NCERT Solutions for Class 8 Maths Chapter 14 Area (PDF) – 2027

Infinitely many ways. Start with any division into 4 equal parts, for example two lines through the centre that make 4 small squares. Now let one part push a bump into its neighbour, and let the neighbour push an equal bump back somewhere else. Each part gains exactly as much area as it loses, so all 4 parts stay equal. The bumps can have any size and shape, so there are infinitely many such divisions.

— each of these divides a square of side s into 4 parts of equal area:

• Four equal strips: three lines parallel to one side, one quarter of the side apart. Each strip is s × s/4.
• Four triangles: join the centre to the four corners. Each triangle is ½ × s × s/2 = s²/4.
• Two perpendicular lines through the centre, tilted at any angle: a quarter-turn about the centre carries each piece onto the next, so the 4 pieces are identical.
• Bumpy pieces: start from any of these and swap equal bumps and dents across a boundary, as in Q1.

The 7 cm × 4 cm rectangle.

14 cm² each.

No. Perimeter measures the boundary; area measures the region enclosed. They are different quantities, and one does not determine the other.

A unit square is the natural yardstick for a region because copies of it can be packed to fill up a region without gaps or overlaps. Counting how many fit tells us exactly “how much surface” there is.

No to both.

Same perimeter, three different areas. So equal perimeters do not force equal areas.

And a larger perimeter does not mean a larger area: a 10 cm × 1 cm rectangle has perimeter 22 cm but area only 10 cm², while a 4 cm × 4 cm square has the smaller perimeter, 16 cm, and the larger area, 16 cm².

We need Region 1 with the larger perimeter but the smaller area.

Take a square of side 6 cm and cut deep, thin notches into it, like the teeth of a comb.

• Region 2 (the plain 6 cm square): perimeter 24 cm, area 36 cm².
• Region 1 (the comb): the same square with 4 slits, each 5 cm deep and 0.6 cm wide. Each slit removes 5 × 0.6 = 3 cm² of area but adds 5 + 5 = 10 cm of boundary.
  Area = 36 − 4 × 3 = 24 cm²; perimeter = 24 + 4 × 10 = 64 cm.

So Region 1 has the larger perimeter but the smaller area.

(i) ? = 2 in. Each rectangle gives a length that the next one needs.

(ii) The top strip is 4 m high.

(i) Measure the sides of the outer rectangle ABCD and of the park EFGH.

(ii) No — we also need the park’s length l and width b. With path widths p, q (left, right) and t, d (top, bottom), break the path into rectangles:

(iii) No. The path is still Area(ABCD) − Area(EFGH); only its shape changes.

You need the widths of the two strips — nothing else.

Let the strip running along the 14 m side have width a, and the strip running along the 12 m side have width b.

Area of the spiral tube = 112 sq. units. (For the hint, the straight tube must be 9 units long.)

Straighten the spiral the same way: add the nine arm lengths (measured along the outer edge) and subtract 1 for each of the 8 bends.

Each region becomes 4 times as large, so the increase in each region is 3 times its original area. With side a:

Reason: doubling the side doubles every length in the figure, so every area is multiplied by 2 × 2 = 4. (Region 3 is half the square; the segment from the corner to the centre — the midpoint of the diagonal — splits the other half into regions 1 and 2 of a quarter each.)

Draw the two perpendicular cuts through the centre of the square, as in the figure, and cut out the 4 identical pieces. Turn each piece so that its right-angled corner (the one that was at the centre) becomes an outer corner, and fit the pieces round in a ring. They make a bigger square with a square hole in the middle.

Neither — the two triangles have exactly the same area, and each is half of the rectangle.

They are equal again. Each triangle is half of its rectangle, even though they sit on different sides.

Since the rectangles are identical, both halves are the same. So Area (∆XDC) = Area (∆YBC).

From Fig. 7.1 the enclosing rectangle ABCD has DC = 5 and AD = 4, and X lies on AB.

Just two: one side (the base) and the height to that side — that is, the perpendicular distance from the opposite vertex to the line of that base.

Draw the line l through the apex A parallel to the base BC. Then drop perpendiculars to l from B and from C.

Because it has been built to have four right angles.

• BE ⊥ BC and CD ⊥ BC by construction, so ∠EBC = ∠DCB = 90°.
• BE and CD are both perpendicular to BC, so BE ‖ CD.
• ED lies along the line l, and l ‖ BC by construction.

A quadrilateral with both pairs of opposite sides parallel and one angle 90° has all four angles 90°. So BCDE is a rectangle, with BC as its base and BE as its height.

X is the foot of the perpendicular from A to BC, so ∠AXB = 90°. Also ∠XBE = 90° (BE ⊥ BC), and EA lies along l which is parallel to BX.

