CBSE Class 8 Mathematics Area Long Answer and Case Study Questions with Answers

Long Answer Questions

Question 1. A rectangular courtyard measures 12 m by 9 m.
(i) Explain how its area is obtained by packing it with unit squares of side 1 m, and write the area.
(ii) A square courtyard is built with the same perimeter as this rectangle. Find the side and the area of the square courtyard.
(iii) Which courtyard gives more floor space? State clearly what this example shows about the relation between perimeter and area.
Answer:
(i) Cover the floor with unit squares of side 1 m, each of area 1 m². One row along the length holds 12 squares, and 9 such rows fit along the width. So the courtyard holds 12 × 9 = 108 unit squares.
Area = length × width = 12 × 9 = 108 m².
(ii) Perimeter of the rectangle = 2(12 + 9) = 42 m. A square with the same perimeter has side = \( \frac{42}{4} \) = 10.5 m and area = 10.5 × 10.5 = 110.25 m².
(iii) The square courtyard gives more space: 110.25 m² against 108 m², a difference of 2.25 m², though both have exactly 42 m of boundary.
This shows that the perimeter of a region does not decide its area. Perimeter measures the boundary; area measures how many unit squares fit inside. Two regions with the same perimeter can have different areas, which is why area is measured by counting unit squares.
In simple words: Same 42 m boundary, but the square holds 110.25 m² and the rectangle only 108 m².

Teacher's Note:
1. Side of square = perimeter ÷ 4.
2. For a fixed perimeter, the square encloses the most area among rectangles.
3. 10.5² = 110.25.

Question 2. In △ABC shown below, ∠B = 90°, AB = 12 cm, BC = 16 cm and AC = 20 cm.
(i) Taking BC as the base, name the matching height and find the area of the triangle.
(ii) Taking AB as the base, name the matching height and find the area again.
(iii) Taking AC as the base, find the length of the altitude drawn from B to AC.
(iv) What do these three results tell you about the base-height pairs of a triangle?
Answer:
(i) With BC as base, the matching height is AB, because ∠B = 90° makes AB perpendicular to BC. Area = \( \frac{1}{2} \) × 16 × 12 = 96 cm².
(ii) With AB as base, the matching height is BC, for the same reason. Area = \( \frac{1}{2} \) × 12 × 16 = 96 cm².
(iii) Let the altitude from B to AC be h. The area is still 96 cm², so \( \frac{1}{2} \) × 20 × h = 96 ⇒ 10h = 96 ⇒ h = 9.6 cm. Check: \( \frac{1}{2} \) × 20 × 9.6 = 96 cm².
(iv) A triangle has three base-height pairs, and all of them give the same area. So a longer base goes with a shorter height: AC is the longest side (20 cm) and has the shortest altitude (9.6 cm). This is a quick way to find an unknown altitude.
In simple words: Every base-height pair gives 96 cm², so the altitude on the 20 cm side is 9.6 cm.

Teacher's Note:
1. In a right triangle, the two legs are heights for each other.
2. Use the known area to find an unknown height.
3. Longest side ↔ shortest altitude.

Question 3. A parallelogram-shaped glass panel is to be studied.
(i) With the help of a rough diagram, describe how a parallelogram can be cut and rearranged into a rectangle of the same area.
(ii) Using your diagram, state the formula for the area of a parallelogram and say why it is correct.
(iii) Find the area of a parallelogram whose base is 14 cm and whose corresponding height is 6.5 cm.
(iv) If its adjacent side measures 13 cm, find the height corresponding to that side.
Answer:
(i) Draw parallelogram ABCD with DC as base. From A drop AX perpendicular to DC. Cut off the right triangle AXD and slide it to the right-hand end, where it fits exactly as triangle BYC (BY perpendicular to DC produced).

