# NCERT Class 7 Maths Practical Geometry

The chapter 10 begins with an introduction to Practical Geometry by explaining the stepwise procedure for Construction of a line parallel to the given line through a point not on the line.After this, the Construction of a Triangle and various categories under it such as construction of a triangle using SSS Criterion [the length of three sides are known], SAS Criterion [the measure of two sides and an angle is known ], and ASA Criterion [the measure of two angles and a side is known ] is described in detail. Constructing a right-angled triangle when the length of one leg and its hypotenuse are given [RHS Criterion] is the last topic discussed in this chapter.

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## Chapter 10 Ex.10.1 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.

**Solution**

**Video Solution**

**What is known?**

A line \(AB\) and a point \(C\) outside it.

**To construct:**

A line through \(C\) parallel to \(AB\) using ruler and compasses.

**Reasoning: **

Draw a line \(AB\) and take a point \(C\) outside it. Draw line \(AB\) by using ruler and compasses, follow the steps given below.

**Steps:**

** Steps of construction** –

- Draw a line, \(AB,\) take a point \(C\) outside this line. Take any point \(P\) on \(AB\). Join \(C\) to \(P\).
- Taking \(P\) as centre and a convenient radius draw an arc intersecting line \(AB\) at \(D\) and \(PC\) at \(E.\)
- Taking \(C\) as the centre and the same radius in previous step, draw an arc \(FG\) intersecting \(PC\) at \(H. \)
- 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 arc \(HG\) at point \(I.\)
- Join the point \(C\) and \(I\) to draw a line \(L.\)

This is the required line \(L\) which is parallel to \(AB.\)

## Chapter 10 Ex.10.1 Question 2

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

**Solution**

**Video Solution**

**What is known?**

A line \(l\) and \(m\) which are \(4\,\rm{cm}\) apart.

**To construct:**

A perpendicular at any point on line \( l\), and then draw a line *\(m\)* parallel to \(l\) through \(X\) on the perpendicular which is \(4\) cm from the line \(l\).

**Reasoning: **

Draw a line *\(l\)* and then 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 *\( l\)*. Follow the steps given below.

**Steps:**

** Steps of construction** –

- Draw a line
*\( l\)*, take a point \(P\) on it. Draw a perpendicular passing trough point \(P.\) - Now adjust the compass up-to the length of \(4\) \(\rm cm.\) Draw an arc to intersect this perpendicular at point \(X\) choose any point \(Y\) on line
*\(l \)*, join \(X\) to \(Y\). - Taking \(Y\) as centre and with a convenient radius, draw an arc intersecting
*\(l\)*at \(A\) and \(XY\) at \(B\). - Taking \(X\) as centre and with the same radius as above, draw an arc \(CD\) cutting \(XY\) at \(E\)Adjust the compass up-to to the length of \(AB\).
- Without changing the opening of compass and taking \(E\) as the centre, draw an arc to intersect the previously drawn arc \(CD\) at \(F\).
- Join the points \(X\) and\( F\) to draw a line \(\rm m.\)

Line *\(\rm m\) *is the required line which is parallel to *\(l\)*.

## Chapter 10 Ex.10.1 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?

**Solution**

**Video Solution**

**What is known?**

A line \(l\) and a point \(P\) not on \(l\) and line m parallel to \(l\)..

**What is unknown?**

Shape formed by two sets of two parallel lines.

**Reasoning: **

As line \(l\) is given and\(\,P\,\)is 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 \(S.\)

**Steps:**

__Steps of construction –__

- Draw a line \(l\), take a point \(A\) on it. Take a point\(\,P\,\)not on \(l\) and join \(A\) to \(P.\)
- Taking \(A\) as centre and with a convenient radius draw an arc cutting \(l\) at \(B\) and \(AP\) at \(C.\)
- Taking\(\,P\,\)as the centre and with the same radius as before, draw an arc \(DE\) to intersect \(AP\) at \(F.\)
- 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 previous drawn arc \(DE\) at point \(G.\)
- Join\(\,P\,\)to any point \(G\) to draw line\(\, m \,\). Line\(\,m\,\)will be parallel to \(l\).
- 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.

Thus, \(PQSR\) is a parallelogram.