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Class 8 Ganita Prakash · Part 2 · Chapter 7
Class 8 Ganita Prakash · Part 2 · Chapter 7
Area
Page-wise textbook solutions (pages 148–171, Part 2) · Every step · Figures · Common mistakes · Tips
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✔ Understood 0/49
๐️ Key ideas of this chapter
- Rectangle = length × width. Perimeter does NOT measure area.
- Triangle = \(\tfrac{1}{2}\) × base × height (works for every triangle). A median splits a triangle into two equal areas.
- Parallelogram = base × height. Rhombus = \(\tfrac{1}{2} d_1 d_2\). Trapezium = \(\tfrac{1}{2} h(a + b)\).
- Any polygon = sum of triangles. Dissection keeps area while changing shape (ลulba-Sลซtras).
- 1 in² = 6.4516 cm², 1 ft² = 144 in², 1 km² = 10⁶ m², 1 acre = 43,560 ft².
๐ Textbook Page 148
๐ Page 148Math Talk
Divide a square into 4 parts of equal area in creative ways.
1Start with 4 equal strips or 4 small squares.
2Push one edge of a part in and the next part's edge out by the same amount — areas stay equal.
3Also: the two diagonals give 4 equal triangles; lines through the centre at any angle (perpendicular pair) give 4 equal parts.
✅ Answer: Infinitely many ways.
๐ Textbook Page 149
๐ Page 149Rangoli
Which rectangle needs more powder: 7 cm × 4 cm or 8 cm × 3 cm? Area of each triangle when the 7 × 4 rectangle is cut by a diagonal?
128 cm² vs 24 cm²
2Diagonal makes two congruent triangles: ½ × 7 × 4 = 14 cm²
✅ Answer: 7 × 4 (28 cm²); each triangle 14 cm²
๐ Textbook Page 150
๐ Page 150Perimeter vs area
Give two rectangles where the one with larger perimeter has the smaller area.
110 × 1: perimeter 22, area 10
24 × 4: perimeter 16, area 16
3Other shapes: a thin wiggly strip vs a fat square.
✅ Answer: e.g. 10 × 1 and 4 × 4 — perimeter does not measure area.
๐ Page 150Figure it Out · Q1
Find the missing side lengths. (i) cross figure of areas 28, 21, 35, 14 in² (ii) 29 m² and 11 m² rectangles of height 4 m above a rectangle of area 50 m².
1(i) 21 ÷ 7 = 3 in tall. The 28 block is 4 + 3 = 7 in tall → 4 in wide. The 35 block is 3 + 4 = 7 in wide → 5 in tall. The 14 block is 5 + 2 = 7 in tall → width = 2 in.
2(ii) Widths: 29 ÷ 4 = 7.25 m and 11 ÷ 4 = 2.75 m (total 10 m).
3The 50 m² rectangle hangs under the 29 m² one (width 7.25 m) → height = 50 ÷ 7.25 ≈ 6.9 m.
4(If instead 50 m² is the whole figure, the bottom part is 10 m² → height 10 ÷ 7.25 ≈ 1.38 m.)
✅ Answer: (i) 2 in (ii) 7.25 m, 2.75 m, about 6.9 m
๐ Textbook Page 151
๐ Page 151Figure it Out · Q2
Path around a park EFGH inside ABCD. (i) Measurements and formula (ii) if widths are given (iii) does moving the outer rectangle change the path area?
1(i) Measure AB, AD (outer) and EF, EH (park). Path = AB × AD − EF × EH. E.g. 20 × 15 − 16 × 11 = 300 − 176 = 124 m².
2(ii) Widths alone are not enough — also need the park's length l and breadth b. With widths p (left), q (right), r (top), s (bottom): outer = (l + p + q)(b + r + s); path = (l + p + q)(b + r + s) − lb.
3(iii) No — the outer and inner rectangles keep the same size, so the difference stays the same.
✅ Answer: Path area = outer area − park area; it doesn't change when the park is shifted.
๐ Page 151Figure it Out · Q3
A 14 m × 12 m plot with a cross-path. What else is needed? Formula?
