Area

7.1 Rectangles and squares

Area counts unit squares covering a region without gaps or overlaps. Rectangle area=length×width; square area=side². A 7 cm by 4 cm rectangle covers 28 cm², while 8 cm by 3 cm covers 24 cm². Doubling both side lengths multiplies area by four. Perimeter measures boundary length and is not interchangeable with area.

Why perimeter cannot determine area

Rectangles 1×9 and 4×6 have equal perimeter 20 units but areas 9 and 24 square units. A 1×10 rectangle has perimeter 22 and area 10; a 4×4 square has perimeter 16 and area 16. Hence a larger perimeter can accompany a smaller area.

Area of paths and composite regions

For an outer rectangle L×W containing an inner rectangular region l×w, path area=LW−lw. If a path of uniform width t surrounds an l×w park, outer dimensions are (l+2t)×(w+2t), so path area=(l+2t)(w+2t)−lw. For two perpendicular paths of widths p,q across a rectangle L×W, combined area=Lp+Wq−pq, assuming each spans the full corresponding dimension.

7.2 Triangles

A diagonal divides a rectangle into two congruent triangles. Triangle area=½×base×perpendicular height, including obtuse triangles whose altitude lies outside the triangle. If a triangle's area is 30 cm² and base is 10 cm, height=2×30/10=6 cm. A median divides a triangle into two equal-area triangles.

Triangles on the same base between parallels

Triangles sharing a base and with third vertices on a line parallel to the base have the same perpendicular height, so equal area, even when their shapes and perimeters differ. Reflecting a point across a line can convert a broken-path minimum problem into a straight-line minimum.

Area of any polygon

Divide a simple polygon into non-overlapping triangles and add their areas. For a quadrilateral split by diagonal d, with perpendicular heights h₁,h₂ of the other vertices to the diagonal, area=½d(h₁+h₂). For a regular polygon with perimeter P and apothem a, area=½Pa.

Parallelogram

Dissect a triangular piece from one end of a parallelogram and move it to the other end to make a rectangle. Therefore area=base×corresponding perpendicular height, not base×sloping side. Opposite sides are equal but the area depends on the angle between adjacent sides.

Rhombus

A rhombus is a parallelogram with all four sides equal. Its area is base×corresponding height. Its diagonals are perpendicular and bisect one another, giving area=½d₁d₂. If diagonals are 10 cm and 24 cm, area=120 cm².

Trapezium

A trapezium has a pair of parallel sides (bases) of lengths a,b and perpendicular height h. Area=½(a+b)h. One derivation places two congruent copies together into a parallelogram of base a+b and height h, so one trapezium has half that area.

Unit conversions and precision

1 m²=10,000 cm²; 1 cm²=100 mm²; 1 hectare=10,000 m²; 1 km²=1,000,000 m². Converting lengths by a factor k converts areas by k². Round only at the end, and label answers with square units.

Formula comparison table

Activities, common errors and exam tips

Use squared paper to create two regions of equal area but different perimeters. Cut a parallelogram and rearrange it into a rectangle. Draw a triangle with a vertex sliding along a line parallel to its base and compare areas. Common errors include using a slanted side as a height, forgetting to subtract overlapping path area, and writing cm instead of cm².

NCERT first Figure it Out

1. Question 1. The first diagram shows rectangles with marked areas and missing side lengths. How can each unknown length be calculated?

   Step-by-step solution: For each rectangle, divide the labelled area by the known perpendicular side. For composite shapes, identify the rectangle sharing that side before dividing. The printed diagram must be consulted to assign each of the values 14,35,28 and 21 in², and 29,11 and 50 m² to the correct segment.

   Final answer: Use missing side=area÷known side; numerical assignments require the original diagram..

2. Question 2. A path surrounds a rectangular park. What measurements and formula determine its area?

   Step-by-step solution: Measure the outer length L and width W and the park length l and width w. Subtract park area from outer area. Example: outer 24×16 m, park 20×12 m gives 384−240.

