MathsCBSE Class 8 · Area

Area of Rectangles, Paths, Triangles and Parallelograms for Class 8

Updated 17 October 2026 · Fact-checked

Area counts unit squares: a rectangle has area length x width. Perimeter does not decide area. A triangle has area half x base x height, a parallelogram has area base x height (any side can be the base), and a line from a vertex to the midpoint of the opposite side halves a triangle. Triangles on the same base with apex on a parallel line have equal area. Exams ask for path areas, equal-area reasoning and dissection.

Notice the triangle fills half of its b × h rectangle, and the parallelogram's slanted end moves across to make a b × h rectangle.

Illustration drawn for QuestionBuddy.

Understand Area of Rectangles, Paths, Triangles and Parallelograms

Area as unit squares. The area of a region is the number of unit squares (which may be fractions) that fill it. A rectangle with sides 7 cm and 4 cm contains $7 \times 4 = 28$ unit squares, so $area = length \times width$ (28 sq cm, written $28 cm^{2}$ ).

Perimeter does not decide area. Regions with the same perimeter can have different areas, and a region with the larger perimeter can have the smaller area. Rectangles $6 \times 2$ and $4 \times 4$ both have perimeter 16 but areas 12 and 16. The boundary length tells you nothing certain about the area, which is why area is measured by counting unit squares.

Paths and frames. The area of a path or border is the difference of two areas, outer minus inner (or a sum of rectangles, taking care not to count an overlap twice). A path of width $w$ round the outside of an $l \times b$ rectangle has area $(l + 2 w) (b + 2 w) - l b$ ; inside it, $l b - (l - 2 w) (b - 2 w)$ . For a cross-path made of two strips, add the two strips and subtract the common square.

Area of a triangle. Draw a triangle with base $b$ and height $h$ (the perpendicular distance from the opposite vertex to the base). A rectangle $b \times h$ can be drawn around it, and the triangle covers exactly half:
$ $area = \frac{1}{2} \times b \times h .$ $
For a triangle where the rectangle cannot be drawn on the chosen base (an obtuse triangle), the area is a difference of two triangles that can be enclosed, so the formula holds for all triangles. Any side can be the base if you use its own height: $\frac{1}{2} b_{1} h_{1} = \frac{1}{2} b_{2} h_{2}$ .

A line from a vertex to the midpoint of the opposite side divides a triangle into two triangles of equal area (equal bases, same height). The diagonals of a rectangle divide it into four triangles of equal area.
Triangles with the same base whose third vertex lies anywhere on a line parallel to the base have the same height, so the same area. Their perimeters differ: reflecting in the line as in a mirror shows that the apex above the midpoint of the base (the perpendicular bisector) gives the smallest perimeter.

Area of a parallelogram. Cut off the right triangle $A X D$ at one end (with $A X$ perpendicular to the base) and move it to the other end; it fits exactly over triangle $B Y C$ because $B Y = A X$ , both have a right angle, and $B C = A D$ (RHS congruence). The parallelogram becomes a rectangle, so
$ $area = base \times height .$ $
Any side may be the base provided its own perpendicular height is used. A rectangle and a parallelogram with the same side lengths do not have the same area: the rectangle has the greater area (its height is the whole side). Moving pieces to form a figure of equal area is called dissection; the Sulba-sutras describe dissections that convert a triangle, a rectangle or an isosceles triangle into one another.

Rhombus, trapezium, polygons and area units are on the next topic page of this chapter.

Area of Rectangles, Paths, Triangles and Parallelograms: formulas to remember

Rectangle: $A = l \times b$; Perimeter 2(l + b) does not determine A.
Triangle: $A = \frac{1}{2} b h$; h is the perpendicular height to the chosen base.
Parallelogram: $A = b h$; Any side as base with its own height.
Path outside / inside a rectangle: $(l + 2 w) (b + 2 w) - l b, l b - (l - 2 w) (b - 2 w)$; Path of uniform width w.
Cross-path: $A = l w_{1} + b w_{2} - w_{1} w_{2}$; Strips of widths w1 and w2; subtract the overlap.

How to solve Area of Rectangles, Paths, Triangles and Parallelograms questions

Routine for area questions.

