Class 8 Maths Ganita Prakash Chapter 14 Solutions (NCERT 2026–27) – Area

These Class 8 Maths Ganita Prakash Chapter 14 solutions cover Area — the areas of triangles, parallelograms, rhombuses, trapeziums and composite figures. This is Chapter 7 of Ganita Prakash Part II (the 14th and final chapter of the Class 8 course). Every “Figure it Out” question, Math Talk and Try This prompt is solved step by step, with every area value verified and written with correct units (cm2, m2, in2, ft2).

Class: 8 Subject: Mathematics Book: Ganita Prakash (Part II) Chapter: 14 (Part II Ch 7) Exercises: 5 “Figure it Out” sets Session: 2026–27
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Chapter 14 Overview

Chapter 14 of Ganita Prakash, Area, is the closing chapter of the Class 8 course (Chapter 7 of Part II). It starts from the familiar idea that the area of a rectangle is length × width, then builds the area of a triangle (½ × base × height) by enclosing it in a rectangle. From there the chapter derives special formulas for the parallelogram, rhombus and trapezium using dissection — cutting a figure into pieces and rearranging them into a rectangle of equal area, a method that traces back to the ancient Indian Śulba-Sūtras. It also shows that any polygon can be split into triangles, and closes with real-life area units (cm2, in2, ft2, m2, acres, km2). The Class 8 Maths Ganita Prakash Chapter 14 solutions below work through every part of the chapter.

Key Concepts & Definitions

Area: the number of unit squares (sidelength 1 unit) that exactly fill a region; written in square units such as cm2 or sq. cm.

Base & height of a triangle: any side can be the base; the height is the perpendicular distance from the opposite vertex to that base.

Diagonal of a rectangle: divides it into two congruent triangles, so each triangle is half the rectangle.

Dissection: cutting a figure into pieces and rearranging them into a different figure of equal area.

Perimeter is not area: two regions can have the same perimeter but different areas, and a larger perimeter does not mean a larger area.

Polygon → triangles: any polygon can be divided into triangles, so its area is the sum of the triangle areas.

Median property: the line joining a vertex to the midpoint of the opposite side splits a triangle into two triangles of equal area.

Important Formulas (Chapter 14)

Rectangle: Area = length × width.

Triangle: Area = ½ × base × height.

Parallelogram: Area = base × height (height = perpendicular distance between the base and its opposite side).

Rhombus: Area = ½ × product of the diagonals = ½ × d1 × d2.

Trapezium: Area = ½ × height × (sum of the two parallel sides) = ½ × h × (a + b).

Unit conversions: 1 in = 2.54 cm, so 1 in2 = 6.4516 cm2; 1 ft = 12 in; 1 acre = 43,560 ft2.

Figure it Out — Rectangles & Composite Figures (Page 150)

1. Identify the missing sidelengths. (i) A staircase-shaped figure made of rectangles labelled with areas 28 in2, 21 in2, 35 in2, 14 in2, with given lengths 4 in, 3 in, 7 in, 2 in. (ii) A figure of total area 50 m2 split into parts of 29 m2 and 11 m2, with a given side 4 m.

SOLUTION (i) Use Area = length × width for each rectangle and the shared sides. • Top rectangle (28 in2) has height 4 in ⇒ its width = 28 ÷ 4 = 7 in; together with the 3 in piece its base = 3 + 7 = 10 in is the column width. • Right rectangle (21 in2) has the marked length 7 in ⇒ its height = 21 ÷ 7 = 3 in. • Lower-right strip (14 in2) has height (3 + 2) = … taking its given base, missing “? in” width = 14 ÷ (height shown 2 in) = 7 in; so the bottom width is 7 in. • Left rectangle (35 in2) shares the 7 in width ⇒ its height = 35 ÷ 7 = 5 in. (ii) The full rectangle has area 50 m2 and one side 4 m ⇒ the other (full) side = 50 ÷ 4 = 12.5 m. The 29 m2 part has height 4 m ⇒ its width = 29 ÷ 4 = 7.25 m. The 11 m2 part has height 4 m ⇒ its width = 11 ÷ 4 = 2.75 m. Check: 7.25 + 2.75 = 10 m, and the lower block makes up the remaining 50 − 29 − 11 = 10 m2 region, consistent with the figure.

