NCERT Solutions Ganita Prakash (Part 1) Chapter 7 –149Introduction · 7.1 Equilateral Triangles · 7.2 Constructing a Triangle When its Sides are Given — In-text Questions
Book page 146 Updated on2026-09-05
Q1.
What happens when the three vertices lie on a straight line?
Answer
No triangle is formed at all. The figure collapses into a single line segment.
When A, B and C lie on one line, AB + BC = AC and the triangle has no inside.
Why it happens: A triangle must enclose some space. If B lies on the segment AC, then the "side" AB and the "side" BC together lie exactly along AC, so AB + BC = AC. Nothing is enclosed, and the angle at B becomes 180° — a straight angle, not a corner. Such a set of three points is called collinear.
Q2.
Construct a triangle in which all the sides are of length 4 cm.
Answer
Use a compass — a ruler alone would need many trials.
Draw AB = 4 cm with a ruler.
With A as centre and radius 4 cm, draw a long arc above AB.
With B as centre and the same radius 4 cm, draw another arc cutting the first one at C.
Join AC and BC. ∆ABC is the required equilateral triangle.
The two arcs of radius 4 cm cross exactly at C, so AC = BC = AB = 4 cm.
Check it yourself: Measure the three angles with a protractor. Each one comes out to be 60°.
Q3.
How did you construct this triangle and what tools did you use? Can this construction be done only using a marked ruler (and a pencil)?
Answer
Yes, it can be done with only a marked ruler and a pencil — but it takes many trials.
Draw AB = 4 cm Mark a point C with AC = 4 cm Measure BC → it is usually not 4 cm Shift C and measure again … and again
Why it happens: A ruler fixes one distance at a time. There are endlessly many points C with AC = 4 cm, and only two of them also have BC = 4 cm. Hunting for them by eye is slow. A compass fixes all the points at distance 4 cm from A in one sweep, which is why it makes the job exact.
Tip: Tools used — ruler for the base, compass for the two arcs, pencil to join. A protractor is not needed at all.
Q4.
How do we make this construction more efficient?
Answer
Replace the guessing with two compass arcs.
Arc from A, radius 4 cm → every point on it is 4 cm from A Arc from B, radius 4 cm → every point on it is 4 cm from B Their meeting point C is 4 cm from both
Why it happens: This is exactly the trick used last year in Playing with Constructions to fix the top point of a 'house'. One arc gives a whole family of correct points; the second arc picks out the one point that satisfies the second condition too. No trial is left.
Q5.
Let C be the point of intersection of the arcs. The construction ensures that both AC and BC are of length 4 cm. Can you see why?
Answer
Because C lies on both arcs at once.
C is on the arc centred at A of radius 4 cm → AC = 4 cm C is on the arc centred at B of radius 4 cm → BC = 4 cm And AB = 4 cm was drawn first So AB = BC = CA = 4 cm
Why it happens: Every point of a circle (or arc) is at the same distance — the radius — from its centre. So membership of an arc is a statement about distance. Being on two arcs means satisfying two distance conditions together.
Q6.
How do we construct triangles that are not equilateral?
Answer
The same two-arc method works — only the two radii are now different.
Choose any one of the three lengths as the base and draw it.
From one end, draw an arc whose radius is the second length.
From the other end, draw an arc whose radius is the third length.
Join the crossing point to both ends.
Tip: Choosing the longest side as the base makes the two arcs cross at a comfortable angle, so the third vertex is easy to mark sharply.
Q7.
Construct a triangle of sidelength 4 cm, 5 cm and 6 cm.
Answer
Take AB = 4 cm as the base, AC = 5 cm and BC = 6 cm.
Step 1: Draw AB = 4 cm.
Step 2: With A as centre, draw a sufficiently long arc of radius 5 cm.
Step 3: With B as centre, draw an arc of radius 6 cm so that it cuts the first arc.
Step 4: Call the crossing point C. Join AC and BC to get ∆ABC.
Base 4 cm; the arc of radius 5 cm from A and the arc of radius 6 cm from B cross at C.
Check it yourself: 4 < 5 + 6, 5 < 4 + 6 and 6 < 4 + 5, so the arcs are certain to meet. This is the triangle inequality you will meet a few pages later.
Q8.
How do we construct this triangle more efficiently?
Answer
Do not draw the full circles — two short arcs are enough.
Points 5 cm from A → the circle centred at A, radius 5 cm Points 6 cm from B → the circle centred at B, radius 6 cm C = a point common to both → only the arcs near the crossing are needed
Why it happens: The full circle carries far more information than we need. The vertex C sits where the two circles cross, so we only have to draw the small pieces of each circle around that region. Fewer lines means a cleaner, more accurate figure.
Did you know? The two circles actually cross at two points, one above AB and one below. Both give a correct triangle — they are mirror images of each other in the line AB.