1Need the widths of the two paths: say x (path parallel to 12 m side) and y (parallel to 14 m side).
2Area = 12x + 14y − xy (the centre square is counted twice).
3Example x = y = 1 m: 12 + 14 − 1 = 25 m².
✅ Answer: 12x + 14y − xy
⚠️ Common mistake: Forgetting to subtract the overlapping middle square.
๐ Textbook Page 152
๐ Page 152Figure it Out · Q4
Area of the spiral tube of width 1.
1Hint idea: an L-bend of 5 and 5 (width 1) has area 5 + 4 = 9 → same as a straight tube of length 9 (each bend loses 1).
2Outer lengths of the 9 straight pieces: 20 + 20 + 20 + 15 + 15 + 10 + 10 + 5 + 5 = 120
38 bends → subtract 8: 112
✅ Answer: 112 square units (a straight tube of length 112)
๐ Page 152Figure it Out · Q5
If the side of the square is doubled, how do regions 1, 2, 3 change?
1Every length doubles, so every area becomes 2 × 2 = 4 times.
2Increase = 4 − 1 = 3 times the original area of each region.
✅ Answer: Each region becomes 4 times (increases by 3 times its area).
๐ Page 152Figure it Out · Q6
Cut a square by two perpendicular lines through the inside and rearrange into a bigger square with a hole.
1The 4 pieces are congruent if the lines pass through the centre.
2Rotate each piece so its square corner points outward; place them around a central gap.
3The outer square's side is the length of a cut; the hole is a small square.
✅ Answer: Big square = 4 pieces + square hole (total area stays the same plus the hole).
๐ Textbook Page 153
๐ Page 153Think
Compare ∆XDC and ∆YDC (or ∆YBC) in identical rectangles.
1Drop the altitude from X: the triangle has base DC and height = rectangle height.
2So each triangle is half of the rectangle.
✅ Answer: All have equal area — half the rectangle.
๐ Textbook Page 154
๐ Page 154Formula
Why is the area of a triangle ½ × base × height, even for an obtuse triangle?
1Enclose the triangle in a rectangle with the same base and height: triangle = half of it.
2Obtuse case: ∆ABC = ∆ADC − ∆ADB = ½h(DC − DB) = ½ h × BC.
✅ Answer: Area = ½ × base × height for every triangle.
๐ Textbook Page 155
๐ Page 155Application
Find BY in ∆ABC with BC = 5, AX = 3, AC = 4.
1Area = ½ × 5 × 3 = 7.5
2Also ½ × 4 × BY = 2BY → BY = 3.75
✅ Answer: 3.75 units
๐ Page 155Diagonals
Do the diagonals of a rectangle make 4 triangles of equal area?
1Triangles 1 and 2 have equal bases OD = OB (diagonals bisect each other) and the same height from A.
2Same for every pair.
✅ Answer: Yes — a median divides a triangle into two equal areas.
๐ Textbook Page 156
๐ Page 156Parallel lines
Triangles on base BC with third vertex on l ∥ BC: max/min area and perimeter?
1All have the same height → same area (no max or min).
2Perimeter: reflect C in l to C′. B → A → C has the same length as B → A → C′, shortest when A is on BC′.
3By symmetry BC′ meets l exactly above the midpoint of BC → A is on the perpendicular bisector (isosceles triangle).
4Moving A far away makes the perimeter as large as we like → no maximum.
✅ Answer: Equal areas; minimum perimeter for the isosceles one; no maximum.
๐ Textbook Page 157
๐ Page 157Figure it Out · Q1
Areas: (i) base BC = 4 cm, height 3 cm (ii) EF = 5 cm with altitude DN = 3.2 cm (iii) right triangle with legs 4 cm and 3 cm.
1(i) ½ × 4 × 3 = 6
2(ii) ½ × 5 × 3.2 = 8
3(iii) ½ × 3 × 4 = 6
✅ Answer: 6 cm², 8 cm², 6 cm²
๐ก Tip: In (i), if 4 cm is only EC, measure BC first. Always pair a base with the height drawn TO it.
