Chapter 7 exercise answers: Area

Open NCERT p. 150Checked by computerReads two ways: both answers shown

(i) Identify the missing sidelengths.

1. 21 in² box: width 7 in, so height = 21 ÷ 7 = 3 in.
2. The 28 in² box stands on the same bottom line and rises 4 in higher, so its height = 3 + 4 = 7 in and its width = 28 ÷ 7 = 4 in.
3. The 35 in² box has the same right edge as the 28 in² box and sticks out 3 in further left, so its width = 4 + 3 = 7 in and its height = 35 ÷ 7 = 5 in.
4. The 14 in² strip starts level with the top of the 35 in² box and ends 2 in lower, so its height = 5 + 2 = 7 in and its width = 14 ÷ 7 = 2 in.

AnswerThe missing side is 2 in.

(ii) Identify the missing sidelengths.

1. The top row is 4 m tall. Top-left side = 29 ÷ 4 = 29/4 m (7¼ m). Top-right side = 11 ÷ 4 = 11/4 m (2¾ m).
2. The arrow from 'Area = 50 m²' ends on the left edge of the lower box, and the page does not say whether 50 m² is the lower box alone or the whole thick-outlined rectangle (the 29 m² box together with the box under it).
3. Reading 1 (50 m² is the whole thick-outlined rectangle). We lead with this because that is the shape drawn with the thick outline, and the drawing fits it: the lower box is drawn about 2.6 m tall. Lower box = 50 − 29 = 21 m², with width 29/4 m, so its height = 21 ÷ 29/4 = 21 × 4/29 = 84/29 m (about 2.9 m).
4. Reading 2 (50 m² is the lower box alone): height = 50 ÷ 29/4 = 200/29 m (about 6.9 m).

AnswerTop sides: 29/4 m and 11/4 m. Right side read with 50 m² as the whole thick rectangle: 84/29 m; read with 50 m² as the lower box alone: 200/29 m.

Watch this explained “Reading a figure backwards”, 4:59 into Area as a count of unit squares

Open NCERT p. 151One way to think about it

(i) What measurements do you need to find the area of the path?

1. The path is what is left of the outer rectangle ABCD once the park EFGH is taken out. So area of path = area of ABCD − area of EFGH.
2. So you need four lengths: the length and width of ABCD (AB and AD), and the length and width of the park (EF and FG).
3. Any values will do as long as the park fits inside ABCD. Here is one choice: EF = 8 m, FG = 5 m, and a path 1 m wide all round, so AB = 10 m and AD = 7 m.
4. Area of ABCD = 10 × 7 = 70 m². Area of EFGH = 8 × 5 = 40 m².
5. Area of the path = 70 − 40 = 30 m².

In shortMeasure AB, AD, EF and FG. Area of path = AB × AD − EF × FG. For example, with AB = 10 m, AD = 7 m, EF = 8 m and FG = 5 m, area of path = 70 − 40 = 30 m².

(ii) If the width of the path along each side is given

1. No. The widths alone are not enough: a path 1 m wide round a small park covers less ground than a path 1 m wide round a big park. You also need the park's length EF = l and width FG = b.
2. Call the path's widths a (left), c (right), d (top) and e (bottom). The values below are one choice; any others work the same way.
3. Break the path into 4 rectangles. The left and right strips run alongside the park, so each is as tall as the park: a × b and c × b.
4. The top and bottom strips run the full outer length, l + a + c: d × (l + a + c) and e × (l + a + c).
5. Formula: area of path = (a + c) × b + (d + e) × (l + a + c).
6. Example: l = 8 m, b = 5 m, a = c = 1 m, d = e = 2 m. Area = (1 + 1) × 5 + (2 + 2) × 10 = 10 + 40 = 50 m².
7. Check with (i): ABCD is 10 m by 9 m, so the path is 90 − 40 = 50 m².

In shortNo. You also need the park's length l and width b. Area of path = (a + c) × b + (d + e) × (l + a + c). For example, l = 8 m, b = 5 m, side widths 1 m and 1 m, top and bottom widths 2 m and 2 m give 10 + 40 = 50 m².

(iii) Does the area of the path change when the outer rectangle is…

1. Area of the path = area of ABCD − area of EFGH.
2. Moving the outer rectangle, with the park still inside it, changes neither rectangle's length or width, so neither area changes.
3. So their difference, the area of the path, stays the same. Only the widths on the separate sides change.

In shortNo. The area of the path stays the same, because it is always area of ABCD − area of EFGH, and moving the rectangle changes neither.

