Question
Come up with a method to construct the angle bisector using a rope.
✓
Answer
Let ∠XOY be the given angle.
Fix a small pole at the point O.
Take a rope and make a loop at one end.
Mark a point at some distance on the rope.
Fix the loop of the rope at the pole at O and rotate the rope from OX to OY.
Mark points A and B at a fixed distance mark on the rope.
Fix small poles at A and B.
Take a rope and make loops on both ends.
Fix the loops of this piece of rope with poles at A and B.
Mark the midpoint of this rope and hold the rope at the midpoint.
Make both ends of the rope tight and mark the point at the midpoint of the rope.
Let this point be M.
Join AM, BM, and OM.
We have OA = OB and AM = BM.
In ∆OAM and ∆OBM, we have OA = OB, AM = BM, and OM is common.
∴ ∆OAM = ∆OBM (Using SSS-rule)
∴ ∠AOM = ∠BOM
∴ ∠XOM = ∠YOM
∴ OM is the bisector of the given angle ∠XOY.
Fix a small pole at the point O.
Take a rope and make a loop at one end.
Mark a point at some distance on the rope.
Fix the loop of the rope at the pole at O and rotate the rope from OX to OY.
Mark points A and B at a fixed distance mark on the rope.
Fix small poles at A and B.
Take a rope and make loops on both ends.
Fix the loops of this piece of rope with poles at A and B.
Mark the midpoint of this rope and hold the rope at the midpoint.
Make both ends of the rope tight and mark the point at the midpoint of the rope.
Let this point be M.
Join AM, BM, and OM.
We have OA = OB and AM = BM.
In ∆OAM and ∆OBM, we have OA = OB, AM = BM, and OM is common.
∴ ∆OAM = ∆OBM (Using SSS-rule)
∴ ∠AOM = ∠BOM
∴ ∠XOM = ∠YOM
∴ OM is the bisector of the given angle ∠XOY.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Construct a $\triangle\text{ABC}$ in which $BC = 3.6\ cm, AB = 5\ cm$ and $AC = 5.4\ cm.$ Draw the perpendicular bisector of the side $BC.$
→The following problem was set by Sridharacharya in his book, Patiganita.''$6 \frac{1}{4}$ is divided by $2 \frac{1}{2}$, and $60 \frac{1}{4}$ is divided by $3 \frac{1}{2}$ Tell the quotients separately.'' Can you try to solve by converting the fractions into decimals?
→Find the values of n in the following:
If $\frac{9^{\text{n}}\times3^2\times3^{\text{n}}-27^{\text{n}}}{(3^3)^5\times2^3}=\frac{1}{27},$ find the value of $n$
→If $\frac{9^{\text{n}}\times3^2\times3^{\text{n}}-27^{\text{n}}}{(3^3)^5\times2^3}=\frac{1}{27},$ find the value of $n$
Construct a right-angled triangle whose hypotenuse measures $5\ cm$ and one of the other sides measures $3.2\ cm.$
→In a "magic square", the sum of the numbers in each row, in each column and along the diagonal is the same. Is this a magic square?
→| $\frac{4}{11}$ | $\frac{9}{11}$ | $\frac{2}{11}$ |
| $\frac{3}{11}$ | $\frac{5}{11}$ | $\frac{7}{11}$ |
| $\frac{8}{11}$ | $\frac{1}{11}$ | $\frac{6}{11}$ |
Take the data giving the minimum and the maximum temperature of various cities given in Table. Plot a double bar graph using the data and answer the question: Name two cities where maximum temperature of one was less than the minimum temperature of the other.
→| Temperatures of Cities as on $20.6.2006$ | ||
| City | Max. | Min. |
| Ahmedabad | $38^{\circ} \mathrm{C}$ | $29^{\circ} \mathrm{C}$ |
| Amritsar | $37^{\circ} \mathrm{C}$ | $26^{\circ} \mathrm{C}$ |
| Banglore | $28^{\circ} \mathrm{C}$ | $21^{\circ} \mathrm{C}$ |
| Chennai | $36^{\circ} \mathrm{C}$ | $27^{\circ} \mathrm{C}$ |
| Delhi | $38^{\circ} \mathrm{C}$ | $28^{\circ} \mathrm{C}$ |
| Jaipur | $39^{\circ} \mathrm{C}$ | $28^{\circ} \mathrm{C}$ |
| Jammu | $41^{\circ} \mathrm{C}$ | $26^{\circ} \mathrm{C}$ |
| Mumbai | $32^{\circ} \mathrm{C}$ | $27^{\circ} \mathrm{C}$ |
Consider an algebraic expression
$8 p^2-2 p q+10 q^2 p$.
Find its terms, factors and the numerical coefficient of $p^2$. Then, find the value of given expression when p = 1 and q = 2.
→$8 p^2-2 p q+10 q^2 p$.
Find its terms, factors and the numerical coefficient of $p^2$. Then, find the value of given expression when p = 1 and q = 2.
Shikha deposited $Rs. 2000$ in a bank which pays $6\%$ simple interest. She withdrew $Rs. 700$ at the end of first year. What will be her balance after $3$ years?
→In the given figure $\text{l}\ ||\ \text{m}$ and t is a transversal. If $\angle5=70^\circ$ find the measure of each of the angles $\angle1,\ \angle3,\ \angle4\ \text{and}\ \angle8.$
→In $\triangle A B C, \angle A=50^{\circ}, \angle B=70^{\circ}$ and bisector of $\angle C$ meets $A B$ in $D$. Find the angles of the triangles $A D C$ and $B D C$.
→