GSEB Class 6 Maths Solutions Chapter 14 પ્રાયોગિક ભૂમિતિ Exercise 14.5

 

Question 1. 7.3 સેમી લંબાઈનો \(\overline{A B}\) દોરો અને તેની સંમિતિનો અક્ષ નિશ્ચિત કરો.
Answer:

1. First, draw a line segment \( \overline{A B} \) that is 7.3 cm long.
2. With A as the center and a radius greater than half the length of \( \overline{A B} \), draw one arc above \( \overline{A B} \) and another arc below \( \overline{A B} \).
3. Next, using B as the center and the same radius, draw two more arcs. These arcs should cut the previous two arcs, one above and one below \( \overline{A B} \).
4. Name the intersection points of these two sets of arcs as X and Y, respectively.
5. Finally, draw a line connecting X and Y, creating \( \overleftrightarrow{\mathrm{XY}} \). This line \( \overleftrightarrow{\mathrm{XY}} \) is the perpendicular bisector of \( \overline{A B} \). So, \( \overleftrightarrow{\mathrm{XY}} \) represents the axis of symmetry for \( \overline{A B} \).
In simple words: Draw a 7.3 cm line. Use a compass to make intersecting arcs from both ends, above and below the line. Connect the arc intersection points to form the perpendicular bisector, which is the line of symmetry.

Exam Tip: Remember that a perpendicular bisector divides a line segment into two equal parts and is at a 90-degree angle to it. Ensure your compass opening is always more than half the segment's length.

 

Question 2. 9.5 સેમી લંબાઈનો રેખાખંડ દોરો અને તેનો લંબદ્વિભાજક રચો.
Answer:

1. First, draw a line segment AB that measures 9.5 cm.
2. With A as the center, open your compass to a radius that is more than half the length of \( \overline{A B} \). Draw an arc above and an arc below \( \overline{A B} \).
3. Now, with B as the center and keeping the same compass opening, draw two more arcs. Make sure these arcs cut the previous two arcs, both above and below \( \overline{A B} \).
4. Label the points where the arcs cross as X and Y.
5. Draw a straight line connecting X and Y. This line \( \overleftrightarrow{\mathrm{XY}} \) is the required perpendicular bisector of \( \overline{A B} \).
In simple words: Draw a 9.5 cm line segment. From each end, draw arcs above and below the line with a compass set to more than half the line's length. Connect the points where the arcs cross to get the perpendicular bisector.

Exam Tip: Always use a sharp pencil and precise compass settings for accurate geometric constructions. The point where the bisector meets the segment is its midpoint.

 

Question 3. 10.8 સેમી લંબાઈના \( \overline{X Y} \) નો લંબદ્વિભાજક દોરો.
(a) દોરેલા લંબદ્વિભાજક પર કોઈક બિંદુP લો. PX = PY થાય છે કે કેમ તે ચકાસો.
(b) જો \( \overline{X Y} \) નું મધ્યબિંદુ M હોય, તો \( \overline{M X} \) અને \( \overline{X Y} \) ની લંબાઈ વિશે તમે શું કહી શકો?
Answer:

1. Draw a line segment \( \overline{X Y} \) with a length of 10.8 cm.
2. Using X as the center and a radius greater than half the length of \( \overline{X Y} \), draw one arc above and one arc below \( \overline{X Y} \).
3. Now, with Y as the center and the same radius, draw two more arcs that cut the previous two arcs, both above and below \( \overline{X Y} \).
4. Label the intersection points of these arcs as C and D.
5. Draw a straight line connecting C and D, which forms \( \overleftrightarrow{\mathrm{CD}} \). This line \( \overleftrightarrow{\mathrm{CD}} \) is the perpendicular bisector of \( \overline{X Y} \). Next, choose any point P on \( \overleftrightarrow{\mathrm{CD}} \). Draw line segments \( \overline{P X} \) and \( \overline{P Y} \).
(a) When you use a ruler to measure \( \overline{P X} \) and \( \overline{P Y} \), you will find that \( P X = P Y \). This is because any point on the perpendicular bisector of a line segment is equidistant from its endpoints.
(b) If M is the midpoint of \( \overline{X Y} \), then by measuring with a ruler, it can be observed that \( M X = M Y = \frac { 1 }{ 2 } X Y \). This means \( \overline{M X} \) is exactly half the length of \( \overline{X Y} \).
In simple words: Draw a 10.8 cm line and its perpendicular bisector. (a) Pick any point P on this bisector. You will see that the distance from P to one end of the line (PX) is the same as the distance from P to the other end (PY). (b) If M is the middle point of the original line, then MX is half the total length of XY.

