NCERT Exemplar Class 10 Maths Chapter 11 Exercise 11.2 Solutions 2026-27

Halve the side before squaring.

• Squeezed inside: since the circle is squeezed inside the square, its diameter is the side a and its radius is a2.
• The 14 factor: squaring that radius brings in a factor of 14, so the true area is π a²4.
• The error: the claim is off by a factor of 4, which is exactly what using the side as the radius produces.

False; the correct area is π a²4 cm².

Side equals the diameter here.

• Touching condition: because the circle of radius a fits exactly between opposite sides of the square, each side is the diameter 2a.
• The perimeter: four such sides give 8a, so the statement is correct.
• Mirror of Q11: there the square was inside and the side became the diameter, and the two ideas are mirror images of the same touching condition.

True; perimeter =8a cm.

Doubling the area needs a 2 scale, not 2.

• Side ratio: the outer square's side is d while the inner square's side is d2, so the sides are in the ratio 2, not 2.
• Areas square it: areas scale as the square of lengths, so the true area ratio is (2)²=2, which is why the outer square is only twice the inner one.
• Why not four: the ``four times'' guess would need the sides in ratio 2, and that happens only if the outer side were the inner side doubled rather than the inner diagonal.

False; the outer square is twice, not four times, the inner square.

The word ``segment'' hides two cases.

• Minor case: for the minor segment you carve the triangle away from the sector, so it shrinks below the sector.
• Major case: for the major segment the same triangle is added on, pushing it above its sector; since the claim does not say which segment, one counterexample is enough to mark it false.
• The picture: draw any chord and you get a small minor cap and a big major piece, where the cap is clearly less than its sector but the big piece visibly bulges past its own sector.

False; the claim only holds for the minor segment.

Pick one form of the circumference.

• One per turn: a rolling wheel lays down exactly one circumference for every full turn it makes.
• Two forms: in radius form that distance is 2π r, and in diameter form it is π d, and the two are the same length.
• The error: writing 2π d mixes the two forms, effectively using r=d, so the correct one-revolution distance is π d cm.

False; one revolution is π d cm.

Counting turns is just division.

• One step: each revolution advances the wheel by its circumference 2π r, so the whole journey s contains s2π r of those steps.
• Units check: the formula is dimensionally clean, since metres divided by metres gives a pure count of turns.
• Verdict: the statement is therefore exactly right with no correction needed.

True; the wheel makes s2π r revolutions.

The crossover is at r=2.

• Turning point: solving π r²=2π r gives r=2 as the single radius where the two numbers match exactly.
• Which side wins: past that radius the squared growth of area wins, and below it the linear circumference is bigger; a check at r=1 (area 3.14 versus circumference 6.28) settles it.
• Units note: the units do not even match, since area is in square units and circumference in plain units, so the comparison is purely numerical and hinges entirely on the size of r.

False; true only for r>2.

Equal arcs trade radius for angle.

• Key product: arc length is proportional to the product rθ, so for equal arcs that product stays fixed.
• Angle halves: if the radius doubles from r to 2r, the angle must halve to keep the product steady, so the smaller circle's angle is twice the larger one's.
• Wording trap: the double-negative phrasing ``is this statement false?'' is the only tricky part, since the statement itself is correct.

True statement; ₁=2₂.

Same arc, bigger radius, bigger slice.

• The formula: using area =12r, two sectors with the same arc length still differ if their radii differ, because a larger radius stretches the slice outward and adds area.
• Verdict: the claim is false in general and becomes true only for the special case of equal radii, which is effectively the same circle.
• From a figure: lay the two arcs as equal bases, and the sector with the longer radius is the taller ``triangle'' with the greater area, since equal bases with unequal heights give unequal areas.

False; areas are equal only if the radii are equal.

One equation, two unknowns.

• Single condition: equal areas give just one condition ₁ r₁=₂ r₂, which two different pairs can satisfy.
• Counterexample: the worked pair (4× 3=6× 2) shows two sectors of equal area with unequal arcs, so equal arc length is certainly not necessary; this is the reverse of Question 19.
• The takeaway: fixing the area fixes only the product r, while fixing the arc fixes only , so neither condition alone pins both quantities and the ``must be equal'' claim breaks down for circles of different sizes.

False; only the product r must match.

The short side caps the circle.

• Short side rules: with a>b, the circle is stopped by the breadth b, so its diameter is b and its radius is b2.
• The 14 factor: squaring that radius brings the factor 14, giving the true area π b²4.
• Why a drops out: the length a never enters the answer, because there is always room to spare along the longer direction.

False; the area is π b²4 cm².

One number controls the whole circle.

• One parameter: the radius is the only parameter a circle has, so everything about the circle is decided once it is known.
• Radius then area: equal circumferences pin the radius, and once the radius is fixed the area π r² has no freedom left, so the areas must match.
• Unlike sectors: where a sector lets arc and angle vary independently, a full circle leaves no room for two different sizes to share a circumference.

True; equal circumference forces equal radius and hence equal area.

Run Question 22 backwards.

• Areas to radius: equal areas give r₁²=r₂², and since radii are positive lengths, this forces r₁=r₂.
• Radius to circumference: equal radii then force equal circumferences, completing the chain in the reverse direction.
• The big idea: the two questions together show that for a full circle every standard measure is locked to the radius, so matching one matches all of them.

True; equal area forces equal radius and hence equal circumference.

Use the diagonal shortcut.

• Diagonal spans: the corners of the square sit on the circle, so the diagonal stretches across as the diameter p.
• Half its square: then area =12 p²=p²2, which is exactly half of the stated p².
• The slip: the mistake in the statement is reading p as the side of the square rather than as its diagonal.

False; the area is p²2 cm².