CLASS 8 · GANITA PRAKASH PART II · CHAPTER 7
Find areas of triangles, polygons, parallelograms, rhombuses and trapeziums, and use area units in practical settings.
62 worked answers · short steps first, explanations when you need them
Textbook pages 149–152
Question equal-area-square-parts
Draw both diagonals; the four triangles have equal area by symmetry.
Alternatively, draw the two midlines to make four equal small squares.
Many curved or zigzag boundaries also work if each region has the same area.
There are infinitely many divisions; the two perpendicular midlines give one simple construction.
Question rangoli-rectangle-area
First area=7×4=28 cm².
Second area=8×3=24 cm².
The 7 cm by 4 cm rectangle has area 28 cm², compared with 24 cm² for the 8 cm by 3 cm one, so it needs more powder if spread evenly.
Question diagonal-triangle-area
Rectangle area=7×4=28 cm².
A diagonal divides a rectangle into two congruent triangles.
Each area=28÷2.
14 cm² each.
Question perimeter-not-area
A 6 cm by 4 cm rectangle has perimeter 20 cm and area 24 cm².
A 9 cm by 1 cm rectangle also has perimeter 20 cm but area 9 cm².
A 10 cm by 1 cm rectangle has perimeter 22 cm but area 10 cm², still less than 24 cm².
No. Equal perimeters can enclose different areas, and a shape with larger perimeter can have smaller area.
Question missing-rectangle-sides
(i) The 7 in side with area 21 in² gives the shared height 21÷7=3 in. The rectangle of area 28 in² then has other side 28÷4=7 in.
Continue across the joined rectangles: 35÷7=5 in and 14÷7=2 in. The marked missing segments are 3 in, 5 in, and 2 in.
(ii) The full rectangle is 50 m² and has height 4 m, so its length is 50÷4=12.5 m. The 29 m² region leaves 21 m²; the 11 m² region leaves 10 m². Their widths are 21÷4=5.25 m and 10÷4=2.5 m.
Question park-surrounding-path
Outer dimensions are l+2t and w+2t.
Path area=outer area−park area.
(l+2t)(w+2t)−lw=2t(l+w)+4t².
Area of path = 2t(l+w)+4t². For l=60 m, w=40 m, t=5 m, the path area is 1,100 m².
Question path-different-widths
Outer length=l+p+q.
Outer width=w+r+s.
Subtract inner area lw.
Path area=(l+p+q)(w+r+s)−lw.
Question path-position-does-not-change-area
Path area=area of outer rectangle−area of park.
Neither rectangle changes dimensions, so neither area changes.
No. The path area stays the same, even if the path is wider in some places and narrower in others.
Question crosspath-area
Vertical strip area=12u.
Horizontal strip area=14v.
The intersection of area uv was counted twice, so subtract it once.
Crosspath area=12u+14v−uv. If u=2 m and v=3 m, area=24+42−6=60 m².
Question spiral-tube-area
Add the areas of the nine 1-unit-wide rectangular arms: 20+20+20+15+15+10+10+5+5=120 square units.
At each of the eight bends, the 1×1 corner square was counted in both neighbouring arms. Subtract these 8 extra square units.
Area=120−8.
112 square units.
Question straight-tube-same-area
The bent tube is made of two 5-by-1 arms. Their combined area is 5+5=10 square units.
The 1-by-1 corner square is counted in both arms, so subtract it once: 10−1=9 square units.
For a straight tube of width 1, area=1×length. So its length must be 9 units.
9 units.
Question double-square-side-area
A square’s area scales with the square of its side.
New side factor=2, so new area factor=2²=4.
Each similar region’s area becomes 4 times as large, an increase of 3 times its original area (300%).
Question square-four-pieces-hole
Make the indicated four congruent pieces from a square cut by perpendicular lines.
Rearrange them around a central opening so their outer edges form a larger square.
Check that no piece is stretched or overlapped; total material area remains unchanged.
This is a hands-on dissection puzzle; exact cut positions follow the textbook diagram. Area is conserved during rearrangement.
Textbook pages 153–159
Question triangle-half-rectangle
Both triangles have the same base length as a rectangle side.
Their perpendicular height equals the rectangle’s width.
Each area=1/2×rectangle area.
The triangles have equal area: each is half of the rectangle.
Question triangle-area-formula
For an acute triangle, complete it to a rectangle or parallelogram of the same base and height; the triangle is half.
For an obtuse triangle, subtract the two right triangles outside it: 1/2h×(long segment)−1/2h×(short segment).
