Area of a parallelogram
Cut a triangle off one slanted side of a parallelogram and move it to the other side: you get a rectangle. So the area is like a rectangle: base × height.
Height is the perpendicular distance to the base, not the slanted side.
In the figure: Base 12, perpendicular height 5 (the slant side 13 is not used): area = 12 × 5 = 60.
Formula: Area = base × height
Area of a parallelogram
In the app you can move this figure with your finger.
Open in the app- Easy: Base 12 cm, height 5 cm: area = 12 × 5 = 60 sq cm.
- Medium: Area 84 sq cm, base 12 cm. Height = 84 ÷ 12 = 7 cm.
- Exam-type: Area 60. With base 10 the height is 6. With the other base 8 the height is 60 ÷ 8 = 7.5.
A parallelogram has the same area as a rectangle with the same base and height.
Do not use the slanted side; use only the perpendicular height.
If the height is missing, find it as area ÷ base. Memory clue: Base × height is "across × up"; the slanting side is never the height.
Spot the difference
| Point | Rectangle | Parallelogram |
|---|---|---|
| Area | length × breadth | base × height |
| Slant side | none | not used in area |
| Height | the breadth itself | perpendicular on the base |
| Rearranging | stays a rectangle | cut a triangle and move it to get a rectangle |
Practice questions
- 1A parallelogram has base 8 cm and height 5 cm. Area?
- 2Area 72 sq cm, base 9 cm. Height?
- 3Base 15 cm, height 4 cm. Area?
- 4Area 96 sq cm, height 8 cm. Base?
- 5A parallelogram-shaped field has base 20 m and height 12 m. Grass costs ₹30 per sq m. Total cost?
- 6Area 120 sq cm. Find the height if the base is 15 cm, and if the base is 12 cm.
- 7Adjacent sides of a parallelogram are 12 cm and 9 cm. The height on the 12 cm side is 6 cm. Find the height on the 9 cm side.
- 8A parallelogram has base 14 cm and area 98 sq cm. Find the height; then find the area if the same base and double the height are taken.
Show answers (try them yourself first)
- 18 × 5 = 40 sq cm.
- 272 ÷ 9 = 8 cm.
- 315 × 4 = 60 sq cm.
- 496 ÷ 8 = 12 cm.
- 5Area = 20 × 12 = 240 sq m; cost = 240 × 30 = ₹7200.
- 6120 ÷ 15 = 8 cm; 120 ÷ 12 = 10 cm.
- 7Area = 12 × 6 = 72; height on the 9 cm side = 72 ÷ 9 = 8 cm.
- 8Height = 98 ÷ 14 = 7 cm; new area = 14 × 14 = 196 sq cm (area doubles).
Area of a rhombus
A rhombus kite has two sticks (diagonals) crossing at a right angle, making four equal right triangles.
The diagonals of a rhombus bisect each other at right angles. Adding the four right triangles gives half of the rectangle formed by the two diagonals.
In the figure: Diagonals 12 and 8 cross at a right angle: area = ½ × 12 × 8 = 48.
Formula: Area = ½ × first diagonal × second diagonal
Area of a rhombus
In the app you can move this figure with your finger.
Open in the app- Easy: Diagonals 10 and 24: area = ½ × 10 × 24 = 120. Side = √(5² + 12²) = √169 = 13 ; perimeter 52.
- Medium: Area 90, one diagonal 15. The other = 90 × 2 ÷ 15 = 12.
- Exam-type: Rhombus side 10, one diagonal 16: half-diagonal 8; other half-diagonal = √(100 − 64) = 6; other diagonal 12; area = ½ × 16 × 12 = 96.
The diagonals of a rhombus bisect each other at a right angle.
Forgetting the ½ and multiplying only the diagonals.
Given a side and one diagonal, use the half-diagonal and a right triangle to find the other diagonal. Memory clue: Rhombus = diagonal × diagonal ÷ 2: "diagonal-diagonal-half".
Spot the difference
| Point | Parallelogram | Rhombus |
|---|---|---|
| Needed | base, height | two diagonals |
| Formula | base × height | ½ × d₁ × d₂ |
| Diagonals | bisect each other | bisect at right angle |
| Example | 12 × 5 = 60 | ½ × 12 × 8 = 48 |
Practice questions
- 1Diagonals of a rhombus are 6 cm and 4