The two triangles are congruent (RHS: BY = AX, ∠BYC = ∠AXD = 90°, BC = AD), so the piece fits exactly and we get the rectangle ABYX.
(ii) Area of parallelogram ABCD = area of rectangle ABYX = AX × XY. Here AX is the height, and since DX = CY, we get XY = DC, the base. Hence area of a parallelogram = base × height.
(iii) Area = 14 × 6.5 = 91 cm².
(iv) Any side can be the base with its own height, and the area stays 91 cm². So 13 × h = 91 ⇒ h = 7 cm.
In simple words: Moving one triangle across turns the parallelogram into a rectangle, so area = base × height = 91 cm² here.

Teacher's Note:
1. Dissection does not change area.
2. The height is perpendicular, not the slant side.
3. Check: 13 × 7 = 14 × 6.5 = 91.

Question 4. A field is in the shape of a trapezium. Its parallel sides measure 20 m and 12 m, and the distance between them is 9 m.
(i) Draw a rough figure and show how the trapezium can be split into one rectangle and two right triangles by dropping perpendiculars from the ends of the shorter parallel side.
(ii) Taking the two parts of the longer parallel side outside the rectangle as 5 m and 3 m, find the area of the field as the sum of the three pieces.
(iii) Now find the area using the formula \( \frac{1}{2} \) × height × sum of parallel sides and compare the two answers.
Answer:
(i) Let WXYZ be the field with WX = 12 m parallel to ZY = 20 m. Drop WM and XN perpendicular to ZY. Then WXNM is a rectangle 12 m by 9 m, with right triangles WMZ and XNY at the two ends.

(ii) Take MZ = 5 m and NY = 3 m (5 + 12 + 3 = 20 m).
Area △WMZ = \( \frac{1}{2} \) × 5 × 9 = 22.5 m²; rectangle WXNM = 12 × 9 = 108 m²; area △XNY = \( \frac{1}{2} \) × 3 × 9 = 13.5 m².
Total = 22.5 + 108 + 13.5 = 144 m².
(iii) Area = \( \frac{1}{2} \) × 9 × (20 + 12) = \( \frac{1}{2} \) × 9 × 32 = 144 m².
The two answers agree. The formula needs only the height and the parallel sides, and works for any split, because the two triangle bases always add up to 20 − 12 = 8 m.
In simple words: Rectangle plus two triangles gives 144 m², exactly what the trapezium formula gives.

Teacher's Note:
1. The rectangle's length equals the shorter parallel side.
2. Both triangles have the trapezium's height.
3. Any split 5 + 3, 4 + 4, ... gives the same total.

Question 5. In △ABC, D is the midpoint of BC, so AD is a median. AE is drawn perpendicular to BC.
(i) Using the formula area = \( \frac{1}{2} \) × base × height, justify that area(△ABD) = area(△ACD).
(ii) If the area of △ABC is 86 cm², write the area of each of the two parts.
(iii) PQRS is a rectangle whose diagonals meet at O. Using the same idea, explain why the four triangles POQ, QOR, ROS and SOP have equal areas.
Answer:
(i) D is the midpoint of BC, so BD = DC. AE ⊥ BC, so AE is the height of both △ABD (base BD) and △ACD (base DC), since both bases lie on the line BC.
Area △ABD = \( \frac{1}{2} \) × BD × AE and area △ACD = \( \frac{1}{2} \) × DC × AE. As BD = DC, the two areas are equal. A median divides a triangle into two triangles of equal area.
(ii) Each part has area \( \frac{86}{2} \) = 43 cm².
(iii) The diagonals PR and QS of the rectangle bisect each other at O, so PO = OR and QO = OS.
In △POQ and △QOR, take PO and OR as bases; both have the same height, the perpendicular from Q to PR. Equal bases and equal heights give equal areas, so QO is a median of △PQR.
Repeating this around the rectangle, all four triangles have equal areas, each one-quarter of the rectangle.
In simple words: A median gives two triangles with equal bases and the same height, so equal areas; the same idea makes the four triangles in a rectangle equal.

Teacher's Note:
1. The same perpendicular serves as the height of both parts.
2. The four triangles in a rectangle are equal in area, not all congruent.
3. 86 ÷ 2 = 43.

