Area | CBSE Class 8 Maths Notes

This note covers area and perimeter, rectangles and squares, triangle areas and heights, equal-area triangles, polygons, dissection, parallelograms, rhombuses, trapeziums, and the conversion of area units.

What does area measure, and how do rectangles help?

Definition: Area measures a region by the number of unit squares whose combined area equals that region. This number can include a fraction of a unit square.

A unit square has side length one unit. With the centimetre, written cm, as the length unit, a square of side 1 cm has area one square centimetre, written cm². The superscript ² in cm² indicates a square unit of area.

A rectangle is a four-sided figure with four right angles. A right angle measures 90 degrees, written 90°. A rectangle’s length and width count how many unit squares fit along each direction. Multiplying these counts gives the number covering the rectangle without overlap.

Result: Rectangle and square area

Area = length × width. The sign × means multiplication, and = means equality. A square is a rectangle with all sides equal. Its area is therefore side length multiplied by itself.

Worked example 1. Two rectangular rangoli regions have side lengths 7 cm and 4 cm, and 8 cm and 3 cm. Which needs more powder if both are coloured evenly?

Answer: Their areas are 7 × 4 = 28 cm² and 8 × 3 = 24 cm². The rectangle measuring 7 cm by 4 cm needs more powder because its area is greater.

The comparison depends on even colouring. It connects the surface covered with the amount of powder needed. Comparing just one side would fail here: the rectangle with an 8 cm side has the smaller area.

A diagonal joins two non-neighbouring corners of a polygon, a closed figure bounded by straight sides. A triangle is a polygon with three sides. In a rectangle, a diagonal makes two congruent triangles. Congruent figures have the same shape and size, so these triangles have equal areas.

Worked example 2. A diagonal divides a rectangle measuring 7 cm by 4 cm into two triangles. Find the area of each triangle.

Answer: The rectangle has area 28 cm². Each triangle occupies half of it, so each area is ½ × 7 × 4 = 14 cm². The symbol ½ means one half.

Why is perimeter different from area?

Perimeter is the length of a region's boundary. It measures the distance around the edge, while area measures the region enclosed. A perimeter uses a length unit, whereas an area uses a square unit.

Regions can have equal perimeters but different areas. They can also have equal areas but different perimeters. A region with a larger perimeter can even have a smaller area than another region. Boundary length alone therefore does not determine the enclosed area.

How can a surrounding path be measured?

Consider a rectangular park inside a larger rectangle, with the region between them forming a path. The outer area includes both the park and the path. Removing the park's area leaves the area occupied by the path.

Path area = outer area − park area. The sign − means subtraction. Find each rectangular area using its own length and width. The width of the path alone does not supply the dimensions of the whole park.

If the rectangles keep their sizes and the inner park remains inside the outer rectangle, moving their relative positions leaves this difference unchanged. The widths of particular parts of the path may change, but its total area stays the same.

How should pieces be added?

A dissection cuts a figure into pieces and rearranges them into a different figure of equal area. The pieces must account for the original region without losing part of it. This explains why changing a shape need not change its area.

For a crosspath formed by two intersecting rectangular strips, simply adding both strip areas counts their common portion twice. Subtract the overlap once. The plot dimensions alone are insufficient: the widths of the crossing strips are also needed.

Note: Decide whether a question concerns the boundary, the whole region, or a remaining region. Then choose perimeter, multiplication of dimensions, or subtraction of areas accordingly.

How is the area formula for a triangle obtained?

A triangle is a polygon with three sides. Choose one side as its base. Its corresponding height is the length of an altitude: a segment from the opposite corner meeting the base line perpendicularly, which means at a right angle.

Perpendicular lines meet at a right angle. The altitude may meet the base itself or an extension of the base. The height is not automatically the length of a sloping side.

Result: Triangle area

Area = ½ × base × height. Enclosing the triangle in a suitable rectangle explains the factor ½. Drawing the altitude splits the triangle into two parts, each occupying half of its corresponding rectangular portion.

Adding those two triangular areas gives half the whole rectangle's area. The rectangle's dimensions are the chosen base and its corresponding perpendicular height. This is why these two measurements are enough to find the area.

The rectangle has corners A, B, C and D. Point X lies on its top side, and triangle XDC joins X to the two bottom corners. The bottom side is labelled 5, the left side 4, and the perpendicular from X meets the bottom side at Y.

