Posted by RBSE Solutions for Class 6 Maths Chapter 14: Practical Geometry helps students understand the concepts of drawing and constructing various geometric shapes using a compass, ruler, and protractor. This chapter emphasizes practical skills that are essential in geometry, including the construction of angles, triangles, and other geometric figures. The solutions provided are step-by-step and easy to understand, helping students master practical geometry for their exams.
Key Concepts Covered in Chapter 14: Practical Geometry
- Introduction to Practical Geometry
- Tools Used in Geometry: Compass, Ruler, and Protractor
- Construction of Angles
- Drawing Triangles Based on Given Measurements
- Constructing Special Quadrilaterals
- Using Practical Geometry in Real-Life Applications
Mastering practical geometry helps students develop the skills required for constructing various geometric shapes and solving real-world problems.
Detailed RBSE Class 6 Solutions for Chapter 14: Practical Geometry
Here are detailed, question-wise solutions for Chapter 14, designed to provide step-by-step guidance on how to construct geometric shapes accurately.
NCERT Solutions for Class 6 Maths Chapter 14: Practical Geometry Exercise 14.1
Table of Contents
1. Draw a circle of radius 3.2 cm.
Solutions:
The required circle can be drawn as follows:
Step 1: For the required radius of 3.2 cm, first open the compasses.
Step 2: For the centre of the circle, mark a point ‘O’.
Step 3: Place a pointer of compasses on ‘O’.
Step 4: Now, turn the compasses slowly to draw the required circle.
2. With the same centre O, draw two circles of radii 4 cm and 2.5 cm.
Solutions:
The required circle may be drawn as follows:
Step 1: For the required radius of 4 cm, first open the compasses.
Step 2: For the centre of the circle, mark a point ‘O’.
Step 3: Place a pointer of compasses on ‘O’.
Step 4: Turn the compasses slowly to draw the circle.
Step 5: Next, open the compasses for 2.5 cm.
Step 6: Again, place a pointer of compasses on ‘O’ and turn the compasses slowly to draw the circle.
3. Draw a circle and any two of its diameters. If you join the ends of these diameters, what is the figure obtained? What figure is obtained if the diameters are perpendicular to each other? How do you check your answer?
Solutions:
We will draw a circle having its centre ‘O’, also of any convenient radius. Let AB and CD be the two diameters of this circle. A quadrilateral is formed when we join the ends of these diameters.
We know that the diameter of a circle is equal in length, hence quadrilateral formed will be having its diagonals of equal length.
Also, OA = OB = OC = OD = radius r, and if a quadrilateral has its diagonals of the same length bisecting each other, it will be a rectangle.
Let DE and FG be the diameters of the circle such that both are perpendicular to each other. Now, we can find that a quadrilateral is formed by joining the ends of these diameters.
We can find that OD = OE = OF = OG = radius r.
In this quadrilateral DFEG, diagonals are equal and perpendicular to each other. Also, they are bisecting each other; hence it will be a square.
To check our answers, we can measure the length of the sides of the quadrilateral formed.
4. Draw any circle and mark points A, B and C such that
(a) A is on the circle.
(b) B is in the interior of the circle.
(c) C is in the exterior of the circle.
Solutions:
We will draw a circle with three required points, A, B, and C, as follows:
5. Let A and B be the centres of two circles of equal radii; draw them so that each one of them passes through the centre of the other. Let them intersect at C and D. Examine whether and are at right angles.
Solutions:
Let us draw two circles having the same radius, which are passing through the centre of the other circle.
Here, points A and B are the centres of these circles, and these circles intersect each other at points C and D, respectively.
Now, in quadrilateral ADBC, we may observe that:
AD = AC [radius of circle centered at A]
BC = BD [radius of circle centered at B]
The radius of both circles is equal.
Therefore, AD = AC = BC = BD.
Hence quadrilateral ADBC is a rhombus, and in a rhombus, diagonals bisect each other at 900. Hence
and are at right angles.