Question 6. In the figure, the line l is parallel to BC, and P, Q, R are three points on l.
(i) Compare the areas of △PBC, △QBC and △RBC, giving a clear reason.
(ii) If BC = 12 cm and the distance between l and BC is 7 cm, find the area of each triangle.
(iii) Are the perimeters of these triangles also equal? Which position of the apex on l gives the smallest perimeter, and how does the mirror-image (reflection) argument justify it?
Answer:
(i) All three triangles stand on the same base BC. Their apexes lie on l, and l ∥ BC, so each apex is at the same perpendicular distance from BC. Equal base and equal height give area △PBC = area △QBC = area △RBC.
(ii) Area = \( \frac{1}{2} \) × 12 × 7 = 42 cm² for each triangle, even though they look different.
(iii) No, the perimeters are not equal. BC is common, so only the sum of the other two sides matters, and it grows as the apex moves far along l.
Treat l as a mirror and reflect C to C′. For any apex A on l, AC = AC′, so the path B → A → C has the same length as B → A → C′. This is shortest when it is the straight segment BC′.
BC′ cuts l at the point directly above the midpoint of BC, so the apex there (here Q) gives the isosceles triangle with the smallest perimeter.
In simple words: All three have area 42 cm², but the triangle with apex above the midpoint of BC has the least perimeter.

Teacher's Note:
1. Equal area does not mean equal perimeter.
2. A straight line is the shortest path between two points.
3. The reflection turns a bent path into a straight one.

Question 7. A rectangular garden measures 24 m by 18 m. Two paths, each 2 m wide, are made inside it: one runs across the full length and the other across the full width, crossing each other at the middle.
(i) Find the area of each path taken separately.
(ii) Find the total area of the crosspath, explaining why the two areas cannot simply be added.
(iii) Find the area left for grass.
(iv) Find the cost of paving the crosspath at ₹60 per m².
Answer:
(i) Path along the length: 24 × 2 = 48 m². Path along the width: 18 × 2 = 36 m².
(ii) The paths overlap in a 2 m × 2 m square of area 4 m² at the middle. Adding 48 + 36 counts this square twice, so subtract it once:
Area of the crosspath = 48 + 36 − 4 = 80 m².
(iii) Area of the garden = 24 × 18 = 432 m², so grass area = 432 − 80 = 352 m².
(iv) Cost = 80 × 60 = ₹4800.
Check: 352 + 80 = 432 m², the whole garden.
In simple words: The crosspath covers 80 m² (overlap counted once), leaving 352 m² of grass; paving costs ₹4800.

Teacher's Note:
1. Watch for the overlapping square.
2. Grass + paths must equal the whole garden.
3. Cost = area × rate.

Question 8. A plot of land is a quadrilateral ABCD in which the diagonal AC measures 26 m. From B and D, perpendiculars BM and DN are dropped on AC, with BM = 8 m and DN = 6 m.
(i) Explain how any polygon can be split into triangles so that its area can be found.
(ii) Find the areas of △ABC and △ACD.
(iii) Find the area of the plot ABCD.
(iv) If the plot had been a pentagon instead, how many triangles would it split into from one vertex, and which lengths would have to be measured?
Answer:
(i) From one vertex of a polygon, draw all the diagonals to the non-adjacent vertices. This cuts the polygon into non-overlapping triangles. Each triangle's area is \( \frac{1}{2} \) × base × height, so the polygon's area is the sum of these. Here AC splits ABCD into △ABC and △ACD, with heights BM and DN on the common base AC.
(ii) Area △ABC = \( \frac{1}{2} \) × 26 × 8 = 104 m²; area △ACD = \( \frac{1}{2} \) × 26 × 6 = 78 m².
(iii) Area of ABCD = 104 + 78 = 182 m². The same follows from \( \frac{1}{2} \) × 26 × (8 + 6) = 13 × 14 = 182 m².
(iv) From one vertex of a pentagon, 2 diagonals can be drawn, giving 3 triangles. We would measure the two diagonals and the perpendicular height on each triangle's base, i.e. the bases and heights of the three triangles.
In simple words: The diagonal splits the plot into triangles of 104 m² and 78 m², so the plot is 182 m².