Watch this explained “Wherever the park sits”, 7:09 into Area as a count of unit squares

Open NCERT p. 151One way to think about it

1. Besides the plot's sides (14 m and 12 m), you need the width of each arm of the crosspath: the width of the strip running across, and the width of the strip running down — these can differ.
2. Example: let the across-strip be 2 m wide and the down-strip be 3 m wide.
3. Across-strip area = 14 × 2 = 28 m². Down-strip area = 12 × 3 = 36 m².
4. The two strips overlap where they cross, in a 2 m × 3 m = 6 m² rectangle, which got counted in both strips — so subtract it once.
5. Area of crosspath = 28 + 36 − 6 = 58 m².

In shortFormula: area of crosspath = (plot length × across-width) + (plot width × down-width) − (across-width × down-width). With a 2 m and 3 m wide crosspath on the 14 m × 12 m plot, the area = 28 + 36 − 6 = 58 m².

Watch this explained “The same mistake, twice”, 8:04 into Area as a count of unit squares

Open NCERT p. 152Checked by computer

1. The small bent tube (arms of 5, width 1) is two 5×1 rectangles overlapping in a 1×1 corner square: area = 5 + 5 − 1 = 9, so the equivalent straight tube is 9 long.
2. The same idea works at every bend of the big spiral: two straight bands of width 1 meeting at a right angle overlap in exactly a 1×1 square.
3. Reading the outer corner-to-corner length of each straight run off the figure gives arms 20, 20, 20, 15, 15, 10, 10, 5, 5 (going around from the outside in), with 8 corners joining them.
4. Sum of the arms = 20+20+20+15+15+10+10+5+5 = 120.
5. Area of the spiral = 120 − 8 (one square lost to overlap at each of the 8 corners) = 112 sq units.

AnswerThe bent tube on the left straightens to a tube 9 long. The whole spiral's area = 112 square units.

Watch this explained “The same mistake, twice”, 8:04 into Area as a count of unit squares

Open NCERT p. 152Checked by computer

1. Let the square's original sidelength be s. The diagonal splits the square into two equal triangles, each s²/2; region 3 is one of them.
2. The other half is split again, at the diagonal's midpoint, into region 1 and region 2 — equal triangles (same base, same height), so each is s²/4.
3. Doubling every length in the figure scales each region's area by 2² = 4, whatever its shape.
4. So each region's new area is 4 times its old one, and the increase is 4−1 = 3 times its old area.
5. Region 1 increase = 3 × s²/4 = 3s²/4. Region 2 increase = 3s²/4. Region 3 increase = 3 × s²/2 = 3s²/2.

AnswerRegion 1 increases by 3s²/4, region 2 by 3s²/4, and region 3 by 3s²/2 (s = the original sidelength) — doubling every length always multiplies a region's area by 4, so its increase is always 3 times its original area.

Watch this explained “Nobody changed the unit”, 8:02 into Units for real areas, and why converting between them squares the length factor

Open NCERT p. 152One way to think about it

1. Draw two lines that cross at a right angle. Each line runs from one side of the square to the opposite side, and both are slanted, not parallel to the square's sides, as in the figure. Cut along them to get 4 pieces.
2. The two cuts are the same length, because each crosses the same square at the same slant. Call that length L.
3. Each piece has two right-angled corners: one where the cuts crossed, and one that was a corner of the square.
4. Slide each piece, without turning it, to the diagonally opposite position, so that its 'crossing' corner is on the outside.
5. Now the cut edges form the outside. Each side is made of the two parts of one cut, so it is L long, and each outer corner is a right angle. The outside is a square of side L.
6. The old corners of the square now point inwards, and the old outer edges fence off a square hole in the middle.
7. No area was created. The pieces still hold exactly the area of the original square, so area of hole = L × L − (area of the original square).
8. Example: a 4 cm square whose two cuts each measure 5 cm. The new square is 5 × 5 = 25 cm², the pieces still make 16 cm², and the hole is 9 cm², a 3 cm by 3 cm square.

In shortSlide the four pieces, without turning them, to the diagonally opposite positions so that the cut edges form the outside. They make a square whose side is the length of one cut, with a square hole in the middle. Area of hole = area of new square − area of old square. For example, a 4 cm square with 5 cm cuts gives 25 − 16 = 9 cm².

Watch this explained “Nothing was created”, 9:03 into Area as a count of unit squares

Open NCERT p. 157Checked by computer

(i) Find the areas of the following triangles:

1. Triangle ABC has base BC = 4 cm, with AE = 3 cm as the perpendicular height onto BC.
2. Area = ½ × 4 × 3 = 6 cm².

Answer6 cm²

(ii) Find the areas of the following triangles:

1. Triangle DEF: DN is the perpendicular dropped from D onto side EF, with EF = 5 cm and DN = 3.2 cm.
2. Area = ½ × 5 × 3.2 = 8 cm².

