NCERT Solutions Ganita Prakash Chapter 8 – 214Section 8.6 Points Equidistant from Two Given Points — Construct: House
Book page 211 Updated on2026-09-05
Q1.
House. Recreate this figure. Note that all the lines forming the border of the house are of length 5 cm.
Answer
First decide the order in which the parts will be drawn: the square wall first, then the apex A, then the two slanting lines, and last the arc.
Steps of construction
Step 1. Draw DE = 5 cm for the floor. At D and at E draw perpendiculars and cut off 5 cm on each to get B and C. Join BC lightly. Also draw the door: a 1 cm wide, 2 cm tall rectangle standing on DE.
Step 2. Open the compass to 5 cm, put the tip at B and draw an arc above the wall.
Step 3. With the same opening, put the tip at C and draw another arc. The two arcs cross at one point — that is A, because it is 5 cm from B and 5 cm from C.
Join AB and AC with straight lines.
Step 4. Keeping the opening at 5 cm, put the tip at A and draw the arc from B to C. This is the curved base of the roof. The house is ready!
The house. The two grey arcs of radius 5 cm, drawn from B and from C, cross at the apex A. The red arc — also of radius 5 cm, drawn from A — is the curved edge of the roof.
Why every length is 5 cm: BD, CE and DE are sides of the square wall; AB and AC are made 5 cm by the compass; so triangle ABC has all three sides equal to 5 cm. That is also why the arc from A of radius 5 cm passes neatly through both B and C.
Q2.
Can you complete the figure? Try! We need to locate the point A that is of distance 5 cm from the points B and C. You might have realised that this can be done using a ruler. However, this leads to a lot of trial and error. How can the point A be located without trial and error?
Answer
Use the compass — two arcs instead of a ruler.
Draw the circle (or just an arc) of radius 5 cm with centre B. Every point on it is 5 cm from B.
Draw the circle (or arc) of radius 5 cm with centre C. Every point on it is 5 cm from C.
The point where the two curves cross above the wall is 5 cm from B and 5 cm from C. That crossing is A.
Why the ruler is a bad tool here: with a ruler you can make a point 5 cm from B, but then you must check its distance from C, shift it, check again… and you never know when you are exactly right. The compass turns the search into a picture: one arc shows all the points 5 cm from B at once, the other shows all the points 5 cm from C, and the answer is simply where the two pictures overlap.
Did you notice? This is the same idea as finding B on page 209, where an arc of radius 7 cm met the line l. There a curve met a line; here two curves meet each other.
Q3.
Think: Was it necessary to draw two full circles to get the point A? We only needed part of both the circles.
Answer
No — two small arcs are enough.
With radius 5 cm and the tip at B, draw a short arc in the region above BC, roughly where the roof top will be.
With the same radius and the tip at C, draw another short arc crossing the first.
Their crossing point is A.
Why the rest of the circles is useless: only the crossing point matters, and we already know roughly where it is — above the wall. Drawing the full circles also produces a second crossing point below BC (inside the house), which is not wanted. Short arcs keep the figure clean and the answer unambiguous.
Q4.
Having obtained point A, what remains is the construction of the remaining arc. How do we do it? Can we use the fact that A is of distance 5 cm from both B and C?
Answer
Yes — that fact is exactly what makes the arc possible.
Keep the compass opening at 5 cm (it is already there).
Place the tip at A.
Swing the pencil from B to C. The arc starts at B and ends at C, dipping down into the roof.
AB = AC = 5 cm → both B and C are on the circle of radius 5 cm centred at A
so one arc, drawn without lifting the compass, touches both
Why it fits so exactly: a circle of radius 5 cm about A contains every point 5 cm away from A — and B and C are two such points. The pencil therefore passes through them on its way, with no adjustment at all. Had AB and AC not been equal, no single arc centred at A could have touched both.