See Fig. 7.1 in your NCERT textbook

Here a pair of capital letters, such as DC, names the segment joining those points or its length, as the context requires. A group such as XDC names the triangle with those three corners, also called vertices.

Worked example 3. Triangle XDC has base DC of length 5 units and perpendicular height XY of length 4 units. Find its area.

Answer: Area = ½ × 5 × 4 = 10 square units. Multiplying base by height gives the enclosing rectangle's area; taking half gives the triangle's area.

Why does an outside altitude still work?

Let triangle ABC have its altitude AD outside the triangle, with D, B and C in that order on one straight line. Write h for the altitude length AD. The larger triangle ADC and smaller triangle ADB share height h.

Subtracting their areas gives ½ × h × DC − ½ × h × DB = ½ × h × (DC − DB). Parentheses group the subtraction. Since DC − DB = BC, this becomes ½ × h × BC. The formula therefore applies to all types of triangles.

How can area reveal an unknown height or equal regions?

A triangle can be measured using any side as its base, provided the corresponding height is used. Choosing another base changes the required height but does not change the triangle's area. Two expressions for that one area can reveal a missing altitude.

How do two base-height pairs give the same area?

Worked example 4. In triangle ABC, BC = 5 units and AC = 4 units. The altitude AX meets BC at X and is 3 units long. The altitude BY meets AC perpendicularly at Y. Find BY.

Answer: Using BC, the area is ½ × 5 × 3 = 15/2 square units. The slash / denotes division. Using AC, the same area is ½ × 4 × BY = 2 × BY. Hence BY = 15/4 = 3.75 units.

Both calculations concern the same triangle. The first uses the known base-height pair to establish its area; the second uses that area to find the unknown height. Pairing BC with BY would be incorrect because BY is perpendicular to AC.

Property: A midpoint divides the area equally

A midpoint divides a segment into two equal lengths. Joining a triangle's vertex to the midpoint of the opposite side creates two smaller triangles. Their bases are equal halves of the same side, and their perpendicular heights are equal.

Applying the triangle area formula to each gives equal areas. This statement does not require the two smaller triangles to be congruent. Equal area describes the amount of region covered, not necessarily the shape or all the side lengths.

What happens when both rectangle diagonals are drawn?

The diagonals of a rectangle bisect one another, meaning that each cuts the other into equal halves. Consider two neighbouring triangles formed at the intersection. Their bases can be chosen as the two halves of one diagonal.

Those two triangles have equal bases and the same altitude from their shared rectangle corner. Their areas are equal. Repeating the argument for neighbouring pairs shows that the four triangular regions have equal areas, although they need not all be congruent.

What changes when a triangle's vertex moves along a parallel line?

Parallel lines are lines in a plane that do not meet. The perpendicular distance between them is constant. Fix a base BC and a line l parallel to it; here l is simply the name of the line.

Triangles with base BC and their third vertex anywhere on l have the same base length and perpendicular height. Therefore, they have equal areas. Moving the vertex sideways can alter the other side lengths without changing that area.

Why can their perimeters differ?

The fixed base contributes the same length to each perimeter. Comparing the perimeters therefore reduces to comparing the sum of the two other sides. A sum is the result of addition. An area calculation cannot settle that comparison, because the areas here are already equal.

A perpendicular bisector is a line passing through a segment's midpoint at a right angle. Intuition might suggest placing the third vertex where this line meets l to obtain the smallest perimeter. A reflection argument supplies the justification.

How does reflection find the shortest route?

A reflection places a point on the opposite side of a mirror line at the same perpendicular distance. Reflect C across l to obtain C′, read “C prime”. Let A be a point on l. Then AC and AC′ have equal lengths.

1. Keep the base BC fixed and consider the route from B through A to C.
2. Replace its part from A to C by the equally long part from A to C′.
3. The shortest route from B to C′ is the straight segment joining them.
4. Choose A where that straight segment crosses l. This choice minimises the sum of the two variable sides.

This construction gives the triangle with minimum perimeter, meaning the smallest perimeter among the triangles considered. It also explains the shortest-route idea behind travelling to a straight river boundary before continuing to another destination.

The two comparisons ask different questions: area depends on base and perpendicular height, while perimeter depends on side lengths. Keep both measurements separate even when the triangles occupy the same strip between parallel lines.

How can the area of a polygon be found from triangles?

A quadrilateral is a polygon with four sides; a pentagon has five sides. Joining suitable non-neighbouring vertices divides a polygon into triangles. Calculating these component areas and adding them gives the area of the original polygon.