Teacher's Note:
1. A polygon with n sides splits into n − 2 triangles from one vertex.
2. Common base: area = ½ × diagonal × sum of the perpendiculars.
3. Check both methods give 182.

Question 9. A farmer’s rectangular plot measures 180 ft by 121 ft. Use 1 acre = 43,560 ft², 1 in = 2.54 cm and 1 in² = 6.4516 cm².
(i) Find the area of the plot in ft² and then in acres.
(ii) A rectangular window pane for the farmhouse measures 24 in by 18 in. Find its area in in² and in cm².
(iii) A glass sheet measuring 150 cm by 100 cm is available. By comparing areas only, estimate how many such panes it could give.
(iv) Now work out how many whole panes can actually be cut from the sheet, and explain why this number is smaller than your estimate in part (iii).
Answer:
(i) Area = 180 × 121 = 21,780 ft². In acres: \( \frac{21780}{43560} \) = 0.5 acre, exactly half an acre.
(ii) Area = 24 × 18 = 432 in²; 432 × 6.4516 = 2787.0912 ≈ 2787.09 cm². Check: 24 in = 60.96 cm, 18 in = 45.72 cm, and 60.96 × 45.72 = 2787.0912 cm².
(iii) Area of the sheet = 150 × 100 = 15,000 cm². \( \frac{15000}{2787.0912} \) ≈ 5.38, so by area alone the sheet seems enough for about 5 panes.
(iv) A pane is 60.96 cm by 45.72 cm. Along 150 cm: 150 ÷ 60.96 ≈ 2.46, so 2 fit; along 100 cm: 100 ÷ 45.72 ≈ 2.19, so 2 fit. That gives 2 × 2 = 4 whole panes. (Turning the panes the other way gives only 3 × 1 = 3.)
Glass pieces cannot be joined: the leftover strips (about 28 cm and 8.6 cm wide) have plenty of area, but neither is wide enough for a whole pane. An area calculation gives only an upper limit; the real number also depends on the shape and arrangement.
In simple words: The plot is half an acre; the sheet has room by area for 5 panes, but only 4 can really be cut.

Teacher's Note:
1. Divide ft² by 43,560 to get acres.
2. Count whole pieces along each side, then multiply.
3. Area estimates are upper limits for cutting.

Question 10. The ancient Śulba-Sūtras describe how one shape can be changed into another of exactly the same area.
(i) Give a method to change a given rectangle into a triangle of equal area, and justify it using the area formulas.
(ii) Give a method to change a given triangle into a rectangle of equal area, and justify it.
(iii) Apply your method of part (i) to a rectangle measuring 12 cm by 5 cm, stating the base and height of the triangle you obtain, and verify the areas are equal.
(iv) Apply your method of part (ii) to a triangle of base 16 cm and height 7.5 cm, stating the sides of the rectangle you obtain, and verify.
Answer:
(i) Let the rectangle be ABCD with base AB and width AD. Produce AD beyond D to A′ so that DA′ = AD, and draw the line m through A′ parallel to AB. Take any point P on m and join PA and PB.
△PAB has base AB and height AA′ = 2AD, so area △PAB = \( \frac{1}{2} \) × AB × 2AD = AB × AD = area of the rectangle.
(ii) Let the triangle have base BC and height h. On BC build a rectangle BCYX whose other side is \( \frac{h}{2} \). Its area = BC × \( \frac{h}{2} \) = \( \frac{1}{2} \) × BC × h = area of the triangle.
(By dissection: cut the triangle along the line halfway up, parallel to BC, and fold the two upper pieces down on either side to fill the rectangle.)
(iii) Rectangle 12 cm × 5 cm has area 60 cm². The method gives a triangle with base 12 cm and height 2 × 5 = 10 cm. Check: \( \frac{1}{2} \) × 12 × 10 = 60 cm².
(iv) The triangle has area \( \frac{1}{2} \) × 16 × 7.5 = 60 cm². The method gives a rectangle 16 cm by \( \frac{7.5}{2} \) = 3.75 cm. Check: 16 × 3.75 = 60 cm².
In simple words: Double the height to turn a rectangle into a triangle; halve the height to turn a triangle into a rectangle.