2. The figure shows a path (the shaded portion) laid around a rectangular park EFGH. (i) What measurements do you need to find the area of the path? Once you identify the lengths to be measured, assign possible values of your choice to these measurements and find the area of the path. Give a formula for the area. [Hint: There is a relation between the areas of EFGH, the path, and ABCD.] (ii) If the width of the path along each side is given, can you find its area? If not, what other measurements do you need? Assign values of your choice to these measurements and find the area of the path. Give a formula for the area using these measurements. [Hint: Break the path into rectangles.] (iii) Does the area of the path change when the outer rectangle is moved while keeping the inner rectangular park EFGH inside it, as shown?

SOLUTION (i) The path = outer rectangle ABCD − inner park EFGH. So you need the length and width of both rectangles. Area of path = Area(ABCD) − Area(EFGH). Example: let ABCD = 12 m × 8 m = 96 m2 and EFGH = 8 m × 4 m = 32 m2. Area of path = 96 − 32 = 64 m2. (ii) Knowing only the width of the path is not enough; you also need the inner park’s length and width (or the outer dimensions). Break the path into 4 corner squares + 4 side rectangles. Example: park = 8 m × 4 m, uniform path width w = 1 m. Outer = (8 + 2)(4 + 2) = 10 × 6 = 60 m2; path = 60 − 32 = 28 m2. Formula with park L × B and width w: Area = (L + 2w)(B + 2w) − LB = 2w(L + B) + 4w2. (iii) If the outer rectangle slides but stays the same size and still contains EFGH, the path area = Area(ABCD) − Area(EFGH) is unchanged, since both areas stay the same. (Only the shape of the strips changes, not the total.)

3. The figure shows a plot with sides 14 m and 12 m, and with a crosspath. What other measurements do you need to find the area of the crosspath? Once you identify the lengths to be measured, assign some possible values of your choice and find the area of the path. Give a formula for the area based on the measurements you choose.

SOLUTION You also need the width of each path (one running the 14 m direction, one the 12 m direction). Let each path be w = 2 m wide. Vertical path = 12 × 2 = 24 m2; horizontal path = 14 × 2 = 28 m2. They overlap in a 2 × 2 = 4 m2 square, counted twice. Area of crosspath = 14w + 12w − w2 = 28 + 24 − 4 = 48 m2 (for w = 2 m). In general, for a plot L × B with both paths of width w: Area = w(L + B) − w2.

4. Find the area of the spiral tube shown in the figure. The tube has the same width throughout. [Hint: There are different ways of finding the area. Here is one method.] What should be the length of the straight tube if it is to have the same area as the bent tube on the left?

SOLUTION The tube has constant width 1 unit. Straighten it out: the area equals (total length of the centre-line of the tube) × (width). Adding the straight segments of the spiral (the marked runs of 20, 20, 15, 15, 10, 10, 5, 5 units along the path) gives a total tube length of 100 units. Area = length × width = 100 × 1 = 100 sq. units. So a straight tube of width 1 must be 100 units long to have the same area as the bent tube. The L-shaped piece shown (two 5-unit arms of width 1) has area 5 + 5 − 1 = 9 sq. units, equal to a straight tube of length 9.

5. In this figure, if the sidelength of the square is doubled, what is the increase in the areas of the regions 1, 2 and 3? Give reasons.

SOLUTION All three regions are parts of the square, so each one’s area is a fixed fraction of the square’s area. When the side is doubled (s → 2s), the square’s area becomes (2s)2 = 4s2 — 4 times the original. Every region scales by the same factor. So each of regions 1, 2 and 3 also becomes 4 times its original area, i.e. the increase is 3 times (300% of) the original area of that region. (The shapes stay similar; doubling lengths multiplies any area by 22 = 4.)

6. Divide a square into 4 parts by drawing two perpendicular lines inside the square as shown in the figure. Rearrange the pieces to get a larger square, with a hole inside. You can try this activity by constructing the square using cardboard, thick chart paper, or similar materials.

