In-text — Triangles — Area

1. 13 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 153

   Two identical rectangles ABCD are drawn. In the first pair of figures, X and Y are different points on side AB and the triangles XDC and YDC are compared. In the second pair, triangle XDC is compared with triangle YBC. Which triangle has the greater area in each case?

   Hint. Drop the altitude from the apex and see what it equals.

   First comparison — ∆XDC against ∆YDC. Both triangles stand on the same base DC, and both apexes X and Y lie on AB, the side opposite DC. Drop a perpendicular from X to DC and another from Y to DC: since AB is parallel to DC, both perpendiculars have the same length, namely the width of the rectangle. So

   Area(∆XDC) = ½ × DC × width = Area(∆YDC) = half the rectangle

   Sliding the apex along AB changes the shape of the triangle but not its height above DC, so the area never moves.

   Second comparison — ∆XDC against ∆YBC. Now the two triangles sit on different sides of the same rectangle, but the reasoning is unchanged: ∆XDC has base DC with its apex on the opposite side, and ∆YBC has base BC with its apex on the opposite side. Each triangle's height is the rectangle's other dimension, so each has area

   ½ × (one side) × (the other side) = half the rectangle

   The rule behind both. A triangle whose base is a full side of a rectangle and whose apex lies anywhere on the opposite side always covers exactly half the rectangle. That is why the answer does not depend on where X and Y are placed.

   ✦ Answer: Neither — they are equal in both cases, each triangle being exactly half the area of the rectangle.

2. 22 marksGanita Prakash Cl-8 Part 2, Fig. 7.1, page 153

   In Fig. 7.1 the rectangle ABCD has sidelengths 4 and 5, and X lies on AB. Find the area of ∆XDC.

   Hint. You have just shown that such a triangle is half its rectangle.

   The rectangle's area is 4 × 5 = 20 sq units.

   Triangle XDC has base DC and its apex X on the opposite side AB, so its height is the rectangle's other dimension. That makes it half the rectangle:

   Area(∆XDC) = ½ × 20 = 10 sq units

   The same number comes out of the formula directly. Taking DC = 5 as the base, the height is 4, so ½ × 5 × 4 = 10 sq units. Taking DC = 4 as the base instead, the height is 5 and ½ × 4 × 5 = 10 — the labelling does not matter, since multiplication does not care about order.

   Notice that the exact position of X on AB was never used. Move X to a corner and the triangle becomes a right triangle, still of area 10.

   ✦ Answer: 10 sq units.

3. 34 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 154

   The formula ½ × base × height was derived by fitting the triangle inside a rectangle. But for some triangles you cannot draw a rectangle on base BC that contains the triangle — the apex leans out past the base. Show that the formula still holds for such a triangle.

   Hint. Write the awkward triangle as the difference of two well-behaved ones.

   The awkward case. Suppose ∆ABC has its angle at B obtuse, so the foot of the perpendicular from A lands outside the segment BC — at a point D on CB extended. Let that perpendicular have length h. No rectangle on base BC contains the triangle, so the earlier picture is unavailable.

   The fix — subtract instead of add. Both ∆ADC and ∆ADB are right-angled at D, and each of them can be put inside a rectangle in the usual way. From the picture,

   ∆ABC = ∆ADC − ∆ADB

   since ∆ADB is the extra sliver that ∆ADC carries beyond ∆ABC. Therefore

   Area(∆ABC) = ½ × h × DC − ½ × h × DB = ½ × h × (DC − DB) = ½ × h × BC

   because DC − DB = BC: walking from D to C and then back from D to B leaves exactly the stretch from B to C.

   What this establishes. The formula ½ × base × height needs no case analysis. Whether the apex sits over the base, over an endpoint of it, or out beyond it, the area is half the base times the perpendicular height. That is why the height must always be measured as a perpendicular distance and not as a side.

   ✦ Answer: Yes, the formula holds for every triangle — write ∆ABC = ∆ADC − ∆ADB and the difference collapses to ½ × h × BC.

4. 43 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 155

   In ∆ABC, AX is perpendicular to BC with AX = 5 units and BC = 3 units, and BY is perpendicular to AC with AC = 4 units. Find BY.

   Hint. Compute the same area twice, using a different base each time.

