In-text and Figure it Out — Area of any Polygon — Area

1. 13 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 159

   How do we find the area of a quadrilateral ABCD? What measurements are needed?

   Hint. One well-chosen line turns a shape you cannot handle into two you can.

   Split it into triangles. Join the diagonal BD. That cuts ABCD into ∆ABD and ∆CBD, and a triangle's area is something we can already compute.

   Area(ABCD) = Area(∆ABD) + Area(∆CBD)

   What to measure. Both triangles stand on the same base BD, so take BD as the base for each. Then you need the perpendicular distance from A to BD, call it h₁, and the perpendicular distance from C to BD, call it h₂:

   Area(ABCD) = ½ × BD × h₁ + ½ × BD × h₂ = ½ × BD × (h₁ + h₂)

   So three measurements suffice: the diagonal and the two perpendicular offsets. You do not need any of the four sides, and you do not need any angles.

   With numbers. If BD = 10 cm, h₁ = 4 cm and h₂ = 3 cm, the area is ½ × 10 × 7 = 35 cm².

   A caution. This works when the diagonal BD lies inside the quadrilateral. For a quadrilateral with a dent, choose the other diagonal — one of the two always lies inside.

   ✦ Answer: join a diagonal and add the two triangles, needing only the diagonal's length and the two perpendicular distances to it: Area = ½ × d × (h₁ + h₂).

2. 23 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 159

   How do we find the area of a pentagon? Can any polygon be divided into triangles?

   Hint. Fix one vertex and join it to all the others.

   The pentagon. Pick one vertex, say A, and join it to the two vertices it is not already joined to. Those two lines cut the pentagon ABCDE into three triangles — ∆ABC, ∆ACD and ∆ADE — and adding their three areas gives the pentagon's.

   The general rule. Yes, any polygon can be cut into triangles. Doing it from a single vertex of an n-sided polygon uses n − 3 extra lines and produces n − 2 triangles:

   Polygon	Sides	Triangles
   Quadrilateral	4	2
   Pentagon	5	3
   Hexagon	6	4
   Decagon	10	8

   For a polygon with a dent, joining everything to one vertex may send a line outside the shape, but a triangulation still exists — you just choose the cutting lines more carefully, or cut from an interior point instead.

   Why this matters. It means the single formula ½ × base × height is enough for every straight-sided figure. Everything else in this chapter — the parallelogram, the rhombus, the trapezium — is a shortcut for a triangulation you could always do by hand.

   ✦ Answer: cut the pentagon into 3 triangles from one vertex and add; in general every polygon can be triangulated, an n-gon giving n − 2 triangles.

3. 13 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 160, Q1

   Find the area of quadrilateral ABCD, given AC = 22 cm, BM = 3 cm, DN = 3 cm, with BM and DN both perpendicular to AC.

   Hint. AC is the diagonal; BM and DN are the two offsets.

   The diagonal AC splits ABCD into ∆ABC and ∆ACD, and the two perpendiculars given are exactly the heights of those triangles above AC.

   Area(∆ABC) = ½ × AC × BM = ½ × 22 × 3 = 33 cm² Area(∆ACD) = ½ × AC × DN = ½ × 22 × 3 = 33 cm²

   Adding,

   Area(ABCD) = 33 + 33 = 66 cm²

   Shortcut. Since both triangles share the base AC, add the heights first:

   Area = ½ × 22 × (3 + 3) = ½ × 22 × 6 = 66 cm² ✓

   Note how little the drawing matters — the quadrilateral is drawn long and flat, but only the diagonal and the two offsets entered the arithmetic.

   ✦ Answer: 66 cm².

4. 24 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 160, Q2

   ABCD is a rectangle, 18 cm wide and 10 cm tall. E lies on AB with AE = 10 cm and EB = 8 cm; F lies on AD with AF = 6 cm and FD = 4 cm. Find the area of the shaded region FECD, where C is the far bottom corner.

   Hint. Subtracting two right triangles from the rectangle is quicker than measuring the shaded shape directly.

   Take the whole and remove the corners. The shaded region is the rectangle with two right triangles cut away — the one at corner A and the one at corner B.