NCERT Solutions for Class 6 Maths Chapter 14: Practical Geometry Exercise 14.2
1. Draw a line segment of length 7.3 cm using a ruler.
Solutions:
Using a ruler, we can draw a line segment of length 7.3 cm as follows:
Step 1: Mark a point A on the sheet.
Step 2: Place the 0 mark of the ruler at point A.
Step 3: At 7.3 cm on the ruler, mark a point B on the sheet.
Step 4: Now join A and B.
Here is the required line segment.
2. Construct a line segment of length 5.6 cm using a ruler and compasses.
Solutions:
By using a ruler and compasses, we may draw a line segment of length 5.6 cm as follows:
Step 1: Draw a line l and mark a point A on this line l.
Step 2: On the zero mark of the ruler, place the compasses. Now extend the compasses to place the pencil up to 5.6 cm and mark.
Step 3: Place the pointer of compasses on point A and draw an arc to cut l at B. Now,
is the line segment of 5.6 cm in length.
3. Construct of length 7.8 cm. From this, cut off of length 4.7 cm. Measure .
Solutions:
Step 1: Draw a line l and mark a point A on it.
Step 2: By adjusting the compasses up to 7.8 cm, while putting the pointer of compasses on point A, draw an arc to cut l on B.
is the line segment of 7.8 cm.
Step 3: By adjusting the compasses up to 4.7 cm, draw an arc to cut l on C while putting the pointer of compasses on point A.
is the line segment of 4.7 cm.
Step 4: Now, place the ruler in such a way that the 0 mark of the ruler will match with point C.
Now read the position of point B. It will be 3.1 cm.
is 3.1 cm.
4. Given of length 3.9 cm, construct , such that the length of is twice that of . Verify by measurement.
Solutions:
We can draw a line segment , such that the length of is twice that of as follows:
(1) Draw a line l and mark a point P on it. Let AB be the given line segment of 3.9 cm.
(2) By adjusting the compass up to the length of AB, place the pointer of the compass at point P and draw an arc to cut the line at X.
(3) Again, place the pointer on point X, and draw an arc to cut line l at point Q.
is the required line segment. Using the ruler, we may measure the length of which comes to 7.8 cm.
5. Given of length 7.3 cm and of length 3.4 cm, construct a line segment , such that the length of is equal to the difference between the lengths of and . Verify by measurement.
Solutions:
(1) Given = 7.3 cm and = 3.4 cm
(2) Adjust the compass up to the length of CD, and put the pointer of compasses at point A and draw an arc to cut AB at point P.
(3) Adjust the compass up to the length of PB. Draw a line l and mark a point X on it.
(4) By placing the pointer of compass at point X, draw an arc to cut the line at Y.
is the required line segment
Now, the difference between the lengths of
and
= 7.3 – 3.4
= 3.9 cm
Using the ruler, we may measure the length of which comes to 3.9 cm.
NCERT Solutions for Class 6 Chapter 14: Practical Geometry Exercise 14.3
1. Draw any line segment. Without measuring, construct a copy of .
Solutions:
(1) Let the given line segment be .
(2) Adjust the compasses up to the length of .
(3) Draw any line l and mark a point A on it.
(4) Place the pointer on point A without changing the setting of compasses, and draw an arc to cut the line segment at B point.
is the required line segment.
2. Given some line segment, whose length you do not know. Construct , such that the length of is twice that of .
Solutions:
The following steps are followed to construct a line segment , such that the length of is twice that of .
(1) Let the given line segment be .
(2) Adjust the compass up to the length of .
(3) Draw a line l and mark a point P on it.
(4) Place the pointer on P, and draw an arc to cut the line segment at point X without changing the setting of the compass.
(5) Again, with the same radius as before, by placing the pointer on point X, draw an arc to cut line l at point Q.
is the required line segment.