Answer8 cm²

(iii) Find the areas of the following triangles:

1. Triangle NAT is right-angled at A, with legs AT = 3 cm and AN = 4 cm — each leg is the height for the other as base.
2. Area = ½ × 3 × 4 = 6 cm².

Answer6 cm²

Watch this explained “Three sides, three heights, one area”, 6:56 into Why the area of a triangle is half base times height

Open NCERT p. 158Checked by computer

1. Using base BC = 6 units and height AX = 4 units: area of △ABC = ½ × 6 × 4 = 12 sq units.
2. Using base AC = 8 units and height BY: area of △ABC = ½ × 8 × BY.
3. A triangle has only one area, so ½ × 8 × BY = 12, giving BY = 24 ÷ 8 = 3 units.

AnswerBY = 3 units.

Watch this explained “Finding a length nobody handed you”, 7:39 into Why the area of a triangle is half base times height

Open NCERT p. 158Checked by computer

1. △SUB is isosceles with SU = SB, and SE is the perpendicular from the apex S onto the base UB — so E is the midpoint of UB, UE = EB.
2. △SEB and △SUE then share the same height (SE) and have equal bases (EB = UE), so they have equal areas.
3. Area of △SUB = area of △SEB + area of △SUE = 24 + 24 = 48 sq units.

Answer48 sq units.

Watch this explained “A special case, and then a payoff”, 2:27 into Triangles with the same base and height have the same area

Open NCERT p. 158One way to think about it

1. Take a rectangle ABCD with AB = l along the bottom and AD = w up the side. Its area is l × w.
2. Extend side AD beyond D to a point P with DP = AD, so AP = 2w. Join P to B.
3. Triangle PAB has base AB = l and height AP = 2w, because the angle at A is a right angle. Its area = ½ × l × 2w = l × w, the same as the rectangle.
4. You can also see it by cutting. PB crosses DC at its midpoint M. Triangle MCB (legs w and l/2, right angle at C) is identical to triangle MDP (legs w and l/2, right angle at D). So cutting MCB off the rectangle and moving it to MDP turns the rectangle into triangle PAB.
5. Example: a 6 × 4 rectangle (24 sq units) becomes a triangle with base 6 and height 8: ½ × 6 × 8 = 24.
6. This is one method among several. You could instead keep the side of 4 and double the other side to make a base of 12: ½ × 12 × 4 = 24.

In shortKeep one side of the rectangle as the base and make the triangle's height twice the other side. Area = ½ × l × 2w = l × w. For example, a 6 × 4 rectangle becomes a triangle with base 6 and height 8, area 24. Other choices work too, such as base 12 and height 4.

Watch this explained “Remaking a shape, and one last comparison”, 8:29 into Why the area of a triangle is half base times height

Open NCERT p. 158One way to think about it

1. Take a triangle with base b and height h. Its area is ½ × b × h.
2. Draw a rectangle on the same base b whose other side is half the triangle's height, h/2. One way to mark it: draw a line parallel to the base halfway up the height, and close the rectangle with perpendiculars from the ends of the base.
3. Area of the rectangle = b × h/2 = ½ × b × h, the same as the triangle.
4. Example: a triangle with base 6 and height 4 has area ½ × 6 × 4 = 12. The rectangle 6 × 2 also has area 12.
5. This is one method among several. You could instead halve the base and keep the full height: a 3 × 4 rectangle also has area 12.

In shortKeep the triangle's base b and make the rectangle's other side half the height, h/2. Area = b × h/2 = ½ × b × h. For example, a triangle with base 6 and height 4 (area 12) becomes a 6 × 2 rectangle. A 3 × 4 rectangle (half the base, full height) works too.

Watch this explained “Remaking a shape, and one last comparison”, 8:29 into Why the area of a triangle is half base times height

Open NCERT p. 158Checked by computer

(i) the area of the red region is 49 sq. units

1. Let each square have side s, so each has area s². D, C and E lie on the bottom line, and H is directly above B, so C, B and H lie on one straight line.
2. The red region is triangle DCH: base DC = s and height CH = CB + BH = 2s. Its area = ½ × s × 2s = s².
3. So s² = 49. Each square has area 49 sq units.
4. Let DH cross AB at P. Triangles PAD and PBH have AD = BH = s, right angles at A and B, and vertically opposite angles at P, so they are congruent. So AP = PB = s/2.
5. The blue region is triangle PAD, right-angled at A. Its area = ½ × AD × AP = ½ × s × s/2 = s²/4.
6. Blue area = 49 ÷ 4 = 49/4 = 12.25 sq units.

