Figure it Out — Triangles and Sulba-Sutra Transformations — Area

1. 1 (i)2 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 157, Q1(i)

   Find the area of ∆ABC, in which AE is the perpendicular from A to BC with AE = 3 cm, and BC = 4 cm.

   Hint. The dashed segment is the height; the label under the base gives the base.

   Here AE is drawn dashed from A straight down to the base, meeting it at E, so AE is the height and BC is the base it belongs to.

   Area = ½ × base × height = ½ × 4 × 3 = 6 cm²

   Note that the foot E is not the midpoint of BC and does not need to be — the formula asks only for the perpendicular distance from A down to the line BC, not for where it lands. The two pieces BE and EC play no part in the calculation.

   ✦ Answer: 6 cm².

2. 1 (ii)2 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 157, Q1(ii)

   Find the area of ∆DEF, in which DN is perpendicular to EF with DN = 3.2 cm and EF = 5 cm.

   Hint. Which side is the perpendicular dropped onto? That side is the base.

   The right-angle mark at N tells you DN ⊥ EF, so the pair to use is base EF with height DN — not the horizontal side DF, which has no height marked against it.

   Area = ½ × EF × DN = ½ × 5 × 3.2 = ½ × 16 = 8 cm²

   The base here is a slanted side, which is exactly the point of the question: a base does not have to be the side that looks like it is at the bottom of the page. Any side may serve as the base provided you use the height measured perpendicular to that side.

   ✦ Answer: 8 cm².

3. 1 (iii)2 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 157, Q1(iii)

   Find the area of ∆NAT, which is right-angled at A, with NA = 4 cm and AT = 3 cm.

   Hint. In a right triangle the two legs are already a base–height pair.

   The right angle sits at A, so the two legs NA and AT are perpendicular to each other. That means one can be taken as the base and the other is automatically its height — no extra construction is needed.

   Area = ½ × AT × NA = ½ × 3 × 4 = 6 cm²

   Equivalently, ∆NAT is half of the 3 cm by 4 cm rectangle you would get by completing it, and ½ × 12 = 6 cm².

   The hypotenuse NT is not needed at all. (It happens to be 5 cm, by the Baudhāyana–Pythagoras relation, but the area calculation never touches it.)

   ✦ Answer: 6 cm².

4. 23 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 158, Q2

   In ∆ABC, AX is perpendicular to CB extended with AX = 4 units, BC = 6 units and AC = 8 units. Find the length of the altitude BY drawn to AC.

   Hint. Find the area with the base–height pair you have, then reuse it with the pair you want.

   The area, first way. Take BC as the base. Its height is AX = 4 units — and notice the foot X lies outside the segment BC, which is fine: the earlier proof showed the formula survives that.

   Area(∆ABC) = ½ × BC × AX = ½ × 6 × 4 = 12 sq units

   The area, second way. Take AC as the base. Its height is the altitude BY:

   Area(∆ABC) = ½ × AC × BY = ½ × 8 × BY = 4 BY

   Putting them together. One triangle, one area:

   4 BY = 12 so BY = 3 units

   Sense check. The base grew from 6 to 8 — a factor of 4/3 — so the height had to shrink by the same factor, from 4 to 3 ✓.

   ✦ Answer: BY = 3 units.

5. 33 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 158, Q3

   ∆SUB is isosceles with SE perpendicular to UB, and the area of ∆SEB is 24 sq units. Find the area of ∆SUB.

   Hint. In an isosceles triangle, where does the perpendicular from the apex land?

   Where E is. ∆SUB is isosceles with apex S, so SU = SB. The perpendicular from the apex of an isosceles triangle to its base lands at the base's midpoint, so UE = EB.

   The two halves match. Compare ∆SEU and ∆SEB:

   • UE = EB (just shown)
   • ∠SEU = ∠SEB = 90° (SE ⊥ UB)
   • SE is common

   By SAS the triangles are congruent, so they have equal areas. (You do not even need congruence — both stand on equal bases UE and EB with the same apex S, hence the same height SE, which by itself gives equal areas.)

   Finishing.

   Area(∆SUB) = Area(∆SEU) + Area(∆SEB) = 24 + 24 = 48 sq units

   ✦ Answer: 48 sq units.