   Area(ABCD) = 18 × 10 = 180 cm²

   The corner at A. ∆AFE is right-angled at A, so its two legs AF = 6 cm and AE = 10 cm are already a base–height pair, which gives

   Area(∆AFE) = ½ × 6 × 10 = 30 cm²

   The corner at B. ∆EBC is right-angled at B with legs EB = 8 cm and BC = 10 cm (the rectangle's height), so the same reasoning gives

   Area(∆EBC) = ½ × 8 × 10 = 40 cm²

   Subtracting.

   shaded = 180 − 30 − 40 = 110 cm²

   Check by triangulating instead. Join FC. Then ∆FDC has base DC = 18 and height FD = 4, giving 36 cm²; and ∆FEC has base FE and the remaining area 110 − 36 = 74 cm². Adding the two removed corners back, 110 + 30 + 40 = 180 ✓ — the rectangle is fully accounted for.

   ✦ Answer: 110 cm².

5. 33 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 160, Q3

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

   Hint. Join the centre to every vertex.

   Cut from the centre. Joining the centre O of a regular hexagon to its six vertices produces six identical triangles. Each has a side of the hexagon as its base and the perpendicular distance from O to that side as its height. Calling the side a and that perpendicular distance d,

   Area = 6 × ½ × a × d = 3ad

   So two measurements are enough: one side, and the distance from the centre to a side.

   In fact one measurement is enough. In a regular hexagon those six triangles are equilateral, because the angle at O is 360° ÷ 6 = 60° and the two sides from O are equal, forcing all three angles to be 60°. So each triangle has side a and height d = (√3/2)a, giving

   Area = 3 × a × (√3/2)a = (3√3/2)a² ≈ 2.598 a²

   With numbers. A regular hexagon of side 4 cm has area (3√3/2) × 16 ≈ 41.6 cm². Measuring instead gives d ≈ 3.46 cm and 3 × 4 × 3.46 ≈ 41.5 cm² ✓

   Other splits work too. Two long diagonals cut the hexagon into a trapezium, an equilateral triangle and a rhombus (question 7 of the last exercise), and adding those three gives the same total.

   ✦ Answer: the sidelength a and the centre-to-side distance d, giving Area = 3ad; and since d = (√3/2)a for a regular hexagon, the side alone suffices, giving (3√3/2)a².

6. 44 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 160, Q4

   A rectangle contains a blue region made of two triangles: one standing on the whole top edge and one standing on the whole bottom edge, both with their apex at the same interior point. What fraction of the rectangle is blue?

   Hint. Both triangles have the full width as their base. What do their two heights add up to?

   Name things. Let the rectangle be W wide and H tall, and let P be the shared apex of the two triangles.

   The upper triangle has the whole top edge as its base, so base = W, and its height is the distance from P down to the top edge — call it h₁.

   upper = ½ × W × h₁

   The lower triangle has the whole bottom edge as its base, so base = W again, and its height is the distance from P up to the bottom edge — call it h₂.

   lower = ½ × W × h₂

   The key observation. P sits between the two horizontal edges, so the two distances together span the full height:

   h₁ + h₂ = H

   Adding.

   blue = ½W h₁ + ½W h₂ = ½W(h₁ + h₂) = ½ × W × H = half the rectangle

   Why the apex position is irrelevant. Nowhere did the horizontal position of P enter, and its vertical position only entered through h₁ + h₂, which is fixed. So you may slide P anywhere inside and the blue area stays at exactly one half. (Measuring the coloured pixels of the printed figure gives 0.4987 — half, to the accuracy of the printing.)

   ✦ Answer: exactly one half of the rectangle, wherever the apex is placed.

7. 55 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 160, Q5

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

   Hint. Join the midpoints of the four sides and account for the four corner triangles.

   The construction. Given quadrilateral ABCD, mark P, Q, R, S — the midpoints of AB, BC, CD, DA. Join them in order. The quadrilateral PQRS has exactly half the area of ABCD.