Answer49/4 sq units (12.25 sq units)

(ii) the total area enclosed by the blue and red regions is 180

1. From (i), red = s² and blue = s²/4, so together they make s² + s²/4 = 5s²/4.
2. 5s²/4 = 180, so s² = 180 × 4 ÷ 5 = 144.
3. Each square has area 144 sq units.

Answer144 sq units

Watch this explained “One piece read off another”, 8:07 into Any polygon is a pile of triangles

Open NCERT p. 159Checked by computer

1. Join NY, as the hint says.
2. In triangle XYZ, N is the midpoint of XZ. So triangles XYN and ZYN have equal bases (XN = NZ) and the same height from Y. They have equal areas, and area XYN = ½ × area XYZ.
3. In triangle XYN, M is the midpoint of XY. So triangles XMN and YMN have equal bases (XM = MY) and the same height from N. Area XMN = ½ × area XYN.
4. So area XMN = ½ × ½ × area XYZ = ¼ × area XYZ.

AnswerArea of ΔXMN = 1/4 of the area of ΔXYZ.

Watch this explained “A special case, and then a payoff”, 2:27 into Triangles with the same base and height have the same area

Open NCERT p. 159One way to think about it

1. The house H and the tank T are on the same side of the river. Gopal reaches the water at the bank nearer to them, so call that bank the line l.
2. Reflect the tank in l: drop a perpendicular from T to l and carry it on the same distance beyond l, to a point T′.
3. Draw a straight line from H to T′. Where it crosses the bank l, mark the point P.
4. The shortest path is H → P → T.
5. Why: take any point Q on the bank. Q is on the mirror line, so QT = QT′. So the walk H → Q → T is as long as H → Q → T′. That walk is shortest when it is a straight line, which happens exactly when Q = P.

In shortReflect the water tank in the near bank of the river to get T′. Join the house to T′ with a straight line. The point P where this line meets the bank is where Gopal should fetch the water, and the shortest path is house → P → tank.

Watch this explained “No triangle, no base, no area”, 8:01 into Triangles with the same base and height have the same area

Open NCERT p. 160Checked by computer

1. Diagonal AC splits ABCD into △ABC and △ACD, both standing on the same base AC = 22 cm.
2. BM = 3 cm is the height of △ABC onto AC, and DN = 3 cm is the height of △ACD onto AC (on the other side).
3. Area of ABCD = ½ × AC × (BM + DN) = ½ × 22 × 6 = 66 cm².

Answer66 cm².

Watch this explained “Three numbers, and nothing else”, 0:59 into Any polygon is a pile of triangles

Open NCERT p. 160Checked by computer

1. Rectangle ABCD has AB = AE + EB = 10 + 8 = 18 cm and AD = AF + FD = 6 + 4 = 10 cm, so its area = 18 × 10 = 180 cm².
2. The two unshaded corners are right triangles: △AEF with legs AE = 10 cm, AF = 6 cm (area ½×10×6 = 30 cm²), and △EBC with legs EB = 8 cm, BC = 10 cm (area ½×8×10 = 40 cm²).
3. Shaded area = 180 − 30 − 40 = 110 cm².

Answer110 cm².

Watch this explained “Taking away is a cut too”, 6:17 into Any polygon is a pile of triangles

Open NCERT p. 160One way to think about it

1. Join the centre of the regular hexagon to all 6 corners. This cuts it into 6 triangles, and because the hexagon is regular, all 6 are equilateral with side equal to the hexagon's side s.
2. So one length, the side s, fixes the whole hexagon and therefore its area. Two regular hexagons with the same side are the same size.
3. To work out the area with ½ × base × height, it is easiest to also measure the distance a from the centre to the middle of a side. That is the height of each triangle.
4. Area = 6 × ½ × s × a = 3 × s × a.
5. Example: s = 2 cm gives a ≈ 1.73 cm, so the area ≈ 3 × 2 × 1.73 ≈ 10.4 cm².
6. Other choices of measurement also work, for example the distance across the hexagon from one corner to the opposite corner, which is 2s.

In shortOne length, the side s, is enough, because it fixes the whole regular hexagon: 6 equilateral triangles of side s. To use ½ × base × height, also measure the distance a from the centre to a side: area = 6 × ½ × s × a = 3 × s × a. Other measurements, such as the corner-to-corner distance 2s, work too.