Teacher's Note:
1. The factor ½ in the triangle formula is balanced by doubling or halving the height.
2. Any apex on the parallel line gives the same area.
3. Both examples come out to 60 cm².

Case Study Based Questions

Question 11. Two rangoli rectangles
For a school festival, Meera draws two rectangular rangoli patches on the floor and fills each one evenly with coloured powder. The first patch is 9 cm long and 4 cm wide; the second one is a 6 cm by 6 cm square patch. Before filling them, she packs each patch with non-overlapping unit squares of side 1 cm drawn with chalk, exactly as her textbook suggests. She then runs a white chalk border along the boundary of each patch. Her friend Rahul guesses that the patch with the longer border will surely need more powder.
(i) How many unit squares fit into the first patch?
(ii) Find the area of each patch.
(iii) Find the length of the chalk border of each patch.
(iv) Is Rahul’s guess correct? What does this example show about perimeter and area?
Answer:
(i) Each row holds 9 unit squares and there are 4 rows, so 9 × 4 = 36 unit squares.
(ii) First patch: 9 × 4 = 36 cm²; square patch: 6 × 6 = 36 cm². Both need the same amount of powder.
(iii) Border of the first patch = 2(9 + 4) = 26 cm; border of the square patch = 4 × 6 = 24 cm.
(iv) No. The first patch has the longer border (26 cm) but needs exactly the same powder, since the areas are equal. Perimeter measures the boundary and area the surface covered, so perimeter is not a measure of area.
In simple words: Both patches cover 36 cm², even though their borders are 26 cm and 24 cm.

Teacher's Note:
1. Powder depends on area, chalk border on perimeter.
2. Equal areas can have different perimeters.
3. Count rows × columns for unit squares.

Question 12. The triangular park
A park in a town is shaped like a triangle. Its longest boundary, which runs along the main road, measures 40 m, and the perpendicular distance from the opposite corner of the park to this road is 25 m. The municipal office plans to cover the whole park with lawn grass costing ₹12 per square metre. A straight walking path is also to be laid from the corner opposite the road to the exact middle point of the road boundary, so that the park is divided into two plots, one for a children’s play area and one for a seating garden.
(i) Find the area of the park.
(ii) Find the cost of covering the park with grass.
(iii) Find the area of each of the two plots formed by the walking path.
(iv) Name the property of triangles that justifies your answer in part (iii).
Answer:
(i) Area = \( \frac{1}{2} \) × base × height = \( \frac{1}{2} \) × 40 × 25 = 500 m².
(ii) Cost = 500 × 12 = ₹6000.
(iii) The path goes from the opposite corner to the midpoint of the road boundary, so each plot has area \( \frac{500}{2} \) = 250 m².
(iv) A median divides a triangle into two triangles of equal area. The two plots stand on equal bases (20 m each) and have the same height 25 m from that corner, so their areas are equal.
In simple words: The park is 500 m², grass costs ₹6000, and the path makes two plots of 250 m² each.

Teacher's Note:
1. The road boundary is the base; the perpendicular distance is the height.
2. The walking path is a median.
3. Check: ½ × 20 × 25 = 250.