1Sketch the figure and mark the base and the perpendicular height (not the slanting side).
2Choose the formula: rectangle l x b, triangle (1/2) b h, parallelogram b h.
3For paths and frames, compute outer area minus inner area, or add strips and subtract overlaps.
4For equal-area claims, compare bases and heights: same base and same height means same area.
5Use one base-height pair to find the area, then the area to find the height for another base.
6Check units (cm2, m2) and that the answer is sensible.

Area of Rectangles, Paths, Triangles and Parallelograms in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Area of a path

Suppose a rectangular school ground is 36 m long and 24 m wide, with a 3 m wide path laid inside along all four sides.

Find the area of the path.

Try it first, then show the solutionSolution

Ground area $= 36 \times 24 = 864$ m $^{2}$ .
The inner rectangle is $(36 - 6) \times (24 - 6) = 30 \times 18 = 540$ m $^{2}$ .
Path $= 864 - 540 = 324$ m $^{2}$ .

Answer: 324 square metres

Memory hook: Outer minus inner, and the width is taken off twice.

Same base, same area

Imagine a plot where several triangular flower beds all use the same 8 m wall as one side, and their opposite corners lie on a straight line parallel to the wall at a distance of 3 m.

What can you say about their areas, and which bed needs the least fencing?

Try it first, then show the solutionSolution

Each has base 8 m and height 3 m, so each has area $\frac{1}{2} \times 8 \times 3 = 12$ m $^{2}$ .
The perimeter is smallest when the apex is above the midpoint of the wall: sides $4^{2} + 3^{2} = 5$ each, perimeter $8 + 5 + 5 = 18$ m. An apex above one end of the wall gives $8 + 73 + 3 \approx 19.54$ m.

Answer: All have area 12 square metres; the symmetric bed has the least perimeter (18 m).

Memory hook: Height fixes the area; symmetry minimises the perimeter.

Tricks to remember Area of Rectangles, Paths, Triangles and Parallelograms

Short memory aids. Say each one aloud once, then use it on a question the same day.

Half of the rectangle
A triangle is half of the rectangle drawn on its base and height.
Use it when: Remembering (1/2) b h.
Slide the end triangle
To turn a parallelogram into a rectangle, move the right triangle at one end to the other end.
Use it when: Why area = b x h.
Parallel lines keep heights
Moving the apex along a line parallel to the base does not change the height or the area.
Use it when: Equal-area questions.
Outer minus inner
Path area is outer rectangle minus inner rectangle; subtract the width on both sides.
Use it when: Path questions.

Area of Rectangles, Paths, Triangles and Parallelograms: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Area of a triangle · 1 mark · MCQBoard level

The area of a triangle with base 12 cm and height 7 cm is: (a) 19 sq cm (b) 42 sq cm (c) 84 sq cm (d) 24 sq cm

Try it first, then show the solutionSolution

$\frac{1}{2} \times 12 \times 7 = 42$ .

Answer: (b) 42 sq cm

Example 2: Area of a path · 2 marks · Short answerBoard level

A rectangular park is 40 m by 30 m. A path 2 m wide runs inside along all four sides. Find the area of the path.

Try it first, then show the solutionSolution

Inner rectangle: $(40 - 2 \times 2) \times (30 - 2 \times 2) = 36 \times 26 = 936$ m $^{2}$ . (1 mark)
Path $= 40 \times 30 - 936 = 1200 - 936 = 264$ m $^{2}$ . (1 mark)

Answer: 264 square metres

Example 3: A cross-path · 3 marks · Short answerHarder board level

A lawn is 20 m long and 15 m wide. Two paths cross in the middle: one 2 m wide runs the length of the lawn, the other 3 m wide runs across its width. Find the area of the paths together.

Try it first, then show the solutionSolution

Path along the length: $20 \times 2 = 40$ m $^{2}$ . (1 mark)
Path across the width: $15 \times 3 = 45$ m $^{2}$ . (1 mark)
They overlap in a $2 \times 3 = 6$ m $^{2}$ rectangle, counted twice: total $= 40 + 45 - 6 = 79$ m $^{2}$ . (1 mark)

Answer: 79 square metres

Example 4: Different bases of a triangle · 3 marks · Short answerHarder board level

In triangle ABC, BC = 18 cm and the height from A to BC is 10 cm. Find its area. If AC = 15 cm is taken as base, find the corresponding height. If D is the midpoint of BC, find the area of triangle ABD.