The triangle formula therefore provides a method for finding the area of any polygon. The needed measurements are the bases and corresponding perpendicular heights of the triangles used in the division. Side lengths visible around the boundary need not provide all these heights.

How does a diagonal help with a quadrilateral?

When a diagonal lies inside a quadrilateral, it divides the region into two triangles. Use that diagonal as a common base. The heights are measured separately from the two remaining vertices to the line of the diagonal.

Worked example 5. In quadrilateral ABCD, diagonal AC = 22 cm lies inside the figure. Vertices B and D lie on opposite sides of AC. Their perpendicular distances BM and DN to AC are both 3 cm. Find the quadrilateral's area.

Answer: Triangle ABC has area ½ × 22 × 3 = 33 cm². Triangle ADC also has area 33 cm². Adding these gives the quadrilateral's area as 66 cm².

When is subtraction more convenient?

A shaded polygon inside a rectangle can sometimes be found by subtracting the unshaded triangles from the rectangle. Identify all the excluded regions first. Their areas must be counted without omission or overlap.

Worked example 6. Rectangle ABCD has AB = DC = 18 cm and AD = BC = 10 cm. Point E lies on AB with AE = 10 cm and EB = 8 cm. Point F lies on AD with AF = 6 cm and FD = 4 cm. Find the area of quadrilateral FECD.

Answer: The rectangle's area is 18 × 10 = 180 cm². The excluded triangles AFE and EBC have areas ½ × 10 × 6 = 30 cm² and ½ × 8 × 10 = 40 cm². The required area is 180 − 30 − 40 = 110 cm².

Both methods use the same principle: account for the whole region using simpler pieces. Choose addition when the desired pieces are easy to measure, and subtraction when the surrounding figure and unwanted pieces have convenient dimensions.

Why is a parallelogram's area base multiplied by height?

A parallelogram is a quadrilateral with both pairs of opposite sides parallel. Choose one side as the base. Its corresponding height is the perpendicular distance to the opposite parallel side. A sloping adjacent side is not generally this height.

How does dissection produce a rectangle?

Consider parallelogram ABCD with base DC. Draw the perpendicular AX from A to DC, with X on DC. Cutting along AX separates triangle AXD from the rest of the figure. Moving that triangle to the opposite end completes a rectangle.

The fit can be checked by comparing the cut triangle with the triangle needed at the other end. They have equal perpendicular sides, equal sloping sides and a right angle. Their congruence confirms that the rearrangement leaves neither a gap nor an overlap.

A perpendicular from A meets DC at X. Triangle AXD is separated from the left end. At the right end, the matching triangle BYC completes rectangle ABYX, with Y on the extended base line.

Reference: NCERT Class 8, page 161, unnumbered diagram

Result: Parallelogram area

The resulting rectangle has the parallelogram's height and a length equal to its original base. Dissection preserves area, so Area = base × height. The shape changes, but the amount of region covered does not.

Worked example 7. A parallelogram has base 7 cm and corresponding perpendicular height 4 cm. Find its area.

Answer: Area = 7 × 4 = 28 cm². No factor of ½ is required because the parallelogram has the same area as the complete rearranged rectangle.

Any side can serve as the base if its corresponding perpendicular height is used. Rotating a drawing does not change this requirement. A height drawn horizontally across the page can be just as valid as one drawn vertically.

Parallelograms with equal bases between the same parallel lines have equal heights and equal areas. Their sloping sides, and hence their perimeters, can differ. This provides another way to understand why equal area does not imply equal boundary length.

How do the diagonals determine a rhombus's area?

A rhombus is a parallelogram whose four sides have equal lengths. It can therefore be measured using base multiplied by corresponding height. Its diagonals provide another useful pair of measurements for calculating the same area.

The diagonals of a rhombus are perpendicular bisectors of each other. They cross at right angles, and each is divided into two equal halves at their intersection. These properties make the resulting triangles suitable for rearrangement into rectangles.

How does the dissection work?

Let the rhombus have vertices A, B, C and D, with diagonals AC and BD meeting at O. The letter O names their intersection. Triangle ABD and triangle CBD are isosceles, meaning each has two equal sides.

Each isosceles triangle can be cut into two right-angled triangles, which are triangles containing a right angle. Rearranging each pair makes a rectangle. Joining the two rectangles gives one rectangle with the area of the original rhombus.

This combined rectangle has one side equal to AC and the other equal to half of BD. The first diagonal contributes its full length; the second contributes half its length. Multiplication gives the rhombus formula.