Question 13. A courtyard with a tiled border
A rectangular courtyard of a house measures 10 m by 8 m. The owner decides to lay plain grey tiles over most of the courtyard, but keeps a decorative border of red tiles of uniform width 1 m running inside the courtyard all along its four edges. The grey tiles therefore cover only the rectangular region inside the red border. Red tiles cost ₹150 per square metre, while grey tiles cost ₹90 per square metre.
(i) Find the total area of the courtyard.
(ii) Find the length and breadth of the grey region inside the border.
(iii) Find the area of the red border.
(iv) Find the cost of the red tiles.
Answer:
(i) Total area = 10 × 8 = 80 m².
(ii) The border takes 1 m off each edge, so the grey region is 10 − 1 − 1 = 8 m long and 8 − 1 − 1 = 6 m wide.
(iii) Grey region = 8 × 6 = 48 m², so red border = 80 − 48 = 32 m².
(iv) Cost of red tiles = 32 × 150 = ₹4800.
In simple words: The red border covers 80 − 48 = 32 m² and costs ₹4800.

Teacher's Note:
1. An inside border reduces each side by twice its width.
2. Border = outer area − inner area.
3. Use the red-tile rate, not the grey one.

Question 14. The kite-shaped flag
A school art club prepares small flags for Sports Day. Each flag is cut in the shape of a rhombus whose two diagonals measure 60 cm and 40 cm. The cloth is bought by area. The club first makes one sample flag, then enough cloth is ordered for a set of 25 flags. On the sample, the longer diagonal is stitched with a golden thread, which divides the flag into two triangular halves of different colours. For a special display flag, the club keeps the shorter diagonal unchanged but stretches the longer diagonal to 120 cm.
(i) Find the area of one ordinary flag.
(ii) Find the total area of cloth needed for 25 flags, in cm² and in m².
(iii) Find the area of each coloured half of the sample flag.
(iv) Find the area of the special display flag and say how it compares with an ordinary flag.
Answer:
(i) Area = \( \frac{1}{2} \) × 60 × 40 = 1200 cm².
(ii) Cloth for 25 flags = 25 × 1200 = 30,000 cm². Since 1 m² = 10,000 cm², this is \( \frac{30000}{10000} \) = 3 m².
(iii) A diagonal divides a rhombus into two congruent triangles, so each half has area \( \frac{1200}{2} \) = 600 cm².
(iv) Display flag: area = \( \frac{1}{2} \) × 120 × 40 = 2400 cm², which is twice an ordinary flag, because one diagonal was doubled while the other stayed the same.
In simple words: One flag is 1200 cm², 25 flags need 3 m², and the display flag is twice as big.

Teacher's Note:
1. 1 m² = 100 × 100 = 10,000 cm².
2. Doubling one diagonal doubles the area.
3. A diagonal halves a rhombus.

Question 15. The gravel path around a lawn
A rectangular lawn in a housing colony measures 25 m by 16 m. The residents decide to lay a gravel walking path of uniform width 1.5 m all around the lawn, on the outside, so that the lawn itself is not disturbed. The path is bounded by a low brick edging on its outer side. The contractor charges ₹85 per square metre for laying the gravel. Before approving the work, the colony secretary wants a clear calculation of how much gravel surface will actually be created, and he reminds everyone that the path is not simply the boundary line of the lawn.
(i) Find the length and breadth of the outer rectangle formed by the path.
(ii) Find the area of this outer rectangle.
(iii) Find the area of the gravel path.
(iv) Find the cost of laying the gravel path.
Answer:
(i) The path is 1.5 m wide on every side, so the outer rectangle is 25 + 1.5 + 1.5 = 28 m long and 16 + 1.5 + 1.5 = 19 m wide.
(ii) Area of the outer rectangle = 28 × 19 = 532 m².
(iii) Area of the lawn = 25 × 16 = 400 m², so area of the path = 532 − 400 = 132 m². The path is a region with area, not just a boundary line.
(iv) Cost = 132 × 85 = ₹11,220.
In simple words: The path covers 532 − 400 = 132 m² and costs ₹11,220.

Teacher's Note:
1. Outside path: add twice the width to each side.
2. Path area = outer − lawn.
3. 132 × 85 = 11,220.