AX is the height of the triangle and BE is the height of the rectangle. They are equal — which is exactly why the rectangle’s height can be used in the triangle’s area formula.

Yes, it still holds. For an obtuse triangle the foot of the perpendicular from A falls outside BC, at D on the line CB extended. Then ∆ABC is the difference of two right-angled triangles, each of which does sit inside a rectangle.

The trick is to compute the same area twice, once with each base–height pair.

Use the base and height that are already marked in the figure.

Yes — all four have equal areas, even though they are not all congruent.

Let the diagonals meet at O. Take two adjacent triangles, say 1 and 2, and choose OD and OB as their bases. Both bases lie on the same straight line BD, and both triangles have the same apex A, so they have the same height — the perpendicular from A to BD.

None of them — every one of these triangles has exactly the same area. There is no maximum and no minimum.

There is no maximum perimeter, but there is a definite minimum.

• No maximum: push the apex A further and further along l and the two slanted sides AB and AC grow without limit, so the perimeter grows without limit.
• Minimum: BC is common to every triangle, so we only have to make AB + AC as small as possible. Reflect C in the line l to get C′. Then AC = AC′, so AB + AC = AB + AC′, which is a path from B to C′ through A. The shortest such path is the straight segment BC′, so choose A where BC′ meets l.

They are equal: AB = AB′.

So the bent path B → A → C and the bent path B → A → C′ have exactly the same length.

Yes, it does — so the triangle of least perimeter is the isosceles one.

Set BC along a line and let l be parallel to it at distance d. Reflecting C in l puts C′ directly above C at height 2d, while B is at height 0.

In each case pick the matching base–height pair and halve the product.

In the figure AX ⊥ BC (with X outside the segment), AX = 4 units, BC = 6 units, AC = 8 units, and BY ⊥ AC.

48 sq. units.

∆SUB is isosceles with SU = SB, and SE is the perpendicular from the apex S to the base UB. In an isosceles triangle that perpendicular also bisects the base, so UE = EB.

Take the rectangle ABCD with DC as base and height h.

1. Extend DC beyond C to a point E so that CE = DC. Now DE = 2 × DC.
2. Mark any point P on the line AB (the side opposite DC) — the vertex A itself will do.
3. Join PD and PE. Then ∆PDE is the required triangle.

Cut the triangle at half its height and move the top piece down.

1. In ∆ABC (base BC, height h), mark M and N, the midpoints of AB and AC, and join MN. MN ‖ BC and lies at height h/2.
2. Draw AP ⊥ MN. Cut along MN and along AP: the top piece splits into two right triangles, ∆AMP and ∆ANP.
3. Turn ∆AMP half a turn about M: A lands on B and the piece fills the corner beside B. Turn ∆ANP half a turn about N: A lands on C.

Let each square have side s.

One quarter.

Join NY, as the hint suggests, and use the midpoint fact twice.

Reflect the water tank in the river, join the house to that image by a straight line, and fetch the water where that line meets the river.

1. Let H be the house and T the water tank. Treat the near bank of the river as a straight line r; H and T are on the same side of it.
2. Draw T′, the mirror image of T in r.
3. Join HT′ by a straight line. Let it cut r at P.
4. The shortest journey is H → P → T.

Split it into two triangles with a diagonal, and add their areas.

For quadrilateral ABCD, join BD. Now ABCD = ∆ABD + ∆CBD, and both triangles stand on the same base BD.

The same way — cut it into triangles.

Pick one vertex of the pentagon and join it to the two non-neighbouring vertices. That gives three triangles. Measure each triangle’s base and height, and add the three areas.

Yes. Every polygon can be cut into triangles by drawing diagonals inside it.

This is why the triangle formula is the key to the whole chapter:

The diagonal AC splits ABCD into ∆ABC and ∆ACD, and both stand on AC.

Or in one step:

The shaded region is what is left of rectangle ABCD after the two corner triangles AFE and EBC are cut off.

Join the centre O to the six vertices: the hexagon splits into 6 identical triangles. So we need only two measurements — the sidelength s and the distance h from the centre to a side (the height of each triangle).

(In a regular hexagon, h is fixed once s is known, so in fact the sidelength alone decides the area.)

Exactly one half — and it does not matter where the meeting point sits.

Take the given quadrilateral ABCD.

1. Draw the diagonal AC and mark M, the midpoint of AC.
2. Join BM and DM.
3. ABMD is the required quadrilateral.

One cut and one slide.

1. In parallelogram ABCD, drop a perpendicular from A to the side DC. Call the foot X, so AX ⊥ DC. AX is a height of the parallelogram.
2. Cut along AX. This separates ∆AXD from the trapezium ABCX.
3. Slide ∆AXD across to the right-hand end and fit it against BC. The result is a rectangle.

Yes. The test is to find the triangle that would complete ABCX into a rectangle, and then check whether ∆AXD is congruent to it.