๐ Textbook Page 158
๐ Page 158Figure it Out · Q2
BC = 6, AX = 4, AC = 8. Find BY.
1Area = ½ × 6 × 4 = 12
2½ × 8 × BY = 12 → BY = 3
✅ Answer: 3 units
๐ Page 158Figure it Out · Q3
∆SUB isosceles, SE ⊥ UB, area ∆SEB = 24. Area ∆SUB?
1In an isosceles triangle the altitude SE bisects UB, so ∆SEU and ∆SEB have equal areas.
✅ Answer: 48 sq units
๐ Page 158Figure it Out · Q4
Transform a rectangle into a triangle of equal area.
1Rectangle ABCD (base b, height h). Extend DC to E with CE = DC.
2Join A to E: triangle ADE has base 2b and height h → area bh.
3(Dissection: cut along AM where M is the midpoint of BC and swing triangle ABM down.)
✅ Answer: Triangle with the same height and double the base.
๐ Page 158Figure it Out · Q5
Transform a triangle into a rectangle of equal area.
1Join the midpoints of the two slanting sides (the 'mid-line').
2Drop perpendiculars from these midpoints to the base and cut off the two small corner triangles at the top, folding them down.
3Result: rectangle with the full base and half the height → ½ bh.
✅ Answer: Rectangle: same base, half the height.
๐ Page 158Figure it Out · Q6
Squares ABCD, BCEF, BFGH. (i) Red = 49, blue? (ii) red + blue = 180, square area?
1Let side s. Red = ∆DCH: base s, height 2s → s².
2Line DH crosses AB at its midpoint → blue = ½ × (s/2) × s = s²/4.
3(i) s² = 49 → blue = 12.25
4(ii) s² + s²/4 = 180 → s² = 144
✅ Answer: (i) 12.25 sq units (ii) 144 sq units
๐ Textbook Page 159
๐ Page 159Figure it Out · Q7
M, N midpoints of XY, XZ. Area ∆XMN as a fraction of ∆XYZ?
1Join NY. YN is a median of ∆XYZ → ∆XNY = ½ ∆XYZ.
2NM is a median of ∆XNY → ∆XMN = ½ ∆XNY.
✅ Answer: ¼
๐ Page 159Figure it Out · Q8
Shortest path from house to river to tank.
1Reflect the tank in the river bank (mirror image).
2Join the house to the reflection with a straight line; where it meets the river is where to fetch water.
3Same idea as the minimum-perimeter triangle.
✅ Answer: Go to the point where the house–(reflected tank) line meets the river.
๐ Page 159Polygons
How do we find the area of a quadrilateral or a pentagon?
1Draw diagonals from one vertex to split it into triangles.
2Add the areas of the triangles.
✅ Answer: Any polygon can be split into triangles.
๐ Textbook Page 160
๐ Page 160Figure it Out · Q1
Quadrilateral with AC = 22 cm, BM = DN = 3 cm (both ⊥ AC).
1½ × 22 × 3 + ½ × 22 × 3
✅ Answer: 66 cm²
๐ Page 160Figure it Out · Q2
Shaded area in rectangle ABCD (18 × 10) with AE = 10, EB = 8, AF = 6, FD = 4.
1Rectangle = 180
2Corner ∆AEF = ½ × 10 × 6 = 30; ∆EBC = ½ × 8 × 10 = 40
✅ Answer: 110 cm²
๐ Page 160Figure it Out · Q3
What measurements are needed for the area of a regular hexagon?
1It splits into 6 equilateral triangles of side s.
2Need the side and the height of one triangle (or the distance from the centre to a side).
✅ Answer: Side length (and the triangle height); area = 6 × ½ × s × height.
๐ Page 160Figure it Out · Q4
What fraction of the rectangle is blue?
1Blue = two triangles with the rectangle's top and bottom sides as bases and a common vertex P inside.
2Their heights add up to the rectangle's height H → ½ W h₁ + ½ W h₂ = ½ W H
✅ Answer: ½
๐ Page 160Figure it Out · Q5
Make a quadrilateral with half the area of a given quadrilateral.
1Join the midpoints of the four sides.