NCERT Solutions for Class 6 Maths Chapter 14: Practical Geometry Exercise 14.4
1. Draw any line segment . Mark any point M on it. Through M, draw a perpendicular to . (Use ruler and compasses.)
Solutions:
(1) Draw a line segment and mark a point M on it.
(2) Taking M as the centre and a convenient radius, construct an arc intersecting the line segment
at points C and D, respectively.
(3) By taking centres as C and D and a radius greater than CM, construct two arcs such that they intersect each other at point E.
(4) Join EM. Now,
is perpendicular to .
2. Draw any line segment . Take any point R, not on it. Through R, draw a perpendicular to . (Use ruler and set-square.)
Solutions:
(1) Draw a given line segment and mark a point R outside the line segment .
(2) Place a set square on , such that one of its right angles arms aligns along .
(3) Now, place the ruler along the edge opposite to the right angle of the set square.
(4) Hold the ruler fixed. Slide the set square along the ruler such that point R touches the other arm of the set square.
(5) Draw a line along this edge of the set square which passes through point R. Now, it is the required line perpendicular to .
3. Draw a line l and a point X on it. Through X, draw a line segment perpendicular to l.
Now, draw a perpendicular to XY at Y. (Use ruler and compasses.)
Solutions:
(1) Draw a line l and mark a point X on it.
(2) By taking X as the centre and with a convenient radius, draw an arc intersecting the line l at points A and B, respectively.
(3) With A and B as centres and a radius more than AX, construct two arcs such that they intersect each other at point Y.
(4) Join XY. Here, is perpendicular to l.
Similarly, by taking C and D as centres and radius more than CY, construct two arcs intersecting at point Z. Join ZY. The line 
is perpendicular to at Y.
NCERT Solutions for Class 6 Maths Chapter 14: Practical Geometry Exercise 14.5
1. Draw 
of length 7.3 cm and find its axis of symmetry.
Solutions:
The following steps are followed to construct 
of length 7.3 cm and to find its axis of symmetry:
(1) Draw a line segment 
of 7.3 cm.
(2) Take A as the centre and draw a circle by using compasses. The radius of the circle should be more than half the length of
.
(3) Now, take B as the centre and draw another circle using compasses with the same radius as before. Let it cut the previous circle at points C and D.
(4) Join CD. Now 
is the axis of symmetry.
2. Draw a line segment of length 9.5 cm and construct its perpendicular bisector.
Solutions:
The following steps are observed to construct a line segment of length 9.5 cm and to construct its perpendicular bisector:
(1) Draw a line segment 
of 9.5 cm.
(2) Take point P as the centre and draw a circle by using compasses. The radius of the circle should be more than half the length of 
.
(3) Taking the centre at point Q, again draw another circle using compasses with the same radius as before. Let it cut the previous circle at R and S, respectively.
(4) Join RS. Now 
is the axis of symmetry, i.e., the perpendicular bisector of the line 
.
3. Draw the perpendicular bisector of whose length is 10.3 cm.
(a) Take any point P on the bisector drawn. Examine whether PX = PY.
(b) If M is the midpoint of , what can you say about the lengths MX and XY?
Solutions:
(1) Draw a line segment of 10.3 cm.
(2) Take point X as the centre and draw a circle by using compasses. The radius of the circle should be more than half the length of .
(3) Now, taking Y as the centre, draw another circle using compasses with the same radius as before. Let it cut at the previous circle at points A and B.
(4) Join AB. Here 
is the axis of symmetry.
(a) Take any point P on 
. We may observe that the measure of lengths of PX and PY are the same,
being the axis of symmetry, any point lying on 
will be at the same distance from both ends of .
(b) M is the midpoint of . Perpendicular bisector 
will be passing through point M. Hence, the length of
is double of 
or 2MX = XY.
4. Draw a line segment of length 12.8 cm. Using compasses, divide it into four equal parts. Verify by actual measurement.
Solutions:
(1) Draw a line segment of 12.8 cm.