Since the two triangles are congruent, ∆AXD can be laid exactly over the gap ∆BYC. So yes — the two pieces fit together into rectangle ABYX.

The fourth angle looks after itself.

So only three right angles need to be arranged; the fourth is forced. That is why constructing BY ⊥ XC is enough to finish the rectangle.

Yes — ∆AXD ≅ ∆BYC by the RHS congruency criterion.

The dissection turns parallelogram ABCD into rectangle ABYX, so their areas are equal — and a rectangle’s area we already know.

Once XY is shown to equal DC (see the next question), this becomes

Yes: XY = DC.

Yes. Any side may be used as the base, provided its own height is used with it.

Yes. Take AD as the base and let CZ be the perpendicular from C to AD (extended if necessary).

Cutting along CZ and sliding the piece across gives a rectangle with base AD and height CZ, by exactly the same congruence argument as before — the triangle cut off at one end is congruent to the gap at the other.

So the choice of side is free. Whichever side you can measure, together with the height to it, will do.

(i) All seven have the same area. Each has a base of the same length and lies between the same pair of parallel lines, so each has the same base and the same height.

(ii) Their perimeters are all different. The base stays fixed, but the two slanting sides get longer as the parallelogram leans over further.

• Minimum perimeter: (a) — the one that leans least, so its slant sides are closest to being upright.
• Maximum perimeter: (g) — the one that leans the most, so its slant sides are the longest.

Each time, multiply a side by its own height.

PQRS is a parallelogram with SR = 12 cm, QM ⊥ SR with QM = 6 cm, PS = 7.6 cm, and QN ⊥ PS.

The rectangle has the greater area.

∆ABC has base BC and height h, so its area is ½ × BC × h. We need a rectangle of area BC × h.

Method 1 — enclosing rectangle. Draw the line through A parallel to BC, and drop perpendiculars to it from B and C. The rectangle formed has base BC and height h, so its area is BC × h.

Method 2 — two copies. Turn a second copy of ∆ABC half a turn about the midpoint of AC and join it on: the two copies make a parallelogram of base BC and height h. Cut it along a height and slide the piece across to get the rectangle.

Cut the triangle at half its height and move the top piece down (the same dissection as Q5 of the earlier exercise).

1. In ∆ABC with base BC (take the longest side), mark M and N, the midpoints of AB and AC, and join MN. It is parallel to BC at height h/2.
2. Draw AP ⊥ MN and cut along MN and AP.
3. Turn ∆AMP half a turn about M (A lands on B) and ∆ANP half a turn about N (A lands on C).

Cut along the axis of symmetry, then fit the two right triangles together along their slant sides.

1. AD ⊥ BC in the isosceles ∆ABC (AB = AC), so ∆ADB ≅ ∆ADC (RHS) and D is the midpoint of BC.
2. Cut along AD: two right triangles, each with legs AD and DC (= ½ BC).
3. Keep ∆ADC. Turn ∆ADB over and lay its slant side AB along AC, so that the two right angles are at opposite corners.

Run Q7 backwards.

1. Take rectangle PQRS with base PQ = p and height QR = q.
2. Cut along the diagonal PR into two right triangles.
3. Keep one of them. Turn the other over and join it to the first along the side of length q, so that the two right angles sit side by side and their bases form one straight line.

The square is greater in both cases.

Let the common sidelength be s.

1. Draw both diagonals of rhombus ABCD. They meet at O and are perpendicular bisectors of each other.
2. BD cuts the rhombus into ∆ABD and ∆CBD, both isosceles (AB = AD, CB = CD).
3. Turn each into a rectangle as for an isosceles triangle (Figure it Out Q7): cut along AO (and CO) and put the two right-triangle halves together along their slant sides. This gives rectangles AO by ½BD and CO by ½BD.
4. Join the two rectangles along their equal sides ½BD to get one rectangle WXYZ.

From the dissection:

So the area of a rhombus is half the product of its diagonals.

The diagonals are perpendicular, so AO and CO are heights on the common base BD.

In each figure the dividing lines have already been drawn. Read off the pieces and add.

No lengths are printed on these figures, so measure the lengths with a ruler and substitute. In every case the pieces add up to ½ × height × (sum of the parallel sides).

Construct WM ⊥ ZY and XN ⊥ ZY, and let MZ = x, WM = XN = h, WX = a, NY = y.

Now bring in the second parallel side. Let ZY = b. Since MN = WX = a,

Yes.

Consequently WM = XN, so both perpendiculars equal the height h, and MN = WX = a.

Yes. Replace x + y using the longer parallel side.

So Area of a trapezium = ½ × height × (sum of the parallel sides).

Yes, the formula still holds. Take AB = a, DC = b and height h.

Approach 1. ABEF is a rectangle with FE = a; ∆AFD and ∆BEC are right triangles.