   The idea. A triangle has three bases and three matching heights, and all three pairs must give the same area. Compute the area once with the pair you know, then read the unknown off the other pair.

   Using base BC. Its matching height is AX, so

   Area(∆ABC) = ½ × AX × BC = ½ × 5 × 3 = 15/2 sq units

   Using base AC. Its matching height is BY, so

   Area(∆ABC) = ½ × BY × AC = ½ × BY × 4 = 2 BY

   Equating. The two expressions describe the same triangle, so

   2 BY = 15/2 which means BY = 15/4 = 3.75 units

   Sense check. AC = 4 is longer than BC = 3, and the longer base carries the shorter height — 3.75 against 5 — which is what you would expect since their products must both come to 15.

   ✦ Answer: BY = 15/4 = 3.75 units.

5. 54 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 155

   Both diagonals of a rectangle are drawn, cutting it into four triangles numbered 1 to 4. Are the four triangles equal in area? What general statement about triangles does this suggest?

   Hint. The triangles are not congruent — compare them using base-and-height pairs instead.

   They are not congruent. Two of them are long and flat, two are tall and narrow, so congruence is not the route. Compare areas directly.

   Comparing two adjacent triangles. Let the diagonals meet at O and look at triangles 1 and 2, which share the vertex A. Take OD and OB as their bases. Both triangles then have the same apex A, so both have the same altitude — the perpendicular from A to the diagonal BD. And the diagonals of a rectangle bisect each other, so OB = OD. Equal bases and equal heights give

   Area(1) = Area(2)

   Going round. The same argument applied to the next adjacent pair gives Area(2) = Area(3), and again Area(3) = Area(4). So all four are equal, each being a quarter of the rectangle.

   The general statement. What actually did the work was this: equal bases along the same line, with a common apex, give equal areas. Applied to any triangle it says —

   In a triangle, the line joining a vertex to the midpoint of the opposite side > (a median) divides the triangle into two triangles of equal area.

   This one fact is used again and again later in the chapter: in the midpoint problem on page 159, in halving a quadrilateral, and in the trapezium proof on page 170.

   ✦ Answer: Yes, all four have equal areas, each one quarter of the rectangle; the general fact is that a median divides a triangle into two triangles of equal area.

6. 65 marksGanita Prakash Cl-8 Part 2, Section 7.1, page 156

   Line l is parallel to BC. Consider all triangles with base BC whose third vertex lies somewhere on l. (i) Which has the maximum area and which the minimum? (ii) Which has the maximum perimeter and which the minimum?

   Hint. For (i) ask what the height is. For (ii) treat the line l as a mirror.

   (i) Area — every one of them ties. Each triangle stands on the same base BC, and each apex lies on l, which is parallel to BC. The distance between two parallel lines is the same everywhere, so every triangle has the same height h. Hence

   Area = ½ × BC × h for all of them

   There is no biggest and no smallest: sliding the apex along l stretches the triangle sideways without ever changing its area. All the areas are equal.

   (ii) Perimeter — one clear winner, no loser. BC is common to all of them, so only AB + AC matters. Treat l as a mirror and reflect the whole picture below it. Let B′ and C′ be the images of B and C.

   Because a point and its mirror image are the same distance from the mirror, ∆AXB ≅ ∆AXB′ for the foot X, so AB = AB′; likewise AC = AC′. Therefore

   AB + AC = AB + AC′ = length of the path B → A → C′

   The shortest path from B to C′ is the straight segment BC′, so the best A is the point where BC′ crosses l. Any other position of A makes the path a bent one, which is longer.

   Where that point is. Put B = (0, 0), C = (c, 0) and l as the line y = h. Then C′ = (c, 2h), and the segment from B(0, 0) to C′(c, 2h) crosses y = h at x = c/2. That is directly above the midpoint of BC, so A lies on the perpendicular bisector of BC — the triangle of least perimeter is the isosceles one, confirming the intuition the book started with.

   There is no maximum perimeter: push A far out along l and AB + AC grows without limit.

   ✦ Answer: (i) all the triangles have the same area, ½ × BC × h — there is no maximum or minimum; (ii) the isosceles triangle (apex on the perpendicular bisector of BC) has the minimum perimeter, and there is no maximum.

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.