
Construct an equilateral triangle whose sides are equal to \[a\] units and justify the construction.
Answer
562.5k+ views
Hint: Draw a horizontal line segment BC of length \[a\] units. Now, take point B as the center and use the compass to draw the arc of length \[a\] units. Similarly, take point C as the center and draw an arc of length \[a\] units. Join A to B and B to C. Now, \[\Delta ABC\] is the required equilateral triangle.
Complete step-by-step answer:
According to the question, we are asked to construct an equilateral triangle whose sides are equal to \[a\] units.
The length of the side of the given equilateral triangle = \[a\] units …………………………………….(1)
We know the property that the sides of an equilateral triangle are equal to each other.
From the above property and equation (1), we can say that the length of each side of the given equilateral triangle is \[a\] units.
First of all, let us draw a horizontal line segment BC of length \[a\] units as shown below …………………………………(2)
Now, take the compass and set it at \[a\] units.
Thereafter, taking point B as the center and using the compass to draw the arc of length \[a\] units.
Similarly, taking point B as the center and using the compass to draw the arc of length \[a\] units.
Now, labeling the point of intersection of arcs drawn using points B and C as A.
At the last, joining A to B and B to C as shown below.
Since we have set our compass at \[a\] units, the point of intersection of arcs drawn from points B and C must be of \[a\] units from B and C.
That is the length of AB and AC is of \[a\] units …………………………………..(3)
From equation (2), we have the length of side BC equal to \[a\] units.
Now, from equation (2) and equation (3), we get
\[BC=AB=AC=a\] units ………………………………………..(4)
So, \[\Delta ABC\] is an equilateral triangle whose sides are equal to \[a\] units.
Therefore, \[\Delta ABC\] is the required equilateral triangle.
Note: For this question, one point must be kept in mind that is the sides of an equilateral triangle are equal to each other. And when there is any requirement to make two sides equal to each other, then always try to use the intersection of arcs.
Complete step-by-step answer:
According to the question, we are asked to construct an equilateral triangle whose sides are equal to \[a\] units.
The length of the side of the given equilateral triangle = \[a\] units …………………………………….(1)
We know the property that the sides of an equilateral triangle are equal to each other.
From the above property and equation (1), we can say that the length of each side of the given equilateral triangle is \[a\] units.
First of all, let us draw a horizontal line segment BC of length \[a\] units as shown below …………………………………(2)
Now, take the compass and set it at \[a\] units.
Thereafter, taking point B as the center and using the compass to draw the arc of length \[a\] units.
Similarly, taking point B as the center and using the compass to draw the arc of length \[a\] units.
Now, labeling the point of intersection of arcs drawn using points B and C as A.
At the last, joining A to B and B to C as shown below.
Since we have set our compass at \[a\] units, the point of intersection of arcs drawn from points B and C must be of \[a\] units from B and C.
That is the length of AB and AC is of \[a\] units …………………………………..(3)
From equation (2), we have the length of side BC equal to \[a\] units.
Now, from equation (2) and equation (3), we get
\[BC=AB=AC=a\] units ………………………………………..(4)
So, \[\Delta ABC\] is an equilateral triangle whose sides are equal to \[a\] units.
Therefore, \[\Delta ABC\] is the required equilateral triangle.
Note: For this question, one point must be kept in mind that is the sides of an equilateral triangle are equal to each other. And when there is any requirement to make two sides equal to each other, then always try to use the intersection of arcs.
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