Approach 2. ABGD is a parallelogram (AB ‖ DG, BG ‖ AD), so DG = a and GC = b − a.

Yes — provided you draw the parallel line from an endpoint of the shorter parallel side.

We get a quadrilateral. The 6-sided possibility is ruled out by an angle count.

A parallelogram.

Its base is a + b and its height is h, and it is made of exactly two copies of the trapezium.

Reverse the rhombus dissection of pages 164–165. Let the rectangle be m by n.

1. Cut the rectangle across the middle of its side m into two rectangles, each ½m by n.
2. Cut each of them along a diagonal: four identical right triangles with legs ½m and n.
3. Put the four right angles together at one point O, with the legs along two perpendicular lines. The four slant sides form a rhombus with diagonals m and 2n.

Each is a trapezium, so use Area = ½ × height × (sum of the parallel sides).

Let ABCD be an isosceles trapezium with AB = a ‖ DC = b, AD = BC, and height h.

1. Drop AF ⊥ DC and BE ⊥ DC. ABEF is a rectangle (a by h), and the end triangles ∆AFD and ∆BEC are congruent (RHS).
2. Cut off both end triangles. Turn one over and join it to the other along the slant side: they make a small rectangle h by ½(b − a).
3. Place it against ABEF. Both have height h, so together they make one rectangle.

Find the midpoints of the two slanting sides and draw the perpendiculars through them.

1. Mark I, the midpoint of AD, and J, the midpoint of BC.
2. Through I draw a line perpendicular to DC. It meets the line AB at H and the line DC at G.
3. Through J draw a line perpendicular to DC. It meets AB at E and DC at F.
4. EFGH is the required rectangle.

A trapezium has the same area as a rectangle of the same height and width ½(a + b). Start from a rectangle of area 144 cm², say 16 cm × 9 cm.

1. Keep the height h = 9 cm and choose parallel sides with ½(a + b) = 16 cm, i.e. a + b = 32 cm — say a = 12 cm and b = 20 cm.
2. Draw DC = 20 cm. Draw a line parallel to DC at a distance of 9 cm, mark AB = 12 cm on it, and join AD and BC.

Trapezium : equilateral triangle : rhombus = 3 : 1 : 2.

Join the centre of the regular hexagon to its six vertices. This gives six congruent equilateral triangles, and every one of the three regions is made up of a whole number of them.

Let each small equilateral triangle have area t, so the hexagon has area 6t.

B is the point where the line ZA, extended, meets the line WX extended. The whole proof is one congruence.

Now build both figures out of the same pieces:

(measure your own table): a school desk of 120 cm by 60 cm.

Laying whole sheets without overlapping: put the 29.7 cm side across the 60 cm width (2 × 29.7 = 59.4 cm) and the 21 cm side along the 120 cm length (5 × 21 = 105 cm) — 10 sheets fit.

Multiply by 2.54, since 1 in = 2.54 cm.

Divide by 2.54, since 2.54 cm = 1 in.

A square of side 1 in is the same square as one of side 2.54 cm.

Every 6.4516 cm² makes one square inch, so divide.

Pace out the length and the breadth, then multiply. For a typical classroom of 8 m by 6 m:

In square feet, since 1 m ≈ 3.28 ft:

Estimate first, then check against the school’s records or a satellite map.

A school with 20 classrooms of about 48 m² each, plus corridors, offices, an assembly area and a playground, might work out like this:

That is roughly 1.2 acres (since 1 acre = 43,560 ft² ≈ 4047 m²).

Ask a farmer, a shopkeeper or the local land records office. Different parts of India use very different units, and even the same name can mean different sizes in different districts.

Estimate by fitting a rectangle around the settlement on a map, then look up the official figure from the panchayat, municipality or census.

Divide the two areas — after writing both in the same unit.

So about 4800 schools of that size would fill the town.

It depends on whether “city” means the area under the municipal body or the whole administrative region, so check current figures.

• (i) India: Delhi is usually named the largest — the National Capital Territory covers about 1,484 km². Bengaluru (about 700–750 km² under its municipal corporation) is also among the largest.
• (ii) World: on paper, some Chinese administrative “cities” are the largest — Chongqing (about 82,000 km²) and Hulunbuir (about 2,64,000 km²) — but these are huge regions, mostly countryside. By continuous built-up area, New York and Tokyo are among the largest, each roughly 8,000–12,000 km².

This is harder to pin down than the largest, because there is no agreed cut-off for when a town becomes a city.

• (i) India. There is no single agreed answer: many small towns and hill stations cover only a few square kilometres. Look up the areas of the smallest towns in your state in the Census town list or on your state’s urban development website.
• (ii) The world. Vatican City is the smallest independent state at about 0.44 km² — small enough to walk across in about a quarter of an hour.