Watch this explained “The cheapest shape of all”, 5:11 into Any polygon is a pile of triangles

Open NCERT p. 160Checked by computer

1. The blue region is two triangles meeting at one interior point: one stands on the whole top edge, the other on the whole bottom edge.
2. Both triangles have the same base (the rectangle's width, w). Their two heights (the point's distance from the top edge and from the bottom edge) add up to the rectangle's full height, h, wherever the point sits.
3. So the two triangles' areas add to ½×w×(h−p) + ½×w×p = ½×w×h — exactly half the rectangle, for any position p of the point.
4. Fraction = (½ wh) / (wh) = ½.

Answer½ of the rectangle, whatever point the two triangles meet at.

Watch this explained “The answer that does not depend on the drawing”, 7:09 into Any polygon is a pile of triangles

Open NCERT p. 160One way to think about it

1. There is more than one way to do this; here is one.
2. Take the quadrilateral ABCD and mark P, Q, R and S, the midpoints of its sides AB, BC, CD and DA.
3. Join P, Q, R and S in order. The quadrilateral PQRS is the one we want.
4. Why it is half: draw the diagonal AC. Joining a corner of a triangle to the midpoint of the opposite side halves the triangle, and doing it twice gives a quarter, so ΔPBQ is one quarter of ΔABC and ΔRDS is one quarter of ΔACD. Together these two corner triangles are one quarter of ABCD.
5. In the same way, using the diagonal BD, the corner triangles ΔSAP and ΔQCR together are another quarter of ABCD.
6. So the four corner triangles cut off add up to half of ABCD, and PQRS, which is what remains, is the other half. (This argument is for a quadrilateral with no corner pointing inwards, like the one in the book.)

In shortJoin the midpoints of the four sides in order: the quadrilateral formed has exactly half the area of the given quadrilateral. Other methods also work.

Watch this explained “Halving it, and the shortcuts that follow”, 9:03 into Any polygon is a pile of triangles

Open NCERT p. 162Checked by computer

(i) What can we say about the areas of all these parallelograms

1. Every parallelogram (a)-(g) is drawn on the same grid, standing on the same base (5 grid units) with the same height (3 grid units) between the top and bottom sides.
2. Area = base × height, and that does not depend on how far the parallelogram leans over, so all seven have the same area (15 square units each).

AnswerYes — all seven parallelograms have equal area.

(ii) Which figure appears to have the maximum perimeter

1. Perimeter = 2 × (base + slant side), and the slant side gets longer the more a parallelogram leans over, even though its area does not change.
2. Reading the amount of lean (the sideways shift between the top and bottom sides) off the grid, (a) leans the least — closest to a rectangle — and (g) leans the most.

AnswerThe perimeters are NOT all equal: (g) has the greatest perimeter and (a) has the least, even though every parallelogram has the same area.

Watch this explained “One area, seven ways round”, 7:11 into Area of a parallelogram, by turning it into a rectangle

Open NCERT p. 163Checked by computer

(i) 7 cm

1. Base = 7 cm, height = 4 cm, marked perpendicular to the base.
2. Area = base × height = 7 × 4.

Answer28 cm²

(ii) 5 cm

1. Base = 5 cm, height = 3 cm (the diagonal drawn in the figure is not needed).
2. Area = 5 × 3.

Answer15 cm²

(iii) 4.8 cm

1. Here the height, 4.8 cm, is drawn perpendicular to the slanted 5 cm side rather than to a level side.
2. Any side can be used as the base as long as its OWN height is used: Area = 5 × 4.8.

Answer24 cm²

(iv) 4.4 cm

1. The height, 4.4 cm, is drawn perpendicular to the slanted 2 cm side.
2. Area = 2 × 4.4 — a height is a distance across the shape, so it is allowed to be longer than the side it stands on.

Answer8.8 cm² (44/5 cm²)

Watch this explained “Four to measure, one to find”, 8:04 into Area of a parallelogram, by turning it into a rectangle

Open NCERT p. 163Checked by computer

1. The area of a parallelogram is the same whichever side we take as the base, as long as we use the height that belongs to that side.
2. Base SR = 12 cm with its height QM = 6 cm: area = 12 × 6 = 72 cm².
3. Base PS = 7.6 cm (the 7.6 cm arrow runs the whole length of side PS) with its height QN, the perpendicular from Q to PS: area = 7.6 × QN.
4. So 7.6 × QN = 72, and QN = 72 ÷ 7.6 = 720/76 = 180/19 cm.

AnswerQN = 180/19 cm (about 9.47 cm).

Watch this explained “Four to measure, one to find”, 8:04 into Area of a parallelogram, by turning it into a rectangle

Open NCERT p. 163Checked by computer

1. Stand both shapes on the 5 cm side as the base.
2. For the rectangle, the 4 cm side meets the base at a right angle, so it IS the height: area = 5 × 4 = 20 cm².
3. For the parallelogram, the 4 cm side is slanted, so the true height (the perpendicular distance to the opposite side) is shorter than 4 cm.
4. A smaller height on the same base gives a smaller area, so the parallelogram holds less than 20 cm².