2Each corner triangle is ¼ of the triangle cut off by a diagonal, so the corners together make half.
✅ Answer: The midpoint quadrilateral (a parallelogram) has half the area.
๐ Textbook Page 161
๐ Page 161Parallelogram
Convert a parallelogram into a rectangle; find its area.
1Drop AX ⊥ DC and cut off ∆AXD.
2Slide it to the other side: ∆BYC ≅ ∆AXD (RHS).
3Rectangle ABYX has sides XY = DC (base) and AX (height).
✅ Answer: Area = base × height (any side with its own height).
๐ Textbook Page 162
๐ Page 162Figure it Out · Q1
Parallelograms with the same base and height: areas and perimeters?
1Same base, same height → same area.
2The more slanted, the longer the slanting sides → larger perimeter. The rectangle (no slant) has the least perimeter.
✅ Answer: Equal areas; the most slanted has the largest perimeter, the rectangle the smallest.
๐ Textbook Page 163
๐ Page 163Figure it Out · Q2
Areas: (i) base 7, height 4 (ii) base 5, height 3 (iii) side 5, height 4.8 (iv) side 2, height 4.4.
17 × 4
25 × 3
35 × 4.8
42 × 4.4
✅ Answer: 28, 15, 24, 8.8 cm²
๐ Page 163Figure it Out · Q3
Parallelogram PQRS: SR = 12, QM = 6, PS = 7.6. Find QN (⊥ PS).
1Area = 12 × 6 = 72
2Also PS × QN = 72 → QN = 72 ÷ 7.6
✅ Answer: QN ≈ 9.47 cm
๐ Page 163Figure it Out · Q4
Rectangle vs parallelogram with sides 5 cm and 4 cm.
1Same base 5; the parallelogram's height is less than its slant side 4.
✅ Answer: The rectangle (20 cm²) is greater.
๐ Page 163Figure it Out · Q5
Rectangle with twice the area of a triangle.
1Method 1: rectangle with the triangle's base and height (b × h = 2 × ½bh).
2Method 2: copy the triangle, rotate and join to make a parallelogram, then convert to a rectangle.
✅ Answer: b × h rectangle
๐ Textbook Page 164
๐ Page 164Figure it Out · Q6
Rectangle with the same area as a triangle.
1Same base, half the height (or half the base, full height).
✅ Answer: b × h/2
๐ Page 164Figure it Out · Q7
Convert an isosceles triangle into a rectangle by dissection.
1Cut along the altitude AD into two congruent right triangles.
2Turn one over and put the two hypotenuses together: a rectangle BD × AD.
✅ Answer: Rectangle with sides BD (half base) and AD (height).
๐ Page 164Figure it Out · Q8
Convert a rectangle into an isosceles triangle by dissection.
1Cut the rectangle along a diagonal into two right triangles.
2Place them back to back along a leg (the height) to make an isosceles triangle with base twice the rectangle's width.
✅ Answer: Reverse of Q7.
๐ Page 164Figure it Out · Q9
Equilateral triangle vs square of the same side; two triangles vs square?
1Triangle height = \(\tfrac{\sqrt{3}}{2}a \approx 0.87a\) < a → area ≈ 0.43a².
2Two triangles ≈ 0.87a² < a².
✅ Answer: The square is bigger in both cases.
๐ Textbook Page 165
๐ Page 165Rhombus
Show the area of a rhombus is ½ × d₁ × d₂.
1Area = ∆ADB + ∆CDB = ½ AO × BD + ½ CO × BD
2= ½ BD (AO + CO) = ½ BD × AC
✅ Answer: ½ × product of diagonals
๐ Textbook Page 167
๐ Page 167Trapezium
Area of a trapezium with parallel sides a, b and height h.
1Split into a rectangle (a × h) and two triangles (½hx, ½hy) with x + y = b − a.
2Area = ha + ½h(b − a) = ½h(a + b)
3Two copies make a parallelogram of base a + b.
✅ Answer: ½ × h × (a + b)
๐ Textbook Page 169
๐ Page 169Figure it Out · Q1
Rhombus with diagonals 20 cm and 15 cm.