SOLUTION This is a hands-on dissection. Draw the two perpendicular cuts so they cross off-centre; this gives 4 quadrilateral pieces. Slide each piece outward by the same small amount and rotate to reassemble — the four pieces fit into a larger square outline with a small square hole in the middle. Key idea: rearranging pieces never changes the total area, so (area of big square) = (area of original square) + (area of the hole). The “extra” size is exactly the hole — no area is created. (This is the classic dissection illusion; the hole accounts for the apparent gain.)

Figure it Out — Triangles (Page 157)

1. Find the areas of the following triangles: (i) ▵ABC with base BC = 4 cm and height AE = 3 cm (AE ⊥ BC). (ii) ▵DEF with DF = 5 cm and the perpendicular EN from E to DF equal to 3.2 cm. (iii) Right triangle NAT, right-angled at A, with AT = 3 cm and AN = 4 cm.

SOLUTION (i) Area = ½ × base × height = ½ × 4 × 3 = 6 cm2. (ii) Take DF = 5 cm as the base and EN = 3.2 cm as the height: Area = ½ × 5 × 3.2 = 8 cm2. (iii) The legs AT = 3 cm and AN = 4 cm meet at the right angle, so they are base and height: Area = ½ × 3 × 4 = 6 cm2.

2. Find the length of the altitude BY. (In ▵ABC, AB = 4 units is one side with foot X on BC, BC = 6 units, and AC = 8 units; BY is the altitude from B to AC.)

SOLUTION Find the area two ways. Using base BC = 6 and height AX = 4: Area(▵ABC) = ½ × 6 × 4 = 12 sq. units. Using base AC = 8 and height BY: Area = ½ × 8 × BY = 4·BY. Equate: 4·BY = 12 ⇒ BY = 3 units.

3. Find the area of ▵SUB, given that it is isosceles, SE is perpendicular to UB, and the area of ▵SEB is 24 sq. units.

SOLUTION In an isosceles triangle SUB (SU = SB), the altitude SE from the apex to the base UB is also the median, so E is the midpoint of UB and UE = EB. Therefore ▵SEU and ▵SEB have equal bases and the same height SE, so Area(▵SEU) = Area(▵SEB) = 24 sq. units. Area(▵SUB) = Area(▵SEU) + Area(▵SEB) = 24 + 24 = 48 sq. units.

4. [Śulba-Sūtras] Give a method to transform a rectangle into a triangle of equal area.

SOLUTION Take rectangle PQRS. On the line of one side PQ (the base), keep the base PQ. A triangle of equal area on the same base must have height = 2 × (rectangle’s height), because triangle area = ½ × base × height while rectangle area = base × height. Method: extend the side and mark a point T at a perpendicular distance 2h from base PQ (where h = PS). Join P–T–Q. Then Area(▵PQT) = ½ × PQ × 2h = PQ × h = Area of the rectangle.

5. [Śulba-Sūtras] Give a method to transform a triangle into a rectangle of equal area.

SOLUTION Take ▵ABC with base BC and height h from A. Mark the midpoint M of the height and draw a line through M parallel to BC; it meets AB and AC at P and Q. Drop perpendiculars from P and Q to BC, meeting it at P′ and Q′. The two small triangles cut off (AP-side and AQ-side) are congruent to the two corner triangles, so they fold down to fill the rectangle PQ Q′P′. This rectangle has base = PQ = ½BC and height = ½h, so its area = ½BC × ½h × … rearranged equals ½ × BC × h = Area of the triangle. (Equivalently: a rectangle on base BC with height h/2 has the same area.)

6. ABCD, BCEF, and BFGH are identical squares. (i) If the area of the red region is 49 sq. units, then what is the area of the blue region? (ii) In another version of this figure, if the total area enclosed by the blue and red regions is 180 sq. units, then what is the area of each square?

SOLUTION Let each identical square have side s, so each square has area s2. The red region is a triangle whose base and height are full square sides, and the blue region is a smaller triangle along the bottom-left square. (i) The red triangle has area = ½ × s × (2s) = s2 (it spans one square wide and two squares tall). Given red = 49 ⇒ s2 = 49 ⇒ s = 7. The blue triangle has area = ½ × s × s = ½ × 49 = 24.5 sq. units. (ii) Blue + red = ½s2 + s2 = (3/2)s2 = 180 ⇒ s2 = 120. So the area of each square is 120 sq. units.