AnswerThe rectangle has the greater area — equal sidelengths do not give equal area unless the parallelogram is also right-angled (i.e. is itself a rectangle).

Watch this explained “The same two sides, and a different amount”, 9:12 into Area of a parallelogram, by turning it into a rectangle

Open NCERT p. 163One way to think about it

1. There are several methods; here is one, followed by two others.
2. Method 1: Choose any side of the triangle ABC as the base, say BC. At B and at C draw lines perpendicular to BC, and through A draw a line parallel to BC. These three lines and BC form a rectangle.
3. The rectangle has the same base BC as the triangle, and its other side is the triangle's height h (the perpendicular distance from A to BC). So its area is BC × h, while the triangle's area is ½ × BC × h: the rectangle has twice the area of the triangle.
4. Method 2: Take a different side of the triangle as the base, with the height to that side. This gives a different rectangle, also with twice the triangle's area.
5. Method 3: Make a second copy of the triangle, turn it half-way round about the midpoint of one side and join it to the first along that side. The two copies form a parallelogram with twice the triangle's area. Cut along a height of the parallelogram and move the triangle piece to the other end to turn it into a rectangle of the same area.

In shortDraw the rectangle that stands on one side of the triangle and passes through the opposite vertex: its area is base × height, twice the triangle's ½ × base × height. Using another side as the base, or joining two copies of the triangle into a parallelogram and turning that into a rectangle, also works.

Watch this explained “Building the box for a triangle you were handed”, 2:05 into Why the area of a triangle is half base times height

Open NCERT p. 164One way to think about it

1. Here is one method (others are possible). In triangle ABC take the longest side BC as the base, so the height from A falls inside BC.
2. Mark M, the midpoint of AB, and N, the midpoint of AC, and join MN. MN is parallel to BC and lies half-way up the triangle.
3. From A drop the perpendicular AP onto MN. Cut along MN and along AP. The top of the triangle comes off as two right triangles, AMP and ANP, and the bottom piece is the trapezium MBCN.
4. Turn ΔAMP half-way round about M: A lands on B, and the piece fills the gap between side BM and the vertical line through B. Turn ΔANP half-way round about N in the same way: A lands on C.
5. The pieces now form a rectangle with base BC and height equal to half the triangle's height.
6. Nothing was added or removed, and indeed BC × (h/2) = ½ × BC × h, the area of the triangle.

In shortCut the triangle along the line joining the midpoints of two sides and along the perpendicular from the top vertex to that line, then turn the two small top pieces half-way round about those midpoints: they complete a rectangle with the same base and half the height, so the same area. (Simply drawing a rectangle on the same base with half the height also gives a rectangle of the same area.)

Watch this explained “Remaking a shape, and one last comparison”, 8:29 into Why the area of a triangle is half base times height

Open NCERT p. 164One way to think about it

1. In the isosceles triangle ABC (AB = AC), AD is perpendicular to BC. Because the triangle is isosceles, D is the midpoint of BC, and ΔADB and ΔADC are congruent right triangles (RHS: AB = AC, AD is common).
2. Cut along AD to separate the two right triangles.
3. A right triangle is half of a rectangle: two copies of it placed together along the longest side make a rectangle whose sides are the two shorter sides.
4. So turn ΔADC over and place its side AC against side AB of ΔADB, with C on A and A on B. Its right angle now sits at the corner opposite D.
5. The two pieces form a rectangle whose sides are AD (the height) and BD (half the base).
6. Its area is AD × BD = AD × ½BC = ½ × BC × AD, which is the area of the triangle.

In shortCut along the altitude AD, turn one of the two right-angled halves over and join the halves along their equal slanted sides AB and AC: they form a rectangle of height AD and width ½BC, with the same area as the triangle.

Watch this explained “Each triangle becomes a rectangle”, 2:33 into Area of a rhombus from its diagonals

Open NCERT p. 164One way to think about it

1. Here is one method (others are possible). Let the rectangle PQRS have length PQ = l and breadth QR = b.
2. Cut along the diagonal PR. This gives two congruent right triangles, PQR and RSP, each with shorter sides l and b.
3. Turn ΔRSP over and place it beside ΔPQR so that its side of length b lies along QR and its side of length l lies along PQ extended beyond Q.
4. The two right angles now sit side by side at Q and together make a straight angle, so the two sides of length l form one straight line of length 2l.
5. The result is a triangle with base 2l and height b. Its two slanted sides are both copies of the diagonal PR, so it is isosceles.
6. Its area is ½ × 2l × b = l × b, the area of the rectangle.