(2) By taking point X as the centre and radius more than half of XY, draw a circle.
(3) Again, with the same radius and centre as Y, draw two arcs to cut the circle at points A and B. Join AB, which intersects at point M.
(4) By taking X and Y as centres, draw two circles with a radius more than half of 
.
(5) Taking M as the centre and with the same radius, draw two arcs to intersect these circles at P, Q and R, S.
(6) Join PQ and RS. These intersect at points T and U.
(7) The 4 equal parts of are 
= 
= 
= 
.
By measuring these line segments with the help of a ruler, we may observe that each is 3.2 cm.
5. With of length 6.1 cm as a diameter, draw a circle.
Solution:
(1) Draw a line segment of 6.1 cm.
(2) Take point P as the centre and radius more than half of , draw a circle.
(3) Again, with the same radius and Q as the centre, draw two arcs intersecting the circle at points R and S.
(4) Join RS, which intersects at T.
(5) Taking the centre as T and radius TP, draw a circle which passes through Q. Now, this is the required circle.
6. Draw a circle with centre C and a radius 3.4 cm. Draw any chord . Construct the perpendicular bisector of and examine if it passes through C.
Solutions:
(1) Mark any point C on the sheet.
(2) Adjust the compass up to 3.4 cm, and by putting the pointer of the compass at point C, turn the compass slowly to draw the circle. This is the required circle with 3.4 cm radius.
(3) Mark any chord in the circle.
(4) Now, taking A and B as centres, draw arcs on both sides of . Let these intersect each other at points D and E.
(5) Join DE. Now DE is the perpendicular bisector of AB.
If 
is extended, it will pass through point C.
7. Repeat Question 6, if happens to be a diameter.
Solutions:
(1) Mark any point C on the sheet.
(2) Adjust the compass up to 3.4 cm, and by putting the pointer of the compass at point C, turn the compass slowly to draw the circle. This is the required circle with 3.4 cm.
(3) Now mark any diameter in the circle.
(4) Now, taking A and B as centres, draw arcs on both sides of with a radius more than . Let these intersect each other at points D and E.
(5) Join DE, which is the perpendicular bisector of AB.
Now, we may observe that 
is passing through the centre C of the circle.
8. Draw a circle of radius 4 cm. Draw any two of its chords. Construct the perpendicular bisectors of these chords. Where do they meet?
Solutions:
(1) Mark any point O on the sheet. Now adjust the compass up to 4 cm, and by placing the pointer of the compass at point O, turn the compass slowly to draw the circle. This is the required circle with 4 cm radius.
(2) Take any two chords and 
in the circle.
(3) By taking A and B as centres and a radius more than half of , draw arcs on both sides of AB. The arcs intersect each other at points E and F. Join EF, which is the perpendicular bisector of AB.
(4) Again, take C and D as centres and a radius more than half of 
, draw arcs on both sides of CD, such that they intersect each other at points G, H. Join GH, which is the perpendicular bisector of CD.
We may observe that when EF and GH are extended, they meet at point O, which is the centre of the circle.
9. Draw any angle with vertex O. Take point A on one of its arms and B on another, such that OA = OB. Draw the perpendicular bisectors of and . Let them meet at P. Is PA = PB?
Solutions:
(1) Draw any angle with vertex as O.
(2) By taking O as the centre and with a convenient radius, draw arcs on both rays of this angle. Let these points be A and B.
(3) Now take O and A as centres, and with a radius of more than half of OA, draw arcs on both sides of OA. Let these intersect at points C and D, respectively. Join CD.
(4) Similarly, we may find 
which is the perpendicular bisector of . These perpendicular bisectors, 
and 
, intersect each other at point P. Now, measure PA and PB. They are equal in length.
NCERT Solutions for Class 6 Chapter 14: Practical Geometry Exercise 14.6
1. Draw ∠POQ of measure 75° and find its line of symmetry.
Solutions:
The following steps are used to construct an angle of 750 and its line of symmetry:
(i) Draw a line l and mark two points, O and Q, on it. Draw an arc of convenient radius while taking centre as O. Let this intersect line l at R.
