Answer :
Explanation:
Representing \(\sqrt{3.5}\) on the number line involves a few steps, including constructing the value geometrically. Here's a way to do it:
1. **Understanding \(\sqrt{3.5}\)**:
- We need to find a point on the number line that corresponds to the value \(\sqrt{3.5}\).
2. **Simplify the task**:
- \(\sqrt{3.5}\) is a bit tricky to represent directly. We can break it down into steps that involve known geometric constructions.
### Geometric Construction Method
Here's a step-by-step method to represent \(\sqrt{3.5}\) on a number line using geometric construction:
1. **Draw a Number Line**:
- Draw a straight horizontal line and mark a point O (origin). From O, mark points at unit distances, 0, 1, 2, 3, and 4.
2. **Construct a Segment of Length 3.5**:
- Extend the number line to mark the point 3.5 units away from the origin O.
- You can do this by marking 3.5 units directly or by marking 3 units and then a half unit further.
3. **Use the Right Triangle Method**:
- Draw a vertical line from the point 3.5 on the number line, making it perpendicular to the number line. Let's call this vertical line segment OB, where OB is of length 1 unit (just for convenience).
- Now, you have a right triangle OAB, where:
- OA = 3.5 units (base)
- OB = 1 unit (height)
4. **Find the Hypotenuse**:
- Using the Pythagorean theorem, calculate the hypotenuse \(OC\) (which will represent \(\sqrt{3.5 + 1}\)):
\[ OC = \sqrt{OA^2 + OB^2} = \sqrt{3.5^2 + 1^2} = \sqrt{3.5 + 1} = \sqrt{4.5} \]
- We need \(\sqrt{3.5}\), so construct the hypotenuse for the corresponding triangle, keeping the vertical segment OB equal to 0.5 units (height).
\[ OC = \sqrt{OA^2 + OB^2} = \sqrt{3.5^2 + 0.5^2} = \sqrt{3.5 + 0.25} = \sqrt{3.75}\]
5. **Mark the Distance**:
- Using a compass, measure the length OC. With the same radius, place the compass point at the origin O and draw an arc that intersects the number line. The intersection point represents \(\sqrt{3.75}\) on the number line.
### Simplified Approach Using Known Construction
If you prefer a simpler way and more accurate representation, you might want to use a calculator to find \(\sqrt{3.5}\) which approximately equals 1.87. Then, manually mark this distance on the number line:
1. **Draw a number line** and mark 0, 1, and 2.
2. **Use a ruler** to measure out 1.87 units from 0 accurately. Mark this point as \(\sqrt{3.5}\).
### Visual Representation
1. Draw the number line.
2. Mark the units 0, 1, 2, 3.
3. Using a ruler, mark 1.87 units from the origin 0.
This should give you a visual representation of \(\sqrt{3.5}\) on the number line.
Answer:
1 :draw a line and mark point A on it.
2:mark a point B on the line drawn in step 1 such that AB=3.5
3:mark a point C on AB produced such that BC=1unit.
4:find midpoint of AC . let the midpoint be O.
5:taking O as the center and OC=OA as radius draw a semi-circle. Also draw a line passing through B perpendicular to OB. suppose it cuts the semi circle at D.
6: taking B as the center and BD as radius draw an arc cutting OC produced at E. Point E so obtained represent root 3.5
Draw a line AB of10.5 units Extend the line making BC = 1unit, and AC = 11.5 units Take the midpoint of AC as M = 11.5/2 = 5.75 Draw a semi-circle from A to C as M as the centre. Draw a line perpendicular to AC at point C until is intersects and bisects with the semicircle forming a new point called D CD=√10.5 Put your compass needle at point M and then the other end at point D. Draw an arc till it reaches the number line or intersects with the number line and the point formed let's call it D2(or any other thing you like) D=D2 1 :draw a line and mark point A on it.
2:mark a point B on the line drawn in step 1 such that AB=3.5
3:mark a point C on AB produced such that BC=1unit.
4:find midpoint of AC . let the midpoint be O.
5:taking O as the center and OC=OA as radius draw a semi-circle. Also draw a line passing through B perpendicular to OB. suppose it cuts the semi circle at D.
6: taking B as the center and BD as radius draw an arc cutting OC produced at E. Point E so obtained represent root 3.5
Draw a line AB of10.5 units Extend the line making BC = 1unit, and AC = 11.5 units Take the midpoint of AC as M = 11.5/2 = 5.75 Draw a semi-circle from A to C as M as the centre. Draw a line perpendicular to AC at point C until is intersects and bisects with the semicircle forming a new point called D CD=√10.5 Put your compass needle at point M and then the other end at point D. Draw an arc till it reaches the number line or intersects with the number line and the point formed let's call it D2(or any other thing you like) D=D2=√10.5