Question 16. A plot shared among heirs
A farmer owns a triangular plot. Its boundary along the canal measures 60 m, and the perpendicular distance from the opposite corner to the canal is 40 m. In his will he asks that the plot first be divided between his two sons by a straight fence drawn from the corner opposite the canal to the midpoint of the canal boundary, so that each son gets an equal share. Each son later divides his own share equally between his two children in exactly the same way, by joining a corner of his plot to the midpoint of the opposite side. The four grandchildren then find that their plots do not look alike at all.
(i) Find the area of the whole plot.
(ii) Find the area of each son’s share.
(iii) Find the area of each grandchild’s plot.
(iv) The four grandchildren’s plots have different shapes and different boundary lengths. Does this mean their areas are unequal? Give a reason.
Answer:
(i) Area = \( \frac{1}{2} \) × 60 × 40 = 1200 m².
(ii) The fence is a median, which halves the area. Each son gets \( \frac{1200}{2} \) = 600 m².
(iii) Each son's plot is halved again by a median, so each grandchild gets \( \frac{600}{2} \) = 300 m².
(iv) No. Equal area does not need equal shape or equal boundary. Every division used a median, which always gives two triangles of equal base and equal height, so all four plots have area 300 m² even though their shapes and perimeters differ.
In simple words: Halving twice gives four plots of 300 m² each, whatever their shapes.

Teacher's Note:
1. Each median halves the area.
2. Same area, different shapes is possible.
3. 1200 ÷ 4 = 300.

Question 17. Panes cut from a glass sheet
A glazier has a rectangular glass sheet measuring 2 m by 1.2 m. A customer first orders window panes each measuring 50 cm by 40 cm. The glazier plans to cut them in neat rows and columns, with the 50 cm side of every pane placed along the 2 m side of the sheet. Later the customer changes the order to panes measuring 45 cm by 40 cm, keeping the same arrangement, with the 45 cm side along the 2 m side. Glass left over as a narrow strip cannot be used for any pane.
(i) Find the area of the glass sheet in cm².
(ii) Find the area of one 50 cm by 40 cm pane, and how many such panes the sheet yields.
(iii) For the second order, find how many 45 cm by 40 cm panes can be cut in the planned arrangement.
(iv) Find the area of the glass wasted in the second order.
Answer:
(i) The sheet is 200 cm by 120 cm, so its area = 200 × 120 = 24,000 cm².
(ii) One pane = 50 × 40 = 2000 cm². Along 200 cm: 200 ÷ 50 = 4 panes; along 120 cm: 120 ÷ 40 = 3 panes. So 4 × 3 = 12 panes, using 12 × 2000 = 24,000 cm², the whole sheet.
(iii) 200 ÷ 45 = 4.44…, so only 4 panes fit along the length; 120 ÷ 40 = 3 along the width. Number of panes = 4 × 3 = 12.
(iv) Area used = 12 × (45 × 40) = 12 × 1800 = 21,600 cm². Glass wasted = 24,000 − 21,600 = 2400 cm², the leftover strip 20 cm wide and 120 cm long.
In simple words: Both orders give 12 panes, but the smaller panes waste a 2400 cm² strip.

Teacher's Note:
1. Change metres to centimetres first.
2. Only whole panes count, so round down.
3. Check: 20 × 120 = 2400.

Question 18. The trapezium-shaped field
A field is shaped like a trapezium. Its two parallel boundaries, one along a village road and one along a canal, measure 48 m and 32 m, and the perpendicular distance between them is 20 m. The owner harvests 3 kg of grain from every square metre of the field. He is also thinking of exchanging this field for a rectangular field of exactly the same area, whose breadth would again be 20 m, so that the same irrigation channel can be used along it.
(i) Find the area of the trapezium-shaped field.
(ii) Find the total mass of grain harvested from the field.
(iii) Find the length of the rectangular field that would have the same area.
(iv) Compare the length found in part (iii) with the two parallel sides of the trapezium. What do you notice, and why does it happen?
Answer:
(i) Area = \( \frac{1}{2} \) × 20 × (48 + 32) = 10 × 80 = 800 m².
(ii) Grain = 800 × 3 = 2400 kg.
(iii) Length × 20 = 800, so length = \( \frac{800}{20} \) = 40 m.
(iv) 40 m is exactly the average of the parallel sides: \( \frac{48 + 32}{2} \) = 40 m. This is because \( \frac{1}{2} \)h(a + b) can be written as h × \( \frac{a + b}{2} \), the area of a rectangle of the same height whose length is the average of the parallel sides.
In simple words: The field is 800 m², gives 2400 kg of grain, and equals a 40 m × 20 m rectangle.