   Final answer: 144 m² for the stated example; general formula LW−lw..

3. Question 3. A rectangular plot is 14 m by 12 m with two perpendicular crosspaths. Find a formula and evaluate for path widths 2 m and 1 m.

   Step-by-step solution: One path covers 14×2=28 m² and the other 12×1=12 m². Their 2×1 overlap is counted twice; subtract it.

   Final answer: 28+12−2=38 m²; general formula 14p+12q−pq..

4. Question 4. A spiral tube of uniform width is shown. How can its area be found?

   Step-by-step solution: Cut the spiral mentally into nonoverlapping rectangular strips or straighten the strip while preserving area. Measure every straight segment and corner correction from the original diagram, then sum length×width for the nonoverlapping pieces.

   Final answer: Exact numerical result requires the printed spiral diagram..

5. Question 5. If the side of a square doubles, how do the areas of its corresponding scaled regions change?

   Step-by-step solution: A similar region scaled by factor 2 in both dimensions has area multiplied by 2²=4. The absolute increase is 3 times its original area.

   Final answer: Each corresponding region becomes four times as large (a 300% increase), if the entire figure is scaled uniformly..

6. Question 6. How can a square be divided into four pieces and rearranged into a larger square with a central hole?

   Step-by-step solution: Cut along the two perpendicular lines indicated in the textbook, then reposition the four pieces to form the outer boundary. Rearrangement preserves the total material area, so area of larger square minus hole equals original area.

   Final answer: Outer square area−hole area=original square area; exact arrangement needs the figure..

NCERT triangles Figure it Out

1. Question 1. Find the areas of triangles with base 4 cm and perpendicular height 3 cm.

   Step-by-step solution: Apply ½×base×height=½×4×3.

   Final answer: 6 cm².

2. Question 2. A triangle has area 15/2 square units and side AC=4 units. Find the altitude BY to AC.

   Step-by-step solution: Area=½×AC×BY, so 15/2=½×4×BY=2BY. Divide by 2.

   Final answer: BY=15/4=3.75 units.

3. Question 3. An isosceles triangle SUB has altitude SE perpendicular to base UB. If area of triangle SEB is 24 square units, find area of SUB.

   Step-by-step solution: In an isosceles triangle the altitude from the apex bisects the base, producing two equal-area triangles. Double 24.

   Final answer: 48 square units.

4. Question 4. How can a rectangle be transformed into a triangle of equal area?

   Step-by-step solution: For a rectangle with base b and height h, choose a triangle of base 2b and height h (or base b and height 2h). Its area is ½×2b×h=bh.

   Final answer: A triangle of base 2b and height h has the same area..

5. Question 5. How can a triangle be transformed into a rectangle of equal area?

   Step-by-step solution: For triangle base b and height h, construct a rectangle of base b and height h/2. Its area=b×h/2.

   Final answer: Rectangle dimensions b by h/2..

6. Question 6. Three identical squares are arranged with red and blue shaded regions. How do you find the areas?

   Step-by-step solution: Identify congruent or complementary coloured pieces and compare them using the printed diagram. Without confirming which triangles are shaded, assigning a numerical blue area from the red 49 square units would be speculative.

   Final answer: Diagram-dependent; check the original shading before solving both subparts..

7. Question 7. M and N are midpoints of XY and XZ in triangle XYZ. What fraction is area of triangle XMN?

   Step-by-step solution: Triangle XMN is similar to XYZ with side scale 1/2, so its area scale is (1/2)².

   Final answer: One quarter of area of triangle XYZ.

8. Question 8. What is the shortest route from a house to a river and then to a water tank?

   Step-by-step solution: Reflect the water-tank point across the straight riverbank. Draw the straight segment from the house to that reflection. Its intersection with the riverbank is the optimal touch point; reflect back to obtain the two-leg route.