6. 44 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 158, Q4 [Sulba-Sutras]

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

   Hint. A triangle needs twice the base of a rectangle to make up for the ½ in its formula.

   The target. A rectangle of length ℓ and width w has area ℓw. A triangle of base b and height h has area ½bh. To match them you need ½bh = ℓw, so keeping the rectangle's width as the triangle's height forces the base to be doubled.

   The construction. Given rectangle ABCD with base DC and height AD:

   1. Extend DC beyond C to a point E with CE = DC. Now DE = 2 × DC.
   2. Join A to E.
   3. ∆ADE is the triangle you want.

   Why it works. ∆ADE has base DE = 2 × DC and its height is the perpendicular distance from A down to the line DE, which is AD, the rectangle's width. So

   Area(∆ADE) = ½ × 2 × DC × AD = DC × AD = Area(ABCD) ✓

   The other way round. You could instead keep the base DC and double the height: extend AD beyond A to F with AF = AD, and join F to C. ∆FDC then has base DC and height 2 × AD, again giving DC × AD. Either construction answers the question, and both use nothing but a straight edge.

   With numbers. A 6 cm × 4 cm rectangle has area 24 cm²; the triangle of base 12 cm and height 4 cm has area ½ × 12 × 4 = 24 cm² ✓

   ✦ Answer: extend the base to twice its length and join the far end to a top corner — the triangle on the doubled base with the rectangle's height has the same area (equally, keep the base and double the height).

7. 55 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 158, Q5 [Sulba-Sutras]

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

   Hint. Cut along the line joining the midpoints of two sides and turn the top piece over.

   The target. A triangle of base b and height h has area ½bh, which is the area of a rectangle of base b and height h/2. So the rectangle to aim for keeps the base and halves the height.

   The dissection. Given ∆ABC with base BC:

   1. Mark M, the midpoint of AB, and N, the midpoint of AC.
   2. Cut along MN. By the midpoint theorem MN is parallel to BC and half its length, and it sits at height h/2.
   3. Rotate the top piece ∆AMN by a half turn about N.
   4. Convert the resulting parallelogram into a rectangle by the usual drop-a-perpendicular-and-slide-the-corner move.

   Why step 3 lands correctly. Put B = (0, 0), C = (b, 0), A = (a, h). Then M = (a/2, h/2) and N = ((a + b)/2, h/2). A half turn about N sends A to (a + b − a, h − h) = (b, 0), which is C exactly. So the rotated piece slots against the bottom piece along NC, and the two together form a parallelogram standing on BC with height h/2.

   Checking the area.

   parallelogram = base × height = b × h/2 = ½bh = Area(∆ABC) ✓

   and the final rectangle has the same area as the parallelogram since dissection never changes area. Its dimensions are b by h/2.

   With numbers. A triangle of base 10 cm and height 6 cm has area 30 cm²; the rectangle 10 cm by 3 cm also has area 30 cm² ✓

   ✦ Answer: cut along the midline joining the midpoints of two sides, half-turn the top triangle about one of those midpoints, then square up the parallelogram — the result is a rectangle of the triangle's base and half its height.

8. 6 (i)4 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 158, Q6(i)

   ABCD, BCEF and BFGH are three identical squares arranged in an L. A line is drawn from D to H, colouring a red region and a blue region. If the area of the red region is 49 sq units, what is the area of the blue region?

   Hint. Put the corner D at the origin and find where the line DH crosses the top of the first square.

   Setting up. Let each square have side s. Place D = (0, 0), C = (s, 0), B = (s, s), A = (0, s); the second square BCEF sits to the right and the third, BFGH, sits above it, so H = (s, 2s).

   Where DH crosses. The segment from D(0, 0) to H(s, 2s) has slope 2, so at height y = s it is at x = s/2. It cuts the top edge of square ABCD at its midpoint.