Exam Tip: Remember the property that any point on the perpendicular bisector of a line segment is equidistant from its endpoints. This is a key concept for such constructions.

 

Question 4. 12.0 સેમી લંબાઈનો રેખાખંડ દોરો. પરિકરનો ઉપયોગ કરીને તેને ચાર સરખા ભાગમાં વહેચો. ખરેખર કરો.
Answer:

1. First, draw a line segment \( \overline{A B} \) with a length of 12.0 cm.
2. Construct the perpendicular bisector \( \overleftrightarrow{\mathrm{PQ}} \) of \( \overline{A B} \). This bisector will cut \( \overline{A B} \) at point O. Point O is the midpoint of \( \overline{A B} \).
3. Next, construct the perpendicular bisector \( \overleftrightarrow{\mathrm{CD}} \) of \( \overline{A O} \). This bisector will cut line segment AO at point M. Point M is the midpoint of \( \overline{A O} \).
4. Similarly, construct the perpendicular bisector \( \overleftrightarrow{\mathrm{EF}} \) of \( \overline{O B} \). This bisector will cut line segment OB at point N. Point N is the midpoint of \( \overline{O B} \).
Thus, the line segment AB is divided into four equal parts: AM, MO, ON, and NB. By measuring with a ruler, we find that \( AM = MO = ON = NB = 3.0 \) cm.
In simple words: Draw a 12.0 cm line. First, divide it into two equal parts using a perpendicular bisector. Then, divide each of those two halves again using another set of perpendicular bisectors. This splits the original line into four equal pieces, each 3.0 cm long.

Exam Tip: To divide a line segment into 'n' equal parts, you generally need to construct 'n-1' bisectors, or repeatedly bisect until the desired number of segments is achieved.

 

Question 5. 6.1 સેમી લંબાઈનો \( \overline{P Q} \) જેનો વ્યાસ છે, તેવું વર્તુળ દોરો.
Answer:

1. First, draw a line segment \( \overline{P Q} \) with a length of 6.1 cm.
2. Construct the perpendicular bisector \( \overleftrightarrow{\mathrm{XY}} \) of \( \overline{P Q} \). This bisector will intersect \( \overline{P Q} \) at point O. Point O is the midpoint of \( \overline{P Q} \) and will be the center of our circle.
3. Take O as the center and a radius equal to \( \overline{O P} \) (or \( \overline{O Q} \)), which is half of 6.1 cm, i.e., 3.05 cm. Draw a circle using this center and radius. This circle will pass through points P and Q. Thus, the circle with \( \overline{P Q} \) as its diameter is created.
In simple words: Draw a 6.1 cm line. Find its exact middle point. Use that middle point as the center and half the line's length as the radius to draw a circle. The line will then be the circle's diameter.

Exam Tip: The center of a circle is always the midpoint of its diameter. The radius is half the diameter. These are fundamental properties to remember for circle constructions.

 

Question 6. કેન્દ્ર C અને ત્રિજ્યા 3.4 સેમીવાળું વર્તુળ રચો. તેની કોઈ પણ જીવા \( \overline{A B} \) દોરો. \( \overline{A B} \) નો લંબદ્વિભાજક રચો અને તે ત્માંથી પસાર થાય છે કે કેમ તે ચકાસો.
Answer:

1. First, select a point C on a piece of paper as the center.
2. With C as the center and a radius of 3.4 cm, draw a circle.
3. Inside this circle, draw any chord, \( \overline{A B} \).
4. Construct the perpendicular bisector \( \overleftrightarrow{\mathrm{MN}} \) of \( \overline{A B} \). Upon observation, it will be clear that \( \overleftrightarrow{\mathrm{MN}} \) passes through the center C of the circle.
In simple words: Draw a circle with center C and radius 3.4 cm. Draw any line segment (chord) inside it. Then, make the perpendicular bisector of that chord. You will see that this bisector line goes straight through the center C of the circle.