1½ × 20 × 15
✅ Answer: 150 cm²
๐ Page 169Figure it Out · Q2
Convert a rectangle into a rhombus of equal area by dissection.
1Rectangle of length l and width w. Cut it along the vertical line through the midpoints of the long sides: two rectangles l/2 × w.
2Cut each along a diagonal: 4 congruent right triangles with legs l/2 and w.
3Put the 4 right angles together at one point, legs touching: the hypotenuses form a rhombus with diagonals l and 2w.
✅ Answer: Rhombus with diagonals equal to the rectangle's length and twice its width (area ½ × l × 2w = lw).
๐ Page 169Figure it Out · Q3
Areas: (i) parallel sides 10 ft and 7 ft, height 16 ft (ii) 24 m and 36 m, height 14 m (iii) 14 in and 6 in, height 10 in (iv) 12 ft and 18 ft, height 8 ft.
1½ × 16 × 17
2½ × 14 × 60
3½ × 10 × 20
4½ × 8 × 30
✅ Answer: 136 ft², 420 m², 100 in², 120 ft²
๐ Page 169Figure it Out · Q4
Convert an isosceles trapezium to a rectangle by dissection.
1Cut the isosceles trapezium along its line of symmetry: two congruent right trapeziums (parallel sides a/2 and b/2, height h).
2Rotate one piece through 180° (half turn).
3Join the two slanting sides: they fit exactly and make a rectangle of width (a + b)/2 and height h.
✅ Answer: Rectangle ½(a + b) × h.
๐ Page 169Figure it Out · Q5
How do we find the rectangle EFGH of equal area from trapezium ABCD?
1Mark I, J = midpoints of AD and BC.
2Draw vertical lines through I and J; they meet the line AB extended at H, E and DC at G, F.
3∆AHI ≅ ∆DGI and ∆BEJ ≅ ∆CFJ, so what is cut off is added back.
✅ Answer: Vertical lines through the midpoints of the slant sides.
๐ Textbook Page 170
๐ Page 170Figure it Out · Q6
Construct a trapezium of area 144 cm².
1Need ½ h(a + b) = 144 → h(a + b) = 288.
2Example: h = 12 cm, a = 10 cm, b = 14 cm.
✅ Answer: e.g. parallel sides 10 cm and 14 cm, 12 cm apart
๐ Page 170Figure it Out · Q7
Ratio of areas: trapezium, equilateral triangle, rhombus in a regular hexagon.
1A regular hexagon = 6 equal equilateral triangles.
2Trapezium (half hexagon) = 3, triangle = 1, rhombus = 2.
✅ Answer: 3 : 1 : 2
๐ Page 170Figure it Out · Q8
A is the midpoint of XY in trapezium ZYXW; ZA meets WX extended at B. Show trapezium = ∆ZWB.
1∆ZYA ≅ ∆BXA (ASA: AY = AX, vertical angles, alternate angles as ZY ∥ XB).
2Trapezium = quadrilateral ZAXW + ∆ZYA; triangle = ZAXW + ∆BXA.
✅ Answer: Equal areas.
๐ Page 170Real life
A4 area; 5 in, 7.4 in in cm; 5.08 cm, 11.43 cm in inches; 161.29 cm² in in²; in² in 1 ft²; m² in 1 km².
1A4: 21 × 29.7 = 623.7 cm²
25 × 2.54 = 12.7 cm; 7.4 × 2.54 = 18.796 cm
35.08 ÷ 2.54 = 2 in; 11.43 ÷ 2.54 = 4.5 in
4161.29 ÷ 6.4516 = 25 in²
51 ft² = 12 × 12 = 144 in²
61 km² = 1000 × 1000 = 10,00,000 m²
✅ Answer: 623.7 cm²; 12.7 cm, 18.796 cm; 2 in, 4.5 in; 25 in²; 144 in²; 1 million m²
๐ Where do we use this?
- Flooring, painting walls, land measurement, rangoli, designing gardens and paths, paper sizes.
๐ Link to higher classes
- Class 9: Heron's formula, areas of parallelograms on the same base. Class 10: areas of circles and sectors, surface area.
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