7. If M and N are the midpoints of XY and XZ, what fraction of the area of ▵XYZ is the area of ▵XMN? [Hint: Join NY]

SOLUTION Join NY. Since N is the midpoint of XZ, the median YN splits ▵XYZ into two equal halves, so Area(▵XYN) = ½ Area(▵XYZ). In ▵XYN, M is the midpoint of XY, so NM is a median and Area(▵XMN) = ½ Area(▵XYN) = ½ × ½ Area(▵XYZ). Therefore Area(▵XMN) = ¼ of the area of ▵XYZ.

8. Gopal needs to carry water from the river to his water tank. He starts from his house. What is the shortest path he can take from his house to the river and then to the water tank? Roughly recreate the map in your notebook and trace the shortest path.

SOLUTION This is the “reflection” shortest-path problem (the same idea used earlier for the minimum-perimeter triangle). Method: Reflect the water tank across the river line to get the image point T′. Draw the straight line from the House to T′; the point P where it crosses the river is the best touch point. The shortest path is House → P (on the river) → Water tank, because House–P–Tank has the same length as the straight segment House–T′, and a straight line is the shortest distance.

Figure it Out — Area of any Polygon (Page 160)

1. Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN = 3 cm, BM is perpendicular to AC, and DN is perpendicular to AC.

SOLUTION Diagonal AC = 22 cm splits ABCD into ▵ABC (height BM = 3 cm) and ▵ACD (height DN = 3 cm), both on base AC. Area(▵ABC) = ½ × 22 × 3 = 33 cm2; Area(▵ACD) = ½ × 22 × 3 = 33 cm2. Area(ABCD) = 33 + 33 = 66 cm2. (Equivalently ½ × AC × (BM + DN) = ½ × 22 × 6 = 66 cm2.)

2. Find the area of the shaded region given that ABCD is a rectangle. (AB = 10 + 8 = 18 cm along the top with E on AB so that AE = 10 cm and EB = 8 cm; the height is 10 cm with F on AD so that AF = 6 cm and FD = 4 cm; DC = 18 cm. The shaded region is triangle/quadrilateral D-E-C-F.)

SOLUTION Rectangle area = 18 × 10 = 180 cm2. The shaded region is the rectangle minus the three unshaded corner triangles ▵AEF, ▵EBC and (the part above). ▵AEF: legs AE = 10 cm, AF = 6 cm ⇒ area = ½ × 10 × 6 = 30 cm2. ▵EBC: legs EB = 8 cm, BC = 10 cm ⇒ area = ½ × 8 × 10 = 40 cm2. ▵FDC (left lower corner, unshaded above F to D): with FD = 4 cm and DC = 18 cm ⇒ area = ½ × 18 × 4 = 36 cm2. Shaded area = 180 − 30 − 40 − 36 = 74 cm2.

3. What measurements would you need to find the area of a regular hexagon?

SOLUTION Split the regular hexagon into 6 identical triangles by joining the centre to all vertices. You need the side length of the hexagon (the base of each triangle) and the apothem — the perpendicular distance from the centre to a side (the height of each triangle). Then Area = 6 × (½ × side × apothem) = 3 × side × apothem. (Equivalently, ½ × perimeter × apothem.)

4. What fraction of the total area of the rectangle is the area of the blue region?

SOLUTION The blue region is the two opposite triangles formed by joining the midpoints of the left and right sides to the centre — together they form a “bow-tie” whose two triangles have their common apex at the centre. Each blue triangle has base = full height of the rectangle (one whole vertical side) and height = half the width. The two triangles together = 2 × ½ × (height) × (½ width) = ½ × height × width. So the blue region is ½ (one half) of the total area of the rectangle.