Result: Rhombus area

Area = ½ × AC × BD. More generally, the area is half the product of the two diagonal lengths. The word product means the result of multiplication.

Worked example 8. Find the area of a rhombus with diagonals 20 cm and 15 cm.

Answer: Area = ½ × 20 × 15 = 150 cm². The given measurements are full diagonal lengths, so the factor ½ is applied once.

Can the formula also be checked by addition?

Use BD as the common base of triangles ABD and CBD. Their corresponding heights are AO and CO. The sign + means addition. Adding their areas gives ½ × BD × AO + ½ × BD × CO = ½ × BD × (AO + CO).

Since AO + CO = AC, this equals ½ × BD × AC. Thus, adding two triangular areas and dissecting into a rectangle lead to the same result. Neither method requires treating a side of the rhombus as one of its diagonals.

Why does the trapezium formula use the sum of parallel sides?

A trapezium is a quadrilateral with a pair of opposite sides parallel. Its height is the perpendicular distance between those parallel sides. The two parallel lengths both contribute to its area, even when one is shorter than the other.

Let a and b represent the lengths of the two parallel sides, and let h represent their perpendicular separation. These letters stand for lengths, not for names of vertices. Use lengths measured in the same unit.

How do a rectangle and two triangles give the formula?

Consider a trapezium where perpendiculars from the ends of the shorter parallel side meet the longer side. The cuts produce a central rectangle and two outer triangles. The rectangle has width a and height h.

Let x and y be the lengths of the two remaining segments on the longer side. Then b = x + a + y, so x + y = b − a. These segments are the bases of the outer triangles, each with height h.

1. The rectangle contributes area a × h.
2. The two triangles together contribute ½ × h × (x + y).
3. Replace x + y with b − a to get a × h + ½ × h × (b − a).
4. Combining the terms gives ½ × h × (a + b).

Result: Trapezium area

Area = ½ × h × (a + b). In words, multiply the height by the sum of the parallel side lengths, then take half. A sloping non-parallel side cannot replace h.

Why does the formula also work for a leaning trapezium?

Take two identical copies of the trapezium. Rotate one and join the copies along matching non-parallel sides. They form a parallelogram whose base is a + b and whose height is h.

The joined boundary becomes straight at the meeting points because the relevant angles between parallel lines add to 180°. Each original trapezium occupies half of the parallelogram. This proves the same formula without requiring both perpendiculars to fall inside the original trapezium.

Worked example 9. A trapezium has parallel sides 24 m and 36 m, with perpendicular height 14 m. Here m denotes metre. Find its area.

Answer: The sum of the parallel sides is 24 + 36 = 60 m. Area = ½ × 14 × 60 = 420 m², where m² means square metres.

The three essential measurements are therefore the two parallel sides and their perpendicular separation. Adding the non-parallel sides would answer a different question about boundary length and would not produce the area.

How are area units used and converted in everyday measurements?

Area units should suit the size of the region. Areas of classrooms and houses are generally measured in square feet or square metres. Larger land areas are also measured in acres, while larger areas are measured in square kilometres.

A square foot, written ft², is the area of a square with side one foot. A foot is written ft and an inch is written in. A square inch is written in². A kilometre is written km, and km² denotes a square kilometre.

How can an A4 sheet help estimate area?

An A4 sheet has side lengths 21 cm and 29.7 cm. You could perhaps visualise how many such sheets fit on a tabletop. This connects a familiar rectangular region with an estimate of a larger surface.

Worked example 10. Find the area of an A4 sheet measuring 21 cm by 29.7 cm.

Answer: Area = 21 × 29.7 = 623.7 cm². Both measurements are lengths in centimetres, so the area is expressed in square centimetres.

Why must a length conversion be squared?

One inch equals 2.54 cm. A square of side one inch therefore has both its length and width equal to 2.54 cm. Its area is 2.54 × 2.54 = 6.4516 cm², not 2.54 cm².

An acre is a land-area unit equal to 43,560 square feet. The table distinguishes length relationships from area relationships. The conversion between square units must account for both dimensions of the square.

Worked example 11. Convert 161.29 cm² into square inches, using 1 in² = 6.4516 cm².

Answer: Divide the given area by the number of square centimetres in one square inch. Thus, 161.29/6.4516 = 25, so the area is 25 in².

Since 1 ft = 12 in, a square foot contains 12 × 12 = 144 square inches. Multiplying by the length conversion only once would count one direction and miss the other.

