NCERT Solutions Class 7 Maths Chapter 10 Practical Geometry – Here are all the NCERT solutions for Class 7 Maths Chapter 10. This solution contains questions, answers, images, explanations of the complete chapter 10 titled Practical Geometry of Maths taught in class 7. If you are a student of class 7 who is using NCERT Textbook to study Maths, then you must come across chapter 10 Practical Geometry. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 7 Maths Chapter 10 Practical Geometry in one place.
NCERT Solutions Class 7 Maths Chapter 10 Practical Geometry
Here on AglaSem Schools, you can access to NCERT Book Solutions in free pdf for Maths for Class 7 so that you can refer them as and when required. The NCERT Solutions to the questions after every unit of NCERT textbooks aimed at helping students solving difficult questions.
For a better understanding of this chapter, you should also see summary of Chapter 10 Practical Geometry , Maths, Class 7.
| Class | 7 |
| Subject | Maths |
| Book | Mathematics |
| Chapter Number | 10 |
| Chapter Name |
Practical Geometry |
NCERT Solutions Class 7 Maths chapter 10 Practical Geometry
Class 7, Maths chapter 10, Practical Geometry solutions are given below in PDF format. You can view them online or download PDF file for future use.
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Question & Answer
Q.1: In the above construction, can you draw any other line through A that would be also parallel to the line l?
Ans : Missing
Q.2: Can you slightly modify the above construction to use the idea of equal corresponding angles instead of equal alternate angles?
Ans : Missing
Q.3: Draw a line, say AB, take a point C outside it. Through C, draw a line parallel to AB using ruler and compasses only.
Ans : 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.
Q.4: 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.
Ans : The steps of construction are as follows. (i) Draw a line I and take a point P on line I. 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 I 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.
Q.5: 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?
Ans : (i)Draw a line I and take a point A on it. Take a point P not on I and join A to P.
(ii) Taking A as centre and with a convenient radius, draw an arc cutting I 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 I at point S. In quadrilateral PQSR, opposite lines are parallel to each other. PQ II RS and PR II QS Thus, &mnSq1 PQSR is a parallelogram.
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