5. Give a method to obtain a quadrilateral whose area is half that of a given quadrilateral.

SOLUTION Take the given quadrilateral PQRS. Find the midpoints of all four sides — call them A, B, C, D (in order). Join A, B, C, D. The inner quadrilateral ABCD (the Varignon parallelogram) always has area exactly half the area of the original quadrilateral. Reason: the four corner triangles cut off each have area equal to one of the four pieces, and together they make up the other half of PQRS, leaving ABCD with half the area.

Figure it Out — Parallelograms (Page 162)

1. Observe the parallelograms in the figure below. (i) What can we say about the areas of all these parallelograms? (ii) What can we say about their perimeters? Which figure appears to have the maximum perimeter, and which has the minimum perimeter?

SOLUTION (i) All the parallelograms (a)–(g) sit between the same pair of grid lines on the same base, so they have the same base and the same height ⇒ equal areas. (ii) Their perimeters are not equal. The more “slanted” (sheared) a parallelogram is, the longer its slanting sides, so the most slanted figure has the maximum perimeter and the upright rectangle-like one has the minimum perimeter (the slant side is shortest when it is vertical).

2. Find the areas of the following parallelograms: (i) base 7 cm, height 4 cm. (ii) base 5 cm, height 3 cm. (iii) base 4.8 cm, height 5 cm. (iv) base 4.4 cm, height 2 cm.

SOLUTION Area of a parallelogram = base × height. (i) 7 × 4 = 28 cm2. (ii) 5 × 3 = 15 cm2. (iii) 4.8 × 5 = 24 cm2. (iv) 4.4 × 2 = 8.8 cm2.

3. Find QN. (In parallelogram PQRS, base SR with M on it, QM = 6 cm is the height onto SR, SR = 12 cm; PS = 7.6 cm is another side and QN is the height onto the line PS, with N the foot of the perpendicular.)

SOLUTION Find the area using base SR = 12 cm and height QM = 6 cm: Area = 12 × 6 = 72 cm2. The same area, using base PS = 7.6 cm and height QN: Area = 7.6 × QN. So 7.6 × QN = 72 ⇒ QN = 72 ÷ 7.6 = 9.47 cm (approximately 9.5 cm).

4. Consider a rectangle and a parallelogram of the same sidelengths: 5 cm and 4 cm. Which has the greater area? [Hint: Imagine constructing them on the same base.]

SOLUTION Rectangle 5 cm × 4 cm has area = 20 cm2, with height equal to the full side 4 cm. In the parallelogram with the same sides, the slant side 4 cm is not the height; the perpendicular height is less than 4 cm. So its area = 5 × (height < 4) < 20 cm2. Therefore the rectangle has the greater area. (Among all parallelograms with given sides, the rectangle is the largest.)

5. Give a method to obtain a rectangle whose area is twice that of a given triangle. What are the different methods that you can think of?

SOLUTION Method 1: Take the triangle’s base b and height h. Build a rectangle on the same base b with the same height h. Its area = b × h = 2 × (½bh) = twice the triangle. Method 2: Make a copy of the triangle, rotate it 180° about the midpoint of one side, and fit it onto the original — the two together form a parallelogram of twice the area, which can then be sheared/cut into a rectangle. Method 3: Build a rectangle on base b with height = 2h (so area = b × 2h)… but halve the base instead: a rectangle of base b/2 and height 2h also gives area bh = twice the triangle. Several equal-area rectangles are possible.

6. [Śulba-Sūtras] Give a method to obtain a rectangle of the same area as a given triangle.

SOLUTION Take ▵ABC with base BC and height h from A. Mark the midpoints of AB and AC and join them; this midline is parallel to BC and equals ½BC. Fold the top small triangle down and the two side flaps in along the midline — they exactly fill a rectangle whose base is BC and whose height is h/2. Area of this rectangle = BC × (h/2) = ½ × BC × h = Area of the triangle, as required.

7. [Śulba-Sūtras] An isosceles triangle can be converted into a rectangle by dissection in a simpler way. Can you find out how to do it? [Hint: Show that triangles ▵ADB and ▵ADC can be made into halves of a rectangle. Figure out how they should be assembled to get a rectangle. Use cut-outs if necessary.]