Different parts of India also use local area units such as bigha, gaj, katha, dhur, cent and ankanam. These are names of local measurement units; a conversion requires the applicable local definition rather than an assumed common value.

Glossary

• Area — A measure of a region using the number of unit squares equivalent to its extent.
• Unit square — A square whose side length is one unit of the chosen length measurement.
• Perimeter — The total length of the boundary around a region or closed figure.
• Base — The selected side used with its corresponding perpendicular height to calculate area.
• Altitude — A perpendicular segment from a triangle's vertex to the line containing its opposite side.
• Diagonal — A line segment joining two vertices that are not neighbours in a polygon.
• Congruent figures — Figures having the same shape and size, and consequently having equal areas.
• Midpoint — The point that divides a line segment into two parts of equal length.
• Dissection — Cutting a figure into pieces and rearranging them into another figure of equal area.
• Polygon — A closed plane figure whose boundary consists of straight line segments.
• Parallelogram — A quadrilateral in which both pairs of opposite sides are parallel.
• Rhombus — A parallelogram whose four sides are equal and whose diagonals bisect each other perpendicularly.
• Trapezium — A quadrilateral with a pair of parallel opposite sides used in its area formula.
• Perpendicular bisector — A line passing through a segment's midpoint and meeting the segment at a right angle.
• Reflection — A mirror transformation placing corresponding points at equal perpendicular distances on opposite sides of a line.

Common errors and misconceptions

• Misconception: A longer boundary means a larger area. Correct: Perimeter does not determine area; a region with a larger perimeter can have a smaller area.
• Misconception: The sloping side is the height of a triangle. Correct: Height is perpendicular to the chosen base or its extension, so a sloping side cannot simply replace it.
• Misconception: The triangle formula fails when an altitude is outside. Correct: Subtracting two triangular areas proves that the same base-height formula still applies in this situation.
• Misconception: Equal-area triangles must be congruent. Correct: Equal bases and heights give equal areas without requiring all corresponding sides or angles to be equal.
• Misconception: A parallelogram's area is half its base times height. Correct: Its area is the full product. The factor ½ belongs to the triangle formula.
• Misconception: Multiply the diagonals of a rhombus without halving. Correct: The area is half their product, using the full lengths of both diagonals.
• Misconception: Any two trapezium sides can be added for the area formula. Correct: Add the parallel sides and use their perpendicular separation as the height.
• Misconception: Since 1 in = 2.54 cm, 1 in² = 2.54 cm². Correct: Convert both dimensions: 1 in² = 2.54 × 2.54 = 6.4516 cm².

Exam-style questions with model answers

Q1. Two rectangular regions measure 7 cm by 4 cm and 8 cm by 3 cm. Which requires more rangoli powder if both are coloured evenly? [2 marks]

1. Their areas are 7 × 4 = 28 cm² and 8 × 3 = 24 cm².
2. The 7 cm by 4 cm rectangle requires more powder because even colouring requires more powder for the larger area.

Q2. A rhombus has diagonals 20 cm and 15 cm. Calculate its area. [2 marks]

1. A rhombus has area equal to half the product of its full diagonal lengths.
2. Its area is therefore ½ × 20 × 15 = 150 cm², expressed in square centimetres.

Q3. In triangle ABC, BC = 5 units and AC = 4 units. X lies on BC, AX is perpendicular to BC and AX = 3 units. Y lies on AC, and BY is perpendicular to AC. Find BY using two expressions for the triangle's area. [3 marks]

1. Using base BC and its corresponding altitude AX, the area of triangle ABC is ½ × 5 × 3 = 15/2 square units.
2. Using base AC and its corresponding altitude BY, the same triangle has area ½ × 4 × BY = 2 × BY square units.
3. Equating the two expressions gives 2 × BY = 15/2. Dividing by 2 gives BY = 15/4 = 3.75 units.

Q4. A quadrilateral ABCD has an internal diagonal AC = 22 cm. Vertices B and D lie on opposite sides of AC, each at a perpendicular distance of 3 cm from it. Find the area of ABCD. [3 marks]

1. The diagonal separates the quadrilateral into triangles ABC and ADC. Use AC as the common base, with the given perpendicular distance as each triangle's height.
2. Triangle ABC has area ½ × 22 × 3 = 33 cm². The same calculation gives 33 cm² for triangle ADC.
3. The two triangles cover the quadrilateral without overlap of their interiors. Adding their areas gives the required area as 33 + 33 = 66 cm².