In shortCut the rectangle along a diagonal, turn one half over and join it to the other along the side of length b so that the two sides of length l make one straight line: this gives an isosceles triangle with base 2l and height b, whose area ½ × 2l × b = l × b equals the rectangle's.

Watch this explained “Remaking a shape, and one last comparison”, 8:29 into Why the area of a triangle is half base times height

Open NCERT p. 164Checked by computer

1. An equilateral triangle's height is one leg of a right triangle whose hypotenuse is its own side, so the height is always shorter than the side.
2. Area(triangle) = 1/2 × side × height, which is therefore less than 1/2 × side × side — well under half of Area(square) = side × side.
3. Doubling the triangle only reaches 2 × (√3/4)×side² = (√3/2)×side² ≈ 0.866 × side², which is still less than side² (the square), since √3/2 is less than 1.
4. So the square has the greater area in both comparisons.

AnswerThe square has the greater area both times — a single equilateral triangle holds well under half the square, and even two of them together still fall short of it.

Watch this explained “Remaking a shape, and one last comparison”, 8:29 into Why the area of a triangle is half base times height

Open NCERT p. 169Checked by computer

1. Area of a rhombus = 1/2 × (product of its two diagonals).
2. Area = 1/2 × 20 × 15.

Answer150 cm².

Watch this explained “Stack them up”, 3:19 into Area of a rhombus from its diagonals

Open NCERT p. 169One way to think about it

1. Here is one method (others are possible). Let the rectangle PQRS have base PQ = l and height QR = b, and let M be the midpoint of the base PQ.
2. Cut from M to the two top corners S and R. This gives a middle triangle SMR and two corner right triangles, PMS and QMR, each with shorter sides l/2 and b.
3. Slide ΔQMR, without turning it, up and to the left until M lands on S. Slide ΔPMS up and to the right until M lands on R. The two pieces meet at a point T, a height b above the midpoint of SR.
4. The four sides SM, MR, RT and TS are all equal, since each is the longest side of a right triangle with shorter sides l/2 and b. So SMRT is a rhombus, with diagonals SR = l and MT = 2b.
5. Area of the rhombus = ½ × l × 2b = l × b, the area of the rectangle.

In shortCut from the midpoint of one side of the rectangle to the two opposite corners, and slide the two corner triangles across to the other side: the three pieces form a rhombus with diagonals l and 2b, whose area ½ × l × 2b = l × b equals the rectangle's.

Watch this explained “Run it backwards”, 9:25 into Area of a rhombus from its diagonals

Open NCERT p. 169Checked by computer

(i) 10 ft

1. This quadrilateral is drawn leaning across the page. Its two parallel sides are 10 ft and 7 ft, and the dashed line shows the perpendicular distance between them, 16 ft.
2. Area = 1/2 × 16 × (10 + 7).

Answer136 sq ft

(ii) 24 m

1. Parallel sides 24 m (top) and 36 m (bottom), 14 m apart.
2. Area = 1/2 × 14 × (24 + 36).

Answer420 sq m

(iii) 14 in

1. Here the two parallel sides are drawn vertically — 14 in on the left and 6 in on the right — with the dashed 10 in line the perpendicular (horizontal) distance between them.
2. Area = 1/2 × 10 × (14 + 6).

Answer100 sq in

(iv) 12 ft

1. Parallel sides 12 ft (top) and 18 ft (bottom), 8 ft apart.
2. Area = 1/2 × 8 × (12 + 18).

Answer120 sq ft

Watch this explained “The two formulae hiding inside”, 9:25 into Area of a trapezium, derived two different ways

Open NCERT p. 169One way to think about it

1. Here is one method. Let the isosceles trapezium ABCD have AB ∥ DC, with AB the shorter parallel side and AD = BC; let h be its height.
2. From B drop the perpendicular BF onto DC and cut along BF. This removes the right triangle BFC.
3. Because the trapezium is isosceles, ΔBFC is congruent to the corner triangle at D that the perpendicular from A would cut off: both have the height h, a slanted side AD = BC, and the overhang (DC − AB)/2.
4. Turn ΔBFC over and fit its slanted side BC against the slanted side AD, with C on A and B on D. Its right angle then sits at the top-left corner, level with AB.
5. The piece fills the gap at that end, and the shape becomes a rectangle of height h and length AB + (DC − AB)/2 = (AB + DC)/2.
6. Nothing was added or removed, so the rectangle has the trapezium's area, ½ × h × (AB + DC).