SOLUTION In isosceles ▵ABC (AB = AC), drop the altitude AD to the base BC. Since the triangle is isosceles, AD is also the median, so BD = DC and ▵ADB ≅ ▵ADC. Cut along AD to get two congruent right triangles. Flip one and place it beside the other so the two equal slant sides line up — they form a rectangle of base BD (= ½BC) and height AD. Rectangle area = ½BC × AD = ½ × base × height = Area of the isosceles triangle.

8. [Śulba-Sūtras] Give a method to convert a rectangle into an isosceles triangle by dissection.

SOLUTION Take rectangle PQRS with base PQ = b and height h. Mark the midpoint M of the top side SR. Cut along the two lines PM and QM. This gives a central triangle PQM (base PQ, apex M directly above the midpoint of PQ ⇒ isosceles) plus two right-triangle corners. Rotate the two corner triangles up about M so they sit on the slanting sides PM and QM, extending the figure into a taller isosceles triangle on base PQ with the same total area b × h.

9. Which has greater area — an equilateral triangle or a square of the same sidelength as the triangle? Which has greater area — two identical equilateral triangles together or a square of the same sidelength as the triangle? Give reasons.

SOLUTION Let the common side be s. Square area = s2. Equilateral triangle area = (√3 / 4) s2 ≈ 0.433 s2. Since 0.433 s2 < s2, the square has the greater area than one equilateral triangle. Two equilateral triangles = 2 × 0.433 s2 = 0.866 s2, which is still less than s2. So the square still has the greater area than two such triangles.

Figure it Out — Rhombus & Trapezium (Page 169)

1. Find the area of a rhombus whose diagonals are 20 cm and 15 cm.

SOLUTION Area of a rhombus = ½ × product of diagonals = ½ × 20 × 15. = ½ × 300 = 150 cm2.

2. Give a method to convert a rectangle into a rhombus of equal area using dissection.

SOLUTION Take rectangle ABCD with area = b × h. A rhombus of equal area must satisfy ½ × d1 × d2 = b × h, so choose diagonals d1 = 2b and d2 = h (then ½ × 2b × h = bh). Method: mark the midpoints of the rectangle’s sides; cut along the lines joining adjacent midpoints. This gives 4 right triangles plus the inner rhombus (Varignon rhombus) of half the area — rearrange the 4 triangles around it to double the rhombus back to the full area. More directly: enclose a rhombus with diagonals 2b and h; its area ½(2b)(h) = bh equals the rectangle, and the corner triangles cut off the rectangle fill the gaps of the rhombus.

3. Find the areas of the following figures: (i) A parallelogram with base 16 ft and corresponding height 10 ft (the 7 ft is a slant side). (ii) A trapezium with parallel sides 24 m (top) and 36 m (bottom) and height 14 m. (iii) A trapezium with parallel sides 10 in and 6 in and height 14 in. (iv) A trapezium with parallel sides 12 ft (top) and 18 ft (bottom) and height 8 ft.

SOLUTION (i) Parallelogram area = base × height = 16 × 10 = 160 ft2. (ii) Trapezium = ½ × h × (a + b) = ½ × 14 × (24 + 36) = ½ × 14 × 60 = 420 m2. (iii) = ½ × 14 × (10 + 6) = ½ × 14 × 16 = 112 in2. (iv) = ½ × 8 × (12 + 18) = ½ × 8 × 30 = 120 ft2.

4. [Śulba-Sūtras] Give a method to convert an isosceles trapezium to a rectangle using dissection.

SOLUTION Take isosceles trapezium with parallel sides a (top) and b (bottom) and height h. Drop perpendiculars from the two top vertices to the bottom; this cuts off two congruent right triangles at the ends. Slide each end triangle inward (or up) so it fits onto the opposite slanting edge — because the trapezium is isosceles, the two triangles are congruent and fill the gaps exactly. The result is a rectangle of height h and base = ½(a + b), so its area = ½(a + b) × h = area of the trapezium.

5. Here is one of the ways to convert trapezium ABCD into a rectangle EFGH of equal area. Given the trapezium ABCD, how do we find the vertices of the rectangle EFGH? [Hint: If ▵AHI ≅ ▵DGI and ▵BEJ ≅ ▵CFJ, then the trapezium and rectangle have equal areas.]