Try it first, then show the solutionSolution

Area $= \frac{1}{2} \times 18 \times 10 = 90$ sq cm. (1 mark)
With base AC: $\frac{1}{2} \times 15 \times h = 90$ , so $h = 12$ cm. (1 mark)
A median halves the area: $90 \div 2 = 45$ sq cm. (1 mark)

Answer: Area 90 sq cm; height 12 cm; area of ABD is 45 sq cm

Example 5: Parallelogram versus rectangle · 4 marks · Long answerHarder board level

A parallelogram has base 14 cm, an adjacent side of 10 cm and a height of 8 cm on the 14 cm base. Find its area, the height on the 10 cm side, and compare with the area of a rectangle with the same side lengths.

Try it first, then show the solutionSolution

Area $= 14 \times 8 = 112$ sq cm. (1 mark)
With base 10: $10 \times h = 112$ , so $h = 11.2$ cm. (1 mark)
The rectangle with sides 14 and 10 has area $140$ sq cm. (1 mark)
$140 > 112$ : the rectangle has the greater area, because the parallelogram's height (8 cm) is less than its slanting side (10 cm). (1 mark)

Answer: 112 sq cm; height 11.2 cm; the rectangle (140 sq cm) is larger.

Where students lose marks in Area of Rectangles, Paths, Triangles and Parallelograms

The slips examiners see most often, and what to write instead.

Mistake: Using the slanting side instead of the perpendicular height
The height is the perpendicular distance, not the side.
Wrong working
"Parallelogram with base 14 and slanting side 10: area = 14 x 10."
Right working
Area $= 14 \times 8 = 112$ using the perpendicular height 8.
Do this: Mark the right angle on the height.
Usually costs: The whole answer.
Mistake: Subtracting the path width only once
A path along both sides of a dimension takes the width off twice.
Wrong working
"Inner rectangle (40 - 2) x (30 - 2)."
Right working
$(40 - 4) \times (30 - 4) = 36 \times 26$ .
Do this: Draw both sides of the path.
Usually costs: 2 marks.
Mistake: Forgetting the half in the triangle formula
The triangle is half of the rectangle.
Wrong working
"Area = 12 x 7 = 84."
Right working
$\frac{1}{2} \times 12 \times 7 = 42$ .
Do this: Write 1/2 first.
Usually costs: 1 mark.
Mistake: Counting the overlap of two paths twice
The crossing rectangle belongs to both strips.
Wrong working
"Cross-path = 40 + 45 = 85."
Right working
$40 + 45 - 6 = 79$ .
Do this: Subtract the common rectangle.
Usually costs: 1 mark.

Exam tips

Sketch the figure and label the height with a right-angle mark before using any formula.
For paths, write area = outer - inner and show both dimensions of the inner rectangle.
For equal-area claims, name the common base and the equal heights.
Give units with every answer (sq cm, sq m).
Use two base-height pairs to cross-check a triangle or parallelogram area.