In shortDrop a perpendicular from one end of the shorter parallel side, cut off the right triangle it makes, turn it over and attach it along the other slanted side (the two slanted sides are equal): the pieces form a rectangle with the trapezium's height and a length equal to the average of the two parallel sides.

Watch this explained “An average width”, 3:58 into Area of a trapezium, derived two different ways

Open NCERT p. 169One way to think about it

1. Mark I, the midpoint of AD, and J, the midpoint of BC. The tick marks in the figure show AI = ID and BJ = JC.
2. Through I draw the line perpendicular to DC. It meets DC at G and the line AB (extended) at H.
3. Through J draw the line perpendicular to DC. It meets DC at F and the line AB (extended) at E.
4. EFGH is a rectangle: HE lies along line AB and GF along DC, and HG and EF are perpendicular to both, since AB ∥ DC.
5. Why the areas are equal: in ΔAHI and ΔDGI, AI = DI, ∠AIH = ∠DIG (vertically opposite) and ∠AHI = ∠DGI = 90°, so ΔAHI ≅ ΔDGI (AAS). In the same way ΔBEJ ≅ ΔCFJ.
6. The rectangle takes in ΔAHI and ΔBEJ, which lie outside the trapezium, and leaves out ΔDGI and ΔCFJ, which lie inside it. These are equal in area, so the rectangle EFGH and the trapezium ABCD have equal areas.

In shortTake I and J, the midpoints of the slanted sides AD and BC, and draw through them lines perpendicular to DC. These lines meet DC at G and F and the line AB (extended) at H and E. This gives the rectangle EFGH, which has the same area as the trapezium.

Watch this explained “An average width”, 3:58 into Area of a trapezium, derived two different ways

Open NCERT p. 170One way to think about it

1. There are many correct trapeziums; here is one.
2. Start with a rectangle of area 144 cm², for example a square 12 cm by 12 cm.
3. A trapezium has the same area as the rectangle with the same height whose length is the average of its two parallel sides (the conversion in Question 5, used in reverse). So we need a height of 12 cm and two parallel sides with an average of 12 cm, that is, a sum of 24 cm, for example 10 cm and 14 cm.
4. Construct it: draw DC = 14 cm. Draw a line parallel to DC at a distance of 12 cm. Mark AB = 10 cm anywhere along that line, and join AD and BC.
5. Check: area = ½ × 12 × (10 + 14) = 6 × 24 = 144 cm².

In shortOne such trapezium has parallel sides 10 cm and 14 cm, 12 cm apart: ½ × 12 × (10 + 14) = 144 cm². Any height h and parallel sides a and b with ½ × h × (a + b) = 144 also work.

Watch this explained “An average width”, 3:58 into Area of a trapezium, derived two different ways

Open NCERT p. 170Checked by computer

1. Joining the centre of a regular hexagon to all six vertices cuts it into 6 congruent equilateral triangles, each built on one side of the hexagon.
2. The single equilateral-triangle piece shown is exactly one of these six unit triangles.
3. The rhombus shown is made of two adjacent unit triangles glued along a shared edge.
4. What is left over — the trapezium — is made up of the remaining three unit triangles.
5. So the three areas are in the ratio 3 : 1 : 2 (trapezium : triangle : rhombus); 3+1+2 = 6 unit triangles, the whole hexagon.

Answer3 : 1 : 2 (trapezium : equilateral triangle : rhombus).

Watch this explained “The cheapest shape of all”, 5:11 into Any polygon is a pile of triangles

Open NCERT p. 170One way to think about it

1. B is the point where ZA, extended, meets WX extended, as in the figure.
2. Compare ΔZYA and ΔBXA: YA = XA (A is the midpoint of XY); ∠YAZ = ∠XAB (vertically opposite angles); ∠ZYA = ∠BXA (alternate angles, since ZY ∥ WB and XY is a transversal).
3. So ΔZYA ≅ ΔBXA by ASA (the side YA lies between the two angles at Y and A), and the two triangles have equal areas.
4. Trapezium ZYXW = quadrilateral ZAXW + ΔZYA.
5. ΔZWB = quadrilateral ZAXW + ΔAXB.
6. Since ΔZYA and ΔAXB have equal areas, the area of trapezium ZYXW equals the area of ΔZWB.

In shortΔZYA ≅ ΔBXA (ASA: YA = XA, the vertically opposite angles at A, and the alternate angles at Y and X), so cutting ΔZYA off the trapezium and putting ΔBXA in its place turns trapezium ZYXW into ΔZWB without changing the area.

Watch this explained “The same idea, wearing a different hat”, 3:24 into Triangles with the same base and height have the same area