Q5. A trapezium has parallel sides 24 m and 36 m and perpendicular height 14 m. Explain the two-copy method and calculate its area. [5 marks]

1. Take a second identical copy of the trapezium. Rotate it and join the copies along matching non-parallel sides to make a parallelogram.
2. The parallelogram's base combines the two original parallel sides. Its length is therefore 24 + 36 = 60 m.
3. The joining does not change the perpendicular separation of the parallel boundaries. The height of the combined parallelogram is still 14 m.
4. A parallelogram has area equal to base multiplied by height. The combined figure therefore has area 60 × 14 = 840 m².
5. The two identical trapeziums have equal areas, so each occupies half of this total. The required area is ½ × 840 = 420 m².

Q6. Explain why joining a vertex of a triangle to the midpoint of the opposite side makes two equal-area triangles. Must these two triangles be congruent? [3 marks]

1. The midpoint divides the opposite side into two equal segments. Choose these segments as the respective bases of the two smaller triangles.
2. The triangles share the original opposite vertex, so their perpendicular distances to the common base line are equal. Thus they have equal bases and equal heights.
3. The formula ½ × base × height gives equal areas. They need not be congruent, because equality of area does not require equality of all corresponding side lengths.

Q7. Using 1 in = 2.54 cm, derive the square-inch conversion and convert 161.29 cm² into square inches. [3 marks]

1. A square of side one inch has both length and width equal to 2.54 cm. Its area must therefore use this conversion in both directions.
2. Multiplying the two lengths gives 1 in² = 2.54 × 2.54 = 6.4516 cm². This is an area conversion, not merely a length conversion.
3. Divide the given area by 6.4516 to count its square inches. Thus, 161.29/6.4516 = 25, and the required area is 25 in².

Q8. Rectangle ABCD has AB = DC = 18 cm and AD = BC = 10 cm. 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 quadrilateral FECD by subtraction. [5 marks]

1. The whole rectangle has area 18 × 10 = 180 cm². This includes the required quadrilateral and the two unwanted corner triangles.
2. Triangle AFE is right-angled at A. Its perpendicular sides are AE = 10 cm and AF = 6 cm, giving area ½ × 10 × 6 = 30 cm².
3. Triangle EBC is right-angled at B. Its perpendicular sides are EB = 8 cm and BC = 10 cm, giving area ½ × 8 × 10 = 40 cm².
4. These two corner triangles do not overlap. Their combined area is 30 + 40 = 70 cm², which must be removed from the rectangle.
5. The required quadrilateral is the remaining region. Its area is 180 − 70 = 110 cm².

Key takeaways

• Area measures the region covered in square units, whereas perimeter measures the length around its boundary.
• A rectangle has area length multiplied by width; a square uses the same length in both directions.
• For every triangle, area is half the product of a chosen base and its corresponding perpendicular height.
• A line from a triangle's vertex to the opposite side's midpoint divides the triangle into two equal-area regions.
• Break a polygon into triangles and add their areas, or subtract unwanted pieces from a convenient surrounding figure.
• A parallelogram's area equals base multiplied by corresponding height; a rhombus also has area half the product of its diagonals.
• A trapezium's area is half its perpendicular height multiplied by the sum of its two parallel sides.
• Area conversions account for both dimensions: converting inches to centimetres requires squaring the length conversion for square units.

Test yourself

What does a unit square measure when its side is 1 cm?

Its area is one square centimetre, written 1 cm².

Can a region with a larger perimeter have a smaller area?

Yes. Boundary length does not determine the area enclosed by that boundary.

Why can an altitude outside a triangle still be used?

Height is the perpendicular distance to the base line. Subtracting two triangular areas proves that the same formula applies.

What stays unchanged when a vertex moves along a line parallel to a fixed triangle base?

The perpendicular height stays unchanged, so the triangle's area also remains the same.

A parallelogram has base 7 cm and perpendicular height 4 cm. What is its area?

Its area is base multiplied by height: 7 × 4 = 28 cm².

Which measurements are needed for the diagonal formula of a rhombus?

The full lengths of both diagonals are needed; the area is half their product.

What figure is made by joining two suitably arranged identical trapeziums?

They make a parallelogram with the same height and a base equal to the sum of the parallel sides.

Given that 1 ft = 12 in, how many square inches are in 1 ft²?

There are 12 × 12 = 144 square inches, because both dimensions must be converted.