The difference of the horizontal segments is the triangle’s base.
Area of any triangle = 1/2×base×perpendicular height.
Question triangle-xdc-area
Use DC as the base and the perpendicular distance from X to DC as height.
Area=1/2×5×4.
10 square units.
Question find-by-altitude
Area=1/2×AX×BC=1/2×3×5=15/2.
The same area is 1/2×AC×BY=1/2×4×BY=2BY.
2BY=15/2, so BY=15/4.
BY=15/4=3.75 units.
Question diagonals-rectangle-triangles
The diagonals bisect one another, so each half-diagonal has equal length.
Adjacent triangles can use these equal half-diagonals as bases and share the same altitude.
Opposite triangles have equal areas by congruence.
Yes. All four triangles have equal area, each one-quarter of the rectangle.
Question parallel-line-triangle-area-perimeter
The perpendicular distance between the parallel lines is fixed, so every triangle has the same height and area.
Perimeter varies with the two sloping sides.
Reflect one endpoint across the parallel line; the shortest broken path becomes a straight line.
All areas are equal. There is no maximum perimeter if the parallel line is unlimited; the minimum occurs when the vertex lies above the midpoint of BC, making an isosceles triangle.
Question triangle-figure-areas
(i) The triangle has base 4 cm and perpendicular height 3 cm: area=1/2×4×3=6 cm².
(ii) Use base 5 cm and perpendicular height 3.2 cm: area=1/2×5×3.2=8 cm².
(iii) Use base 3 cm and perpendicular height 4 cm: area=1/2×3×4=6 cm².
Question altitude-by
First find the triangle’s area using AX and BC: area=1/2×3×5=15/2 square units.
Using AC as the base, the same area is 1/2×4×BY=2BY.
So 2BY=15/2 and BY=15/4=3.75 units.
BY=3.75 units.
Question isosceles-triangle-area-double
In an isosceles triangle, the perpendicular from the apex to the base bisects the base.
Triangles SEB and SEU have equal bases and common height, so their areas are equal.
Total area=24+24.
48 square units.
Question rectangle-to-equal-triangle
Rectangle area=lw.
A triangle with base l and perpendicular height w has area 1/2lw.
Use base 2l and height w, or base l and height 2w, to match lw.
For example, construct a triangle with base 2l and height w. Its area is lw, equal to the rectangle.
Question triangle-to-equal-rectangle
Triangle area=bh/2.
A rectangle with sides b and h/2 has area b(h/2)=bh/2.
Use a rectangle of dimensions b by h/2. A dissection can rearrange the two halves of the triangle to form it.
Question squares-midpoint-triangle-area
XM=XY/2 and XN=XZ/2.
The included angle at X is the same, so the small triangle is similar to the large one with scale 1/2.
Area scale=(1/2)².
Triangle XMN has 1/4 of the area of triangle XYZ.
Question shortest-house-river-tank
Reflect the tank across the straight river bank (or reflect the house).
Draw one straight line from the house to the reflected point.
Where this line meets the bank is the shortest crossing point.
Reflect one endpoint across the river line, join it to the other endpoint with a straight line, and use the intersection as the crossing point.
Textbook pages 159–164
Question quadrilateral-by-diagonal
Area=area(ABC)+area(ADC).
=1/2×22×3+1/2×22×3.
=22×3.
66 cm².
Question shaded-rectangle-region
The rectangle height AD=AF+FD=6+4=10 cm and its width AB=18 cm. So rectangle area=18×10=180 cm².
The two unshaded right triangles have areas 1/2×10×6=30 cm² and 1/2×8×10=40 cm².
Shaded area=180−30−40=110 cm².
110 cm².
Question regular-hexagon-measurements
Divide it into 6 congruent equilateral triangles; one side length determines each triangle.
Alternatively use perimeter P and apothem a: area=Pa/2.
Knowing the side length is sufficient for a regular hexagon; perimeter and apothem are another convenient pair.
Question blue-rectangle-fraction
The blue region is made of two triangles whose bases are the rectangle’s full length.
If their heights are h₁ and h₂ and the rectangle height is H, then h₁+h₂=H. Their total area is 1/2×length×h₁ + 1/2×length×h₂ = 1/2×length×H.
The rectangle area is length×H, so the blue fraction is (1/2×length×H)/(length×H)=1/2.
One-half of the rectangle.
Question quadrilateral-half-area
Mark the midpoint of each side.
Join the four midpoints in order.