   The blue region is the part of square ABCD lying above DH: a triangle with vertices D(0, 0), A(0, s) and (s/2, s). Taking the side DA as its base,

   blue = ½ × s × s/2 = s²/4

   The red region is everything else the line encloses — the rest of square ABCD plus the triangle above it:

   • inside ABCD: the trapezium D(0,0), C(s,0), B(s,s), (s/2, s), which has parallel sides s and s/2 and height s, giving ½ × (s + s/2) × s = 3s²/4

   • above ABCD: the triangle (s/2, s), H(s, 2s), B(s, s), giving ½ × s/2 × s = s²/4

     red = 3s²/4 + s²/4 = s²

   So the red region is exactly one whole square, and red = 4 × blue.

   Answering. red = 49 gives s² = 49, so

   blue = s²/4 = 49/4 = 12.25 sq units

   ✦ Answer: 12.25 sq units (the red region is always four times the blue one).

9. 6 (ii)3 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 158, Q6(ii)

   In another version of the same figure, the blue and red regions together enclose 180 sq units. What is the area of each square?

   Hint. You already know both regions in terms of s. Add them.

   From the previous part, with s the side of each square:

   blue = s²/4 and red = s²

   Adding them,

   blue + red = s²/4 + s² = 5s²/4

   Setting that equal to the given total:

   5s²/4 = 180 5s² = 720 s² = 144 sq units

   Since s² is precisely the area of one square, each of the three identical squares has area 144 sq units, so each has side 12 units.

   Check. blue = 144/4 = 36 and red = 144; 36 + 144 = 180 ✓

   ✦ Answer: each square has area 144 sq units (side 12 units).

10. 74 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 159, Q7

    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]

    Hint. Use the median fact twice — once in ∆XYZ and once in the triangle it creates.

    Join NY, as the hint says. That splits ∆XYZ into ∆XYN and ∆YNZ.

    First cut. In ∆XYZ, the point N is the midpoint of XZ, so YN is a median. A median divides a triangle into two of equal area, so

    Area(∆XYN) = ½ × Area(∆XYZ)

    Second cut. Now work inside ∆XYN. The point M is the midpoint of XY, so NM is a median of that triangle. By the same fact,

    Area(∆XMN) = ½ × Area(∆XYN)

    Combining.

    Area(∆XMN) = ½ × ½ × Area(∆XYZ) = ¼ × Area(∆XYZ)

    A second route. ∆XMN and ∆XYZ share the angle at X, and XM = ½XY with XN = ½XZ. Halving both sides that enclose the angle quarters the area, which gives the same ¼ — and shows the answer does not depend on the shape of the triangle.

    Check with numbers. X = (0, 0), Y = (8, 0), Z = (0, 6): area 24. Then M = (4, 0), N = (0, 3), and ∆XMN has area ½ × 4 × 3 = 6 = 24/4 ✓

    ✦ Answer: one quarter of the area of ∆XYZ.

11. 84 marksGanita Prakash Cl-8 Part 2, Figure it Out, page 159, Q8

    Gopal starts at his house, must carry water from the river, and then take it to his water tank. What is the shortest path from the house to the river and then to the tank? Describe how to find it.

    Hint. The same mirror trick that found the triangle of least perimeter.

    Restating it. Gopal's path is house H → some point P on the river → tank T. The river is a straight line; the only freedom is where to put P. So the job is to make HP + PT as small as possible.

    The construction.

    1. Reflect the tank T across the river line to get its mirror image T′.
    2. Join H to T′ with a straight line.
    3. Let P be the point where that line crosses the river.
    4. Gopal's shortest route is H → P → T.

    Why it is shortest. P lies on the river, and the river acts as the mirror, so PT = PT′ for every choice of P. That means

    HP + PT = HP + PT′

    and the right-hand side is the length of a path from H to T′ that bends at P. The shortest path between two points is the straight segment HT′, so the bend must be removed — that is, P must lie on HT′. Any other point on the river gives a genuinely bent path, which is longer.

    Same trick as before. This is exactly the argument used on page 156 for the triangle of least perimeter, with the parallel line l replaced by the river and the two fixed points B, C replaced by the house and the tank. It is also the law you meet in Science as reflection: light takes the path of least length, which is why the angle of incidence equals the angle of reflection — and indeed the two angles Gopal's path makes with the river are equal.

    ✦ Answer: reflect the tank in the river, join the house to that image with a straight line, and fetch the water where that line meets the river — the resulting path makes equal angles with the river and is the shortest possible.

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.