(ii) Taking R as the centre and with the same radius as before, draw an arc such that it intersects the previously drawn arc at S.
(iii) By taking the same radius as before and S as the centre, draw an arc intersecting the arc at point T, as shown in the figure.
(iv) Take S and T as the centre, and draw an arc of the same radius such that they intersect each other at U.
(v) Join OU. Let it intersect the arc at V. Now, take S and V as centres draw arcs with a radius of more than 1/2 SV. Let these intersect each other at P, then Join OP. Now, OP is the ray making 750 with the line l.
(vi) Let this ray intersect the major arc at point W. By taking R and W as centres, draw arcs with a radius of more than 1/2 RW in the interior angle of 750. Let these intersect each other at point X. Join OX.
OX is the line of symmetry for the ∠POQ = 750.
2. Draw an angle of measure 147° and construct its bisector.
Solutions:
The following steps are used to construct an angle of measure 1470 and its bisector:
(i) Draw a line l and mark point O on it. Place the centre of the protractor at point O and the zero edge along line l.
(ii) Mark point A at an angle of measure 1470 and join OA. Now, OA is the required ray making 1470 with line l.
(iii) By taking point O as the centre, draw an arc of the convenient radius. Let this intersect both rays of angle 1470 at points A and B.
(iv) By taking A and B as centres, draw arcs of radius more than 1/2 AB in the interior angle of 1470. Let these intersect each other at point C. Join OC.
OC is the required bisector of 1470 angle.
3. Draw a right angle and construct its bisector.
Solutions:
The following steps are used to construct a right angle and its bisector:
(i) Draw a line l and mark a point P on it. Draw an arc of the convenient radius by taking point P as the centre. Let this intersect line l at R.
(ii) Draw an arc by taking R as the centre and with the same radius as before such that it intersects the previously drawn arc at S.
(iii) Take S as the centre and with the same radius as before, draw an arc intersecting the arc at T, as shown in the figure.
(iv) By taking S and T as centres, draw arcs of the same radius such that they intersect each other at U.
(v) Join PU. PU is the required ray making a right angle with the line l. Let this intersect the major arc at point V.
(vi) Now, take R and V as centres, and draw arcs with a radius of more than 1/2 RV to intersect each other at point W. Join PW.
PW is the required bisector of this right angle.
4. Draw an angle of measure 153° and divide it into four equal parts.
Solutions:
The following steps are used to construct an angle of measure 1530 and its bisector:
(i) Draw a line l and mark a point O on it. Place the centre of the protractor at point O and the zero edge along line l.
(ii) Mark a point A at the measure of angle 1530, and join OA. Now, OA is the required ray making 1530 with line l.
(iii) Draw an arc of the convenient radius by taking point O as the centre. Let this intersect both rays of angle 1530 at points A and B.
(iv) Take A and B as centres and draw arcs of radius more than 1/2 AB in the interior of an angle of 1530. Let these intersect each other at C. Join OC.
(v) Let OC intersect the major arc at point D. Draw arcs of radius more than 1/2 AD with A and D as centres and also D and B as centres. Let these are intersecting each other at points E and F, respectively. Now join OE and OF.
OF, OC and OE are the rays dividing the 1530 angle into four equal parts.
5. Construct with ruler and compasses angles of the following measures:
(a) 60°
(b) 30°
(c) 90°
(d) 120°
(e) 45°
(f) 135°
Solutions:
(a) 600
The following steps are followed to construct an angle of 600:
(i) Draw a line l and mark a point P on it. Take P as the centre, and with a convenient radius, draw an arc of a circle such that it intersects the line l at Q.
(ii) Take Q as the centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point R.
(iii) Join PR. PR is the required ray making 600 with the line l.