The midpoint quadrilateral is a parallelogram with half the area of the original (Varignon’s theorem).
Join the four side midpoints; the inner parallelogram has half the original quadrilateral’s area.
Question parallelogram-to-rectangle-dissection
Drop a perpendicular from one vertex to the opposite base to form a right triangle.
Cut off that triangle and move it to the other end.
The result is a rectangle with the same base and height.
Cut the end triangle and shift it to the opposite end; the rectangle has area base×height.
Question parallelogram-area-formula
A parallelogram can be dissected into a rectangle with the same base and perpendicular height.
Rectangle area=base×height.
Area of a parallelogram = base × perpendicular height.
Question parallelogram-same-area-family
Each has base 5 units and perpendicular height 3 units, so area=5×3=15 square units for every one.
The slanted side gets longer as the shape leans more, so the perimeter changes even though the area does not.
From the pictured choices, (d) has the least perimeter and (g) has the greatest perimeter.
All seven have area 15 square units. Figure (d) has the minimum perimeter; figure (g) has the maximum.
Question parallelogram-numeric-areas
(i) Area=7×4=28 cm².
(ii) Area=5×3=15 cm². (iii) Area=5×4.8=24 cm².
(iv) Area=2×4.4=8.8 cm².
Question find-qn-parallelogram
The area using base SR is 12×6=72 cm².
Use PS as the base for the same parallelogram: 7.6×QN=72.
QN=72÷7.6=180/19≈9.47 cm.
QN≈9.47 cm.
Question rectangle-vs-parallelogram-area
Use 5 cm as base.
The rectangle height is 4 cm, so area=20 cm².
The parallelogram’s perpendicular height is less than its sloping side length 4 cm.
The rectangle has the greater area. The parallelogram would match 20 cm² only if its angle were 90° and it became a rectangle.
Question rectangle-double-triangle
For a triangle with base b and height h, area=bh/2.
A rectangle b×h has area bh, exactly twice as large.
Use a rectangle with the same base and height as the triangle. Other dimensions with product bh also work.
Question isosceles-dissection-rectangle
Draw the altitude from the apex; it bisects both the base and the triangle.
Cut along the altitude to make two congruent right triangles.
Rearrange the halves so they form a rectangle of base equal to half the original base and height equal to the triangle height.
The rectangle has dimensions (base/2)×height and therefore the same area as the triangle.
Question rectangle-to-isosceles-triangle
A rectangle of width w and height h has area wh.
An isosceles triangle with base 2w and altitude h has area 1/2×2w×h=wh.
Cut and rearrange the rectangle’s two halves to form the two equal sloping sides.
Use an isosceles triangle with base twice the rectangle width and the same height; the pieces can be rearranged by matching congruent halves.
Question triangle-square-same-side
Square area=s².
Equilateral triangle area=√3/4 s²≈0.433s².
Two triangles have area=√3/2 s²≈0.866s².
The square is larger than one equilateral triangle and also larger than two of them.
Textbook pages 164–171
Question rhombus-diagonal-formula
The diagonals are perpendicular and bisect one another.
They divide the rhombus into four right triangles.
Adding their areas gives 1/2×d1×d2.
Area of a rhombus = 1/2×(diagonal 1)×(diagonal 2).
Question rhombus-triangle-sum
Area(ADB)=1/2×AO×BD.
Area(CDB)=1/2×CO×BD.
AO+CO=AC.
Total=1/2×AC×BD.
Area=1/2×AC×BD.
Question rhombus-20-15
Area=1/2×20×15.
150 cm².
Question rectangle-to-rhombus
The diagonal of a rectangle divides it into two congruent triangles.
Cut along the diagonal and shift/rotate the triangular pieces so their equal edges form a rhombus.
Keep all pieces and avoid overlap; total area is unchanged.
Cut the rectangle into two equal half-height rectangles, then cut each into two right triangles. The four triangles have legs L and W/2; arrange them around their right angles to form a rhombus with diagonals 2L and W, whose area is (2L×W)/2=LW.
Question trapezium-area-formula
Extend the trapezium into a rectangle and two triangles, or join two congruent copies into a parallelogram.
The resulting parallelogram has base a+b and height h.
One trapezium has half its area.
Area = 1/2×h×(a+b).
Question trapezium-slanted-case
Use a rectangle-and-triangles decomposition; the two triangles account for the offset lengths.
Their horizontal pieces plus the top base total the bottom base.
The total simplifies to 1/2h(a+b).
