ii) At point B, draw a ray BX making an angle of 90º with BC.

(iii) Taking C as centre, draw an arc of 6 cm radius to intersect ray BX at point A.

(iv) Join A to C to obtain the required ΔABC.

#### Page No 203:

#### Question 3:

Construct an isosceles right-angled triangle ABC, where, m∠ACB = 90° and AC = 6 cm.

#### Answer:

In an isosceles triangle, the lengths of any two sides are equal.

Let in ΔABC, AC = BC = 6 cm. A rough sketch of this ΔABC is as follows.

The steps of construction are as follows.

(i) Draw a line segment AC of length 6 cm.

(ii) At point C, draw a ray CX making an angle of 90º with AC.

(iii) Taking point C as centre, draw an arc of 6 cm radius to intersect CX at point B.

(iv) Join A to B to obtain the required ΔABC.

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