Exam Tip: A crucial property of circles is that the perpendicular bisector of any chord always passes through the center of the circle. This fact is often tested in geometry problems.

 

Question 7. \( \overline{A B} \) ને વ્યાસ લઈને પ્રશ્ન 6 ફરીથી કરો.
Answer:

1. First, mark a point C on your paper.
2. With C as the center and a radius of 3.4 cm, draw a circle.
3. Now, draw a diameter \( \overline{A B} \) within this circle. This means \( \overline{A B} \) must pass through the center C.
4. Construct the perpendicular bisector \( \overleftrightarrow{\mathrm{MN}} \) of \( \overline{A B} \). When you observe it, you will find that \( \overleftrightarrow{\mathrm{MN}} \) passes directly through the center C of the circle. This happens because the diameter is itself a chord, and its perpendicular bisector must pass through the center.
In simple words: Draw a circle with center C and a 3.4 cm radius. Draw a line (diameter) across the circle, passing through C. Then make the perpendicular bisector of this diameter. You'll see that this bisector also goes through the circle's center, C.

Exam Tip: A diameter is the longest chord of a circle. Its perpendicular bisector is the diameter perpendicular to it, and it always passes through the center of the circle, as is the case for any chord.

 

Question 8. 4 સેમી ત્રિજ્યાવાળું વર્તુળ દોરો. તેની કોઈ પણ બે જીવા દોરો. આ બંને જીવાના લંબદ્વિભાજકોની રચના કરો. તે બંને (પરસ્પર) ક્યાં છેદે છે?
Answer:

1. First, select a point O on a piece of paper.
2. With O as the center and a radius of 4 cm, draw a circle.
3. Inside this circle, draw any two chords, for example, \( \overline{A B} \) and \( \overline{C D} \).
4. Construct the perpendicular bisector (let's call it 'l') for \( \overline{A B} \), and the perpendicular bisector (let's call it 'm') for \( \overline{C D} \). Extend them if necessary. You will see that both 'l' and 'm' intersect at the center O of the circle.
In simple words: Draw a circle with a 4 cm radius. Draw two random lines (chords) inside it. Then, draw the perpendicular bisector for each chord. Both of these bisector lines will cross each other exactly at the center of the circle.

Exam Tip: This construction demonstrates that the intersection point of the perpendicular bisectors of any two chords of a circle is always the center of that circle. This is a common method to find the center of a given circle.

 

Question 9. O શિરોબિંદુવાળો કોઈ ખૂણો દોરો. તેના એક (કિરણ) ભુજ પર બિંદુ A લો અને બીજા કિરણ (ભુજ) પર બિંદુ B એવી રીતે લો કે જેથી OA = OB થાય. \( \overline{O A} \) અને \( \overline{O B} \)ના લંબદ્વિભાજકો દોરો, જે બંને Pમાં છે. PA = PB થાય છે?
Answer:

1. First, select a point O on the paper.
2. Draw an angle \( \angle MON \) with vertex O, of any desired measure.
3. On one ray \( \overrightarrow{\mathrm {OM}} \), take a point A, and on the other ray \( \overrightarrow{\mathrm {ON}} \), take a point B, such that \( O A = O B \).
4. Construct the perpendicular bisector 'l' for \( \overline{O A} \).
5. Construct the perpendicular bisector 'm' for \( \overline{O B} \).
6. Let the intersection point of 'l' and 'm' be P.
7. Draw line segments \( \overline{P A} \) and \( \overline{P B} \), and measure them with a compass. You will observe that \( P A = P B \). This is because P lies on the perpendicular bisector of \( \overline{O A} \), so \( P A = P O \). Also, P lies on the perpendicular bisector of \( \overline{O B} \), so \( P B = P O \). Therefore, \( P A = P B \).
In simple words: Draw an angle. Mark points A and B on its arms so that they are the same distance from the angle's corner (O). Draw the perpendicular bisectors for the lines OA and OB. Where these bisectors cross (point P), you will find that the distance from P to A is equal to the distance from P to B.

Exam Tip: Any point on the perpendicular bisector of a line segment is equidistant from the segment's endpoints. This principle is fundamental to understanding why PA = PB in this construction. This concept is vital for constructing circumcircles of triangles.