NCERT Solutions Ganita Prakash (Part 1) Chapter 7 Visualising the construction of circles — In-text Questions
Book page 155 Updated on2026-09-05
Q1.
Does a triangle exist with sidelengths 4 cm, 5 cm and 8 cm? This satisfies the triangle inequality: 8 < 4 + 5 = 9. Why do we not need to check the other two sides?
Answer
Because the other two comparisons are already guaranteed by the size order.
4 < 5 + 8 = 13 ✓ (4 is smaller than 8 alone) 5 < 4 + 8 = 12 ✓ (5 is smaller than 8 alone) 8 < 4 + 5 = 9 ✓ — the only comparison that could have failed
Why it happens: Each of the smaller lengths is already less than the longest length by itself; adding a third positive length can only widen the gap. So checking the longest against the sum of the other two settles the whole question.
Q2.
Now, suppose that a circle of radius 5 cm is constructed, centred at B. Can you draw a rough diagram of the resulting figure?
Answer
Take AB = 8 cm as the base. Draw the circle of radius 4 cm about A and the circle of radius 5 cm about B — they overlap in a lens-shaped region.
AB = 8 cm, AX = 4 cm so BX = 4 cm, which is less than the 5 cm radius of circle B.
Tip: Draw the base first and mark X, the point where circle A cuts AB. Then you can see at a glance whether circle B swallows X or not.
Q3.
Note that in the figure below, AX = 4 cm and AB = 8 cm. So, what is BX? Does this length help in visualising the resulting figure?
Answer
BX = 4 cm, and it settles the whole picture.
BX = AB – AX = 8 cm – 4 cm = 4 cm
Radius of the circle centred at B = 5 cm BX = 4 cm < 5 cm So X lies inside circle B
Why it happens: X is the point of circle A that is nearest to B. If even this nearest point is inside circle B, then part of circle A is inside circle B and part of it (the far side) is outside — so the two circles must cross, and they cross at exactly two points. Either of those points together with A and B gives a triangle. Hence a triangle with sides 4 cm, 5 cm and 8 cm does exist.