Practice questions on Area of Rectangles, Paths, Triangles and Parallelograms

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · Assertion-reasonBoard level
Assertion (A): Two rectangles with the same perimeter can have different areas.
Reason (R): The rectangles 6 cm by 2 cm and 4 cm by 4 cm both have perimeter 16 cm, but their areas are 12 and 16 square cm.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Compute perimeter and area for both rectangles.
Answer
(a). $2 (6 + 2) = 16 = 2 (4 + 4)$ , while $6 \times 2 = 12$ and $4 \times 4 = 16$ . The example shows A.
Q2 · 2 marks · Very short answerBoard level
A triangular signboard has area 90 square cm and one side of 15 cm taken as its base. Find the height of the signboard on that base.
Hint
Use area = (1/2) base x height.
Answer
$\frac{1}{2} \times 15 \times h = 90$ , so $7.5 h = 90$ and $h = 12$ cm.
Q3 · 3 marks · Short answerHarder board level
A parallelogram has area 180 square cm. One side is 15 cm with the corresponding height 12 cm. Find the height corresponding to the adjacent side of 20 cm, and the area of the rectangle on sides 15 cm and 20 cm.
Hint
Area = base x height for each side.
Answer
$15 \times 12 = 180$ confirms the area. For the side of 20 cm: $20 \times h = 180$ , so $h = 9$ cm. The rectangle on sides 15 and 20 has area $15 \times 20 = 300$ sq cm, which is larger than 180.
Q4 · 4 marks · Long answerHOTS / competency
BC = 8 cm and a line l is parallel to BC at a distance of 3 cm. Triangles ABC are formed with A on l. (i) Show that all such triangles have the same area and find it. (ii) Find the perimeter when A is directly above the midpoint of BC. (iii) Find the perimeter when A is directly above C. (iv) Explain which perimeter is smaller.
Hint
Use the Baudhayana-Pythagoras theorem for the slanting sides.
Answer
(i) Each has base 8 and height 3, so area $= \frac{1}{2} \times 8 \times 3 = 12$ sq cm. (ii) Sides $4^{2} + 3^{2} = 5$ and 5: perimeter $= 8 + 5 + 5 = 18$ cm. (iii) With A above C: $A C = 3$ and $A B = 8^{2} + 3^{2} = 73 \approx 8.54$ : perimeter $\approx 8 + 8.54 + 3 = 19.54$ cm. (iv) The midpoint position gives the smaller perimeter: reflecting the apex in the mirror line shows that the shortest path B to A to C is a straight line through the line l at the point above the midpoint of BC.
Q5 · 4 marks · Case-basedHOTS / competency
Case study: a rectangular field is 50 m long and 30 m wide. A path 2 m wide is built around the outside of the field. Laying the path costs Rs 45 per square metre.
(i) Find the area of the path. (1 mark)
(ii) Find the cost. (1 mark)
(iii) If the path of the same width were instead inside the field, find its area. (1 mark)
(iv) Write formulas for the outside and inside path areas in terms of l, b and w. (1 mark)
Hint
Outside: (l + 2w)(b + 2w) - lb. Inside: lb - (l - 2w)(b - 2w).
Answer
(i) Outer rectangle $54 \times 34 = 1836$ ; path $= 1836 - 1500 = 336$ m $^{2}$ . (ii) $336 \times 45 = 15120$ rupees. (iii) Inner rectangle $46 \times 26 = 1196$ ; path $= 1500 - 1196 = 304$ m $^{2}$ . (iv) Outside: $(l + 2 w) (b + 2 w) - l b = 2 w (l + b) + 4 w^{2}$ (check $2 \cdot 2 \cdot 80 + 16 = 336$ ). Inside: $l b - (l - 2 w) (b - 2 w) = 2 w (l + b) - 4 w^{2}$ (check $320 - 16 = 304$ ).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What is the area of a triangle?Show answer
Half x base x height.
2.What is the area of a parallelogram?Show answer
Base x height.
3.What does a line from a vertex to the midpoint of the opposite side do to the triangle?Show answer
Splits it into two triangles of equal area.
4.Do rectangles with equal perimeters have equal areas?Show answer
No.
5.How do you find the area of a path?Show answer
Outer area minus inner area.
6.What is dissection?Show answer
Cutting a figure into pieces and rearranging them into a different figure of equal area.

Area of Rectangles, Paths, Triangles and Parallelograms: frequently asked questions

Why is the parallelogram area base x height and not base x side?

The slanting side is longer than the height, so base x side would be the area of a rectangle on the same sides, which is bigger; dissection shows the parallelogram equals a base x height rectangle.

Does the triangle formula hold for obtuse triangles?

Yes. The chapter shows the area is the difference of two triangles that can be enclosed in rectangles.

Which triangle on a given base and between two parallel lines has the least perimeter?

The isosceles one, with its apex above the midpoint of the base.

Can I use any side of a parallelogram as the base?

Yes, as long as you use the perpendicular height measured to that side.