(b) 300
The following steps are followed to construct an angle of 300:
(i) Draw a line l and mark a point P on it. By taking P as the centre and with a convenient radius, draw an arc of a circle such that it intersects the line l at Q.
(ii) Take Q as the centre and with the same radius as before, draw an arc intersecting the previously drawn arc at point R.
(iii) By taking Q and R as centres and with a radius of more than 1/2 RQ draw arcs such that they intersect each other at S. Join PS, which is the required ray making 300 with the line l.
(c) 900
The following steps are used to construct an angle of measure 900:
(i) Draw a line l and mark a point P on it. Take P as the centre, and with a convenient radius, draw an arc of a circle such that it intersects the line l at Q.
(ii) Take Q as the centre, and with the same radius as before, draw an arc intersecting the previously drawn arc at R.
(iii) By taking R as the centre and with the same radius as before, draw an arc intersecting the arc at S, as shown in the figure.
(iv) Now, take R and S as the centre, and draw an arc of the same radius to intersect each other at T.
(v) Join PT, which is the required ray, making 900 with line l.
(d) 1200
The following steps are used to construct an angle of measure 1200:
(i) Draw a line l and mark a point P on it. Taking P as the centre and with a convenient radius, draw an arc of the circle such that it intersects the line l at Q.
(ii) By taking Q as the centre and with the same radius as before, draw an arc intersecting the previously drawn arc at R.
(iii) Take R as the centre and with the same radius as before, draw an arc such that it intersects the arc at S, as shown in the figure.
(iv) Join PS, which is the required ray making 1200 with the line l.
(e) 450
The following steps are used to construct an angle of measure 450:
(i) Draw a line l and mark a point P on it. Take P as the centre, and with a convenient radius, draw an arc of a circle such that it intersects the line l at Q.
(ii) Take Q as the centre, and with the same radius as before, draw an arc intersecting the previously drawn arc at R.
(iii) By taking R as the centre and with the same radius as before, draw an arc such that it intersects the arc at S, as shown in the figure.
(iv) Take R and S as centres and draw arcs of the same radius such that they intersect each other at T.
(v) Join PT. Let this intersect the major arc at point U.
(vi) Now, take Q and U as centres and draw arcs with a radius of more than 1/2 QU to intersect each other at point V. Join PV.
PV is the required ray making 450 with the line l.
(f) 1350
The following steps are used to construct an angle of measure 1350:
(i) Draw a line l and mark a point P on it. Taking P as the centre and with a convenient radius, draw a semicircle which intersects the line l at Q and R, respectively.
(ii) By taking R as the centre and with the same radius as before, draw an arc intersecting the previously drawn arc at S.
(iii) Taking S as the centre and with the same radius as before, draw an arc such that it intersects the arc at T, as shown in the figure.
(iv) Take S and T as centres, and draw arcs of the same radius to intersect each other at U.
(v) Join PU. Let this intersect the arc at V. Now, take Q and V as centres and with a radius of more than 1/2 QV, draw arcs to intersect each other at W.
(vi) Join PW, which is the required ray making 1350 with the line l.
6. Draw an angle of measure 45° and bisect it.
Solutions:
The following steps are used to construct an angle of measure 450 and its bisector:
(i) Using the protractor, a ∠POQ of 450 measure may be formed on a line l.
(ii) Draw an arc of the convenient radius with the centre as O. Let this intersect both rays of angle 450 at points A and B.
(iii) Take A and B as centres and draw arcs of radius more than 1/2 AB in the interior of an angle of 450. Let these intersect each other at C. Join OC.
OC is the required bisector of 450 angle.
7. Draw an angle of measure 135° and bisect it.
Solutions:
The following steps are used to construct an angle of measure 1350 and its bisector:
(i) By using a protractor, a ∠POQ of 1350 measure may be formed on a line l.
(ii) Draw an arc of the convenient radius by taking O as the centre. Let this intersect both rays of angle 1350 at points A and B, respectively.