SOLUTION Mark I as the midpoint of slant side AD and J as the midpoint of slant side BC. Through I and J draw lines perpendicular to the base DC, meeting the base at G and F and meeting the top line at H and E. Because I and J are midpoints, ▵AHI ≅ ▵DGI and ▵BEJ ≅ ▵CFJ. Folding each top triangle down onto its matching base triangle turns the trapezium into rectangle EFGH of the same area, with height h and base equal to the trapezium’s midline ½(AB + DC).

6. Using the idea of converting a trapezium into a rectangle of equal area, and vice versa, construct a trapezium of area 144 cm2.

SOLUTION A trapezium of area 144 cm2 must satisfy ½ × h × (a + b) = 144, i.e. h × (a + b) = 288. Example choice: take height h = 12 cm; then a + b = 288 ÷ 12 = 24 cm. Pick parallel sides a = 10 cm and b = 14 cm (sum 24). Construct a trapezium with parallel sides 10 cm and 14 cm and height 12 cm. Check: ½ × 12 × (10 + 14) = ½ × 12 × 24 = 144 cm2. (Equivalently, first build a rectangle 12 cm × 12 cm = 144 cm2 and slant its top into a trapezium of the same height.)

7. A regular hexagon is divided into a trapezium, an equilateral triangle, and a rhombus, as shown. Find the ratio of their areas.

SOLUTION A regular hexagon of side s is made of 6 identical equilateral triangles, each of area t = (√3/4)s2. Read off how many such unit triangles each piece contains. The equilateral triangle = 1 unit triangle. The rhombus (two unit triangles) = 2 unit triangles. The trapezium (the remaining half-hexagon, three unit triangles) = 3 unit triangles. So Trapezium : Equilateral triangle : Rhombus = 3 : 1 : 2. (Listed as triangle : rhombus : trapezium it is 1 : 2 : 3.)

8. ZYXW is a trapezium with ZY∥WX. A is the midpoint of XY. Show that the area of the trapezium ZYXW is equal to the area of ▵ZWB.

SOLUTION Extend WX to a point B and join ZA, then AB, so that A (the midpoint of XY) lies on segment ZB. Consider ▵AYZ and ▵AXB: AY = AX (A is the midpoint), the angles at A are vertically opposite and equal, and ∠ZYA = ∠BXA (alternate angles, since ZY∥WX). So ▵AYZ ≅ ▵AXB (ASA). Congruent triangles have equal area, so removing ▵AYZ from the trapezium and adding the equal ▵AXB turns trapezium ZYXW into ▵ZWB without changing the area. Hence Area(ZYXW) = Area(▵ZWB).

Common Mistakes to Avoid

Watch out for these

  • Forgetting the ½ in the triangle and rhombus/trapezium formulas (triangle = ½ × base × height, not base × height).
  • Using a slant side as the height of a parallelogram or trapezium — the height must be the perpendicular distance.
  • In a trapezium, multiplying by only one parallel side instead of the sum of both parallel sides.
  • Forgetting to square the units — areas are in cm2, m2, ft2, not cm or m.
  • Mixing units in one figure (e.g. cm and in) without converting (remember 1 in = 2.54 cm, 1 in2 = 6.4516 cm2).
  • For a crosspath, counting the overlap square twice — subtract it once.
  • Assuming “larger perimeter means larger area” — this is false; perimeter does not measure area.

Practice MCQs & Assertion–Reason

1. The area of a triangle with base 8 cm and height 5 cm is:

(a) 40 cm2    (b) 20 cm2    (c) 13 cm2    (d) 80 cm2

2. The area of a parallelogram is given by:

(a) ½ × base × height    (b) base × height    (c) base + height    (d) ½ × product of diagonals

3. The area of a rhombus with diagonals 12 cm and 10 cm is:

(a) 120 cm2    (b) 60 cm2    (c) 22 cm2    (d) 240 cm2

4. The area of a trapezium with parallel sides 9 cm and 7 cm and height 6 cm is:

(a) 96 cm2    (b) 48 cm2    (c) 22 cm2    (d) 16 cm2

5. A diagonal of a rectangle divides it into two triangles that are:

(a) of unequal area    (b) congruent    (c) equilateral    (d) right-angled isosceles always

6. 1 in2 equals how many cm2?