   Final answer: House→intersection with riverbank→tank; precise route depends on the map..

NCERT polygons Figure it Out

1. Question 1. A quadrilateral has diagonal AC=22 cm and perpendicular distances BM=DN=3 cm to AC. Find its area.

   Step-by-step solution: Split along AC into triangles ABC and ADC. Sum=½×22×3+½×22×3=33+33.

   Final answer: 66 cm².

2. Question 2. A shaded region lies inside a rectangle with several labelled segments. How is its area determined?

   Step-by-step solution: Find rectangle area and subtract unshaded triangles or rectangles using their printed bases and heights. The numbers 18,10,10,8,6,4 cm must be associated with the correct segments in the original figure.

   Final answer: Diagram-dependent; use area subtraction after verifying each label..

3. Question 3. What measurements are sufficient to calculate the area of a regular hexagon?

   Step-by-step solution: Measure its side length s and apothem a. Its perimeter is 6s; divide into six congruent triangles, so area=½×6s×a=3sa. Alternatively use s alone with area=(3√3/2)s².

   Final answer: Side and apothem, or side alone for a regular hexagon.

4. Question 4. What fraction of a rectangle is occupied by its blue region in the printed figure?

   Step-by-step solution: Identify the blue polygon and partition it into triangles or rectangles. Compute its area and divide by the full rectangle area. The diagram's coloured boundaries are needed to determine the fraction.

   Final answer: Requires the textbook figure; fraction=blue area÷rectangle area..

5. Question 5. How can a quadrilateral of half the area of a given quadrilateral be constructed?

   Step-by-step solution: Join the midpoints of the four sides in order. The resulting Varignon parallelogram has exactly half the original quadrilateral's area, since the four corner triangles occupy the other half.

   Final answer: Join consecutive side midpoints; resulting quadrilateral has half the area..

Original practice

1. Question 1. A parallelogram has base 18 cm and perpendicular height 7 cm. Find area.

   Step-by-step solution: Area=base×height=18×7.

   Final answer: 126 cm².

2. Question 2. A rhombus has diagonals 16 cm and 30 cm. Find area.

   Step-by-step solution: Area=½×16×30.

   Final answer: 240 cm².

3. Question 3. A trapezium has parallel sides 12 cm and 20 cm and height 9 cm. Find area.

   Step-by-step solution: Area=½×(12+20)×9=16×9.

   Final answer: 144 cm².

4. Question 4. A rectangle is 10 m by 6 m. A 1 m wide uniform path lies outside it. Find path area.

   Step-by-step solution: Outer dimensions 12 m by 8 m. Subtract inner 10×6 from outer 12×8: 96−60.

   Final answer: 36 m².

5. Question 5. Convert 2.5 m² to cm².

   Step-by-step solution: 1 m=100 cm, so 1 m²=100²=10,000 cm². Multiply 2.5×10,000.

   Final answer: 25,000 cm².

6. Question 6. Two rectangles have equal perimeter 24 cm. One is 4×8 cm and the other is 5×7 cm. Compare their areas.

   Step-by-step solution: Both have perimeter 2(4+8)=24 and 2(5+7)=24. Areas are 32 and 35 cm².

   Final answer: Same perimeter, but areas 32 cm² and 35 cm².

7. Question 7. A triangle's area is 45 cm² and its base is 15 cm. Find the perpendicular height.

   Step-by-step solution: 45=½×15×h, so h=90/15.

   Final answer: 6 cm.

Official reference and exercise transparency

Source: NCERT Ganita Prakash Part II Chapter 7 (Area). The first exercises and selected polygon problems have been checked against the official PDF text. Some numerical questions depend on drawings, shaded regions and exact placement of labels; these are presented with methods but without invented numeric answers. Additional parallelogram, rhombus and trapezium exercise subparts later in the chapter still require a full diagram-level audit. The original practice section is not part of the official question set.