Teacher's Note:
1. A trapezium equals a rectangle with the average length.
2. Grain = area × yield per m².
3. Divide the area by the breadth to get the length.

Question 19. Triangular plots between two roads
Two straight roads in a new township run parallel to each other and are 18 m apart. Along the first road a stretch BC of length 30 m is marked out as a common boundary. Three triangular plots are then laid out, each having BC as its base, with the third corner at a point P, Q or R on the second road. The point Q is chosen exactly opposite the middle of BC, while P lies above a point near one end of BC and R lies above a point far beyond the other end. A buyer insists on the plot with the largest area, while the township office wants to fence the plot that needs the least fencing.
(i) Find the area of the plot PBC.
(ii) Compare the areas of the three plots, with a reason.
(iii) Which of the three plots needs the least fencing, and why?
(iv) If a fourth corner S is taken very far along the second road, what happens to the area and to the perimeter of the plot SBC?
Answer:
(i) Area of PBC = \( \frac{1}{2} \) × 30 × 18 = 270 m².
(ii) All three plots have the same area, 270 m². They share the base BC, and their third corners lie on a road parallel to the first, so each height is the same 18 m. The buyer gains nothing by choosing.
(iii) Plot QBC needs the least fencing. BC is common, so only the other two sides matter. Reflect C in the second road to C′; for any corner A on that road, AC = AC′, so the fence B → A → C equals the path B → A → C′. The shortest such path is the straight segment BC′, which meets the road exactly above the midpoint of BC, that is, at Q. So △QBC is isosceles and has the least perimeter.
(iv) The area of SBC stays 270 m², since the base and height do not change. The perimeter keeps growing without limit, because SB and SC both get longer. Area and perimeter are independent measures.
In simple words: All plots have 270 m²; the plot with its corner opposite the middle of BC needs the least fencing.

Teacher's Note:
1. Same base, same parallels ⇒ same area.
2. The reflection trick finds the shortest path.
3. Area can stay fixed while perimeter grows.

Question 20. The craft worker’s paper square
A craft worker takes a square sheet of paper of side 24 cm. She marks the midpoints P, Q, R and S of its four sides and joins them in order, so that an inner square PQRS is formed. She colours the inner square and leaves the four corner triangles white, as shown.
(i) Find the area of the original square sheet.
(ii) Find the area of one white corner triangle.
(iii) Find the area of the coloured inner square PQRS, and express it as a fraction of the original square.
(iv) She repeats the same design on a square sheet of side 48 cm. Find the area of the new coloured inner square and state how many times the earlier coloured area it is.
Answer:
(i) Area of the sheet = 24 × 24 = 576 cm².
(ii) Each corner triangle is right-angled with its two shorter sides 12 cm each (half of 24 cm). Area = \( \frac{1}{2} \) × 12 × 12 = 72 cm².
(iii) The four corner triangles cover 4 × 72 = 288 cm². So the coloured square PQRS = 576 − 288 = 288 cm², which is \( \frac{288}{576} \) = \( \frac{1}{2} \) of the original square.
(iv) New sheet: 48 × 48 = 2304 cm²; coloured square = \( \frac{2304}{2} \) = 1152 cm². Since 1152 = 4 × 288, it is 4 times the earlier coloured area. Doubling the side multiplies the area by 4, not 2.
In simple words: The coloured square is always half the sheet: 288 cm² for side 24 cm and 1152 cm² (4 times) for side 48 cm.

Teacher's Note:
1. Joining midpoints of a square's sides gives half its area.
2. Doubling a length makes the area 4 times.
3. Subtract the corners from the whole.