   Why. Joining the midpoints leaves four corner triangles: ∆APS at A, ∆BQP at B, ∆CRQ at C and ∆DSR at D. Each of them is a quarter of a triangle formed by a diagonal, because halving both sides that enclose a vertex quarters the area:

   ∆BQP = ¼ ∆ABC ∆DSR = ¼ ∆ACD ∆APS = ¼ ∆ABD ∆CRQ = ¼ ∆BCD

   Now note that ∆ABC + ∆ACD is the whole of ABCD (split by diagonal AC), and ∆ABD + ∆BCD is also the whole of ABCD (split by diagonal BD). Writing the total area as T and adding the four corners:

   corners = ¼(∆ABC + ∆ACD) + ¼(∆ABD + ∆BCD) = ¼T + ¼T = ½T

   Therefore

   Area(PQRS) = T − ½T = ½T ✓

   Check with a square. Take a 4 × 4 square, T = 16. The midpoint quadrilateral has diagonals 4 and 4 and is a square of side 2√2, area 8 = ½ × 16 ✓

   A second method. Join a diagonal, say AC, and let M be the midpoint of BD. Replace B by the point on the perpendicular from B that halves ∆ABC and do the same on the other side — but the midpoint construction above is simpler, always works, and needs nothing but a ruler.

   ✦ Answer: join the midpoints of the four sides — the resulting quadrilateral has exactly half the original area, since the four corner triangles removed add up to half of it.

Solutions written by the tuition.in editorial team and checked against NCERT Ganita Prakash Grade 8 Part 2 (hegp207.pdf), Chapter 7 'Area', pages 148-171. HAND-WRITTEN throughout. Part 2 books carry NO printed answer key, so every answer was derived from first principles and independently recomputed in Python. FIGURES READ OFF HIGH-DPI RENDERS AND RE-SOLVED: the p.150 pinwheel, whose four rectangles all turn out to have a side of 7 in (areas 14, 21, 28, 35), giving the missing width 2 in; the p.150 step figure, where the BOLD outline encloses the dotted region together with the region below it - 50 sq m in all - so the widths are 29/4 = 7.25 m and 11/4 = 2.75 m (summing to exactly 10 m) and the missing height is (50-29)/7.25 = 84/29 = 2.90 m, which matches the 2.74 m measured off the printed drawing; the p.152 spiral tube, whose nine arms total 120 and whose eight corners are each double-counted once, giving 112 sq units - confirmed by rasterising the nine arms at 20 cells per unit and counting - with the hint's L-tube (arms 5 and 5) coming to 5+5-1 = 9; the p.152 square whose regions are s^2/4, s^2/4 and s^2/2, so doubling the side raises each by 3 times its own area; the p.157 triangles, whose '4 cm' label is centred on BC (not EC) giving areas 6, 8 and 6 sq cm; the p.158 obtuse triangle giving BY = 3 units; the p.158 three-square figure, where the line from D to H crosses the top of the first square at its midpoint, making the red region exactly s^2 and the blue exactly s^2/4, hence 12.25 sq units and 144 sq units; the p.160 quadrilateral (66 sq cm) and shaded region (180-30-40 = 110 sq cm); the p.160 blue 'bowtie', whose two triangles share the full width and whose heights sum to the rectangle's, so it is exactly half - measured as 0.4987 of the printed rectangle; the seven p.162 parallelograms, measured by connected-component analysis to have identical filled areas to within 0.02% (all base 5, height 3, area 15) with leans of 0.4, 1.2, 2.5, 0.6, 0.9, 2.0 and 2.8 grid units, so (g) has the maximum perimeter and (a) the minimum; the p.163 parallelograms (28, 15, 24 and 8.8 sq cm) and QN = 72/7.6 = 180/19 = 9.47 cm; the p.169 trapezia (136 sq ft, 420 sq m, 100 sq in and 120 sq ft); the p.170 hexagon, whose long diagonal and radius cut it into 3, 1 and 2 of the six unit equilateral triangles, giving the ratio 3:1:2; and the p.170 trapezium ZYXW, where ASA gives triangle ZYA congruent to triangle BXA. Unit work checked: A4 = 21 x 29.7 = 623.7 sq cm; 1 sq in = 2.54^2 = 6.4516 sq cm; 161.29 / 6.4516 = 25 sq in exactly; 5 in = 12.7 cm; 7.4 in = 18.796 cm; 5.08 cm = 2 in; 11.43 cm = 4.5 in; 1 sq ft = 144 sq in; 1 sq km = 1,000,000 sq m; 1 acre = 43,560 sq ft.. Questions are referenced from the NCERT textbook for identification.