Yes. The formula depends only on the parallel side lengths and their perpendicular separation.
Question trapezium-two-copies
The two copies make a parallelogram with base a+b and height h.
Parallelogram area=(a+b)h.
One trapezium has half this area.
Area of one trapezium = 1/2h(a+b).
Question trapezium-figure-areas
(i) Parallel sides 10 ft and 7 ft, height 16 ft: area=1/2×(10+7)×16=136 ft².
(ii) Parallel sides 24 m and 36 m, height 14 m: area=1/2×60×14=420 m². (iii) Parallel sides 14 in and 6 in, height 10 in: area=1/2×20×10=100 in².
(iv) Parallel sides 18 ft and 12 ft, height 8 ft: area=1/2×30×8=120 ft².
Question isosceles-trapezium-dissection
Drop perpendiculars from the ends of the shorter base to the longer base.
The two end triangles are congruent because the trapezium is isosceles.
Move one end triangle to the opposite side to make a rectangle.
Cut off the two congruent end triangles and move one to the other end. The rectangle has the same height and the average of the parallel sides as its width.
Question trapezium-equal-rectangle-vertices
Draw the rectangle’s top and bottom parallel to the trapezium’s bases, with the rectangle’s height equal to the perpendicular distance between the bases.
Choose its width as the average of the trapezium’s parallel sides: (AB+DC)/2. This makes rectangle area equal to 1/2×height×(AB+DC).
For the dissection shown, position the cuts so the corner triangles match in pairs: construct ∆AHI congruent to ∆DGI and ∆BEJ congruent to ∆CFJ, as the hint states.
The rectangle has height equal to the trapezium’s height and width equal to half the sum of its parallel sides. The paired congruent triangles make the cut-and-move construction preserve area.
Question construct-trapezium-144
Choose parallel sides a=10 cm and b=14 cm.
Choose height h=12 cm.
Area=1/2×12×(10+14)=6×24.
Parallel sides 10 cm and 14 cm with height 12 cm give area 144 cm². Many other choices are possible.
Question hexagon-partition-ratio
A regular hexagon can be divided into 6 equal small equilateral triangles.
In the diagram, the triangle is 1 such unit, the rhombus is 2, and the trapezium is 3.
Trapezium : equilateral triangle : rhombus = 3:1:2.
Question trapezium-midpoint-equal-triangle
Let the parallel sides have lengths ZY=a and WX=b, and let their perpendicular separation be h.
Because A is the midpoint of XY, line ZA meets the parallel line WX at B so that WB=a+b. This follows from similar triangles: the midpoint on XY makes the two parallel-line gaps from Z to A and from A to B equal.
Area of triangle ZWB=1/2×WB×h=1/2×(a+b)×h. That is exactly the trapezium area formula, 1/2×(ZY+WX)×h.
The trapezium and triangle ZWB have equal area: both are 1/2×(a+b)×h.
Textbook pages 170–171
Question a4-sheet-area
Area=21×29.7.
=623.7.
623.7 cm².
Question inch-centimetre-lengths
5×2.54=12.7 cm; 7.4×2.54=18.796 cm.
5.08÷2.54=2 in; 11.43÷2.54=4.5 in.
5 in=12.7 cm; 7.4 in=18.796 cm; 5.08 cm=2 in; 11.43 cm=4.5 in.
Question square-inch-centimetre
1 in=2.54 cm in each direction.
1 in²=2.54×2.54=6.4516 cm².
10 in²=10×6.4516.
1 in²=6.4516 cm²; 10 in²=64.516 cm².
Question square-cm-to-square-in
1 in²=6.4516 cm².
161.29÷6.4516=25.
25 in².
Question square-feet-inches
1 ft=12 in.
1 ft²=12×12 in².
144 in².
Question square-km-metres
1 km=1,000 m.
1 km²=1,000×1,000 m².
1,000,000 m².
Question land-area-local-project
Choose a suitable unit: m² for rooms, hectares or km² for larger areas.
Break an irregular boundary into rectangles/triangles or use a map grid.
State assumptions and compare your estimate with a measured or published source.
This is a local research activity; the answer depends on the chosen place and the boundary definition.
Question largest-city-area-research
First decide what “city” means: municipal boundary, urban area or metropolitan region.
Use the same kind of boundary and a reliable current source for each comparison.
Record the date and source because administrative boundaries and rankings differ.
There is no single answer without defining “city” consistently. Treat this as a research question: compare like-for-like administrative or urban boundaries and cite the source/date.