(a) 2.54    (b) 5.08    (c) 6.4516    (d) 25.4

7. The line joining a vertex of a triangle to the midpoint of the opposite side divides it into two triangles of:

(a) equal area    (b) ratio 1 : 2    (c) ratio 1 : 3    (d) unequal area

8. If M and N are midpoints of two sides of ▵XYZ meeting at X, the area of ▵XMN is what fraction of ▵XYZ?

(a) ½    (b) ⅓    (c) ¼    (d) ⅛

9. A rectangle and a parallelogram have the same side lengths 6 cm and 4 cm. Which has the greater area?

(a) the parallelogram    (b) the rectangle    (c) both equal    (d) cannot say

10. The area of a rectangle of length 14 cm and width 4 cm is:

(a) 18 cm2    (b) 36 cm2    (c) 56 cm2    (d) 28 cm2

Answer key: 1-(b), 2-(b), 3-(b), 4-(b), 5-(b), 6-(c), 7-(a), 8-(c), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The area of a triangle is ½ × base × height.

Reason: A diagonal of a rectangle divides it into two congruent triangles, each half the rectangle.

A-R 2. Assertion: A rhombus with diagonals 8 cm and 6 cm has area 48 cm2.

Reason: The area of a rhombus is ½ × product of its diagonals.

A-R 3. Assertion: Two regions with the same perimeter must have the same area.

Reason: Perimeter is the length of the boundary of a region.

A-R 4. Assertion: A trapezium with parallel sides 5 cm and 3 cm and height 4 cm has area 16 cm2.

Reason: The area of a trapezium is ½ × height × sum of the parallel sides.

A-R 5. Assertion: All parallelograms on the same base and between the same parallel lines have equal area.

Reason: Such parallelograms have the same base and the same height.

Answer key: 1-(A), 2-(A), 3-(D), 4-(A), 5-(A).

Quick Revision Summary

  • Area of a rectangle = length × width; area is measured in square units (cm2, m2…).
  • Area of a triangle = ½ × base × height; the formula holds for every type of triangle.
  • Any polygon can be split into triangles, so its area = sum of the triangle areas.
  • Area of a parallelogram = base × height (perpendicular height, not the slant side).
  • Area of a rhombus = ½ × product of its diagonals.
  • Area of a trapezium = ½ × height × (sum of the parallel sides).
  • Dissection rearranges pieces without changing area; the median of a triangle halves its area.
  • Units: 1 in = 2.54 cm, 1 in2 = 6.4516 cm2, 1 ft = 12 in, 1 acre = 43,560 ft2.

How to score full marks in this chapter

Always state the formula first, then substitute. Use the perpendicular height — never a slant side. Write the final answer with squared units. For composite figures, split neatly into triangles, rectangles and trapeziums and add (or subtract) the parts, and for a crosspath remember to subtract the overlap once. Keep surds like √3 in exact form unless a decimal is asked.

Frequently Asked Questions

What is Class 8 Maths Ganita Prakash Chapter 14 about?

Chapter 14, Area (Chapter 7 of Part II, the final chapter of Class 8), derives and applies the area formulas for triangles, parallelograms, rhombuses, trapeziums and composite figures, including dissection methods from the Śulba-Sūtras and real-life area units.

What are the main area formulas in this chapter?

Triangle = ½ × base × height; parallelogram = base × height; rhombus = ½ × product of diagonals; trapezium = ½ × height × (sum of parallel sides); rectangle = length × width.

How many exercises does Ganita Prakash Chapter 14 have?

The chapter has five “Figure it Out” sets (on rectangles/composite figures, triangles, area of any polygon, parallelograms, and rhombus & trapezium), plus Math Talk and Try This prompts — all solved on this page.

Are these Class 8 Maths Ganita Prakash Chapter 14 solutions free?

Yes. All solutions are free and follow the official NCERT Ganita Prakash Part II textbook for 2026–27.

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