(iii) Take A and B as centres and draw arcs of a radius of more than 1/2 AB in the interior of an angle of 1350. Let these intersect each other at C. Join OC.
OC is the required bisector of 1350 angle.
8. Draw an angle of 700. Make a copy of it using only a straight edge and compasses.
Solutions:
The following steps are used to construct an angle of measure 700 and its copy:
(i) Draw a line l and mark a point O on it. Now, place the centre of the protractor at point O and the zero edge along line l.
(ii) Mark a point A at an angle of measure 700. Join OA. Now, OA is the ray making 700 with line l. With point O as the centre, draw an arc of a convenient radius in the interior of 700 angle. Let this intersect both rays of angle 700 at points B and C, respectively.
(iii) Draw a line m and mark a point P on it. Again, draw an arc with the same radius as before and P as the centre. Let it cut the line m at point D.
(iv) Adjust the compasses up to the length of BC. With this radius, draw an arc taking D as the centre, which intersects the previously drawn arc at point E.
(v) Join PE. Here, PE is the required ray which makes the same angle of measure 700 with the line m.
9. Draw an angle of 400. Copy its supplementary angle.
Solutions:
The following steps are used to construct an angle of measure 450 and a copy of its supplementary angle:
(i) Draw a line segment
and mark a point O on it. Place the centre of the protractor at point O and the zero edge along the line segment.
.
(ii) Mark a point A at an angle of measure 400. Join OA. Here. OA is the required ray, making 400 with
. ∠POA is the supplementary angle of 400.
(iii) With point O as the centre, draw an arc of convenient radius in the interior of ∠POA. Let this intersect both rays of ∠POA at points B and C, respectively.
(iv) Draw a line m and mark a point S on it. Again draw an arc by taking S as the centre with the same radius as used before. Let it cut the line m at point T.
(v) Now, adjust the compasses up to the length of BC. Taking T as the centre, draw an arc with this radius which will intersect the previously drawn arc at point R.
(vi) Join RS. Here, RS is the required ray which makes the same angle with the line m as the supplementary of 400 [i.e., 1400].
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Benefits of RBSE Class 6 Chapter 14 Solutions
- Hands-On Learning: Encourages students to practice geometry through constructions, building hands-on skills.
- Clear Instructions: The solutions provide step-by-step instructions, making the construction process easy to understand.
- Improved Accuracy: By practicing these solutions, students will improve their ability to draw accurate geometric shapes.
- Real-Life Applications: Practical geometry is widely used in fields such as architecture, engineering, and design, helping students understand its real-world relevance.
FAQs on RBSE Solutions for Class 6 Maths Chapter 14: Practical Geometry
Q1: What is practical geometry?
A1: Practical geometry involves drawing and constructing various geometric shapes using basic tools like a ruler, compass, and protractor. It is a hands-on approach to learning geometry.
Q2: How do I construct a triangle using given side lengths?
A2: To construct a triangle, you need to draw three sides with the given lengths using a ruler and connect the ends. The angle measurements are verified using a protractor if needed.
Q3: Why is practical geometry important?
A3: Practical geometry helps in understanding the geometric properties of shapes and figures, which is essential in fields like design, engineering, and architecture.
Q4: Where can I find RBSE solutions for Class 6 Maths Chapter 14?
A4: You can access detailed solutions for Class 6 Maths Chapter 14 on rbsesolution.in, where all questions are answered step-by-step for better understanding.
Q5: What tools are used in practical geometry?
A5: The main tools used in practical geometry are a compass, ruler, and protractor. These tools help in accurate construction of angles, lines, and shapes.
Conclusion
The RBSE Solutions for Class 6 Maths Chapter 14: Practical Geometry provide students with a comprehensive understanding of how to construct geometric shapes accurately. These step-by-step solutions help students gain confidence in using geometric tools like the compass, ruler, and protractor. By practicing these constructions, students will enhance their skills in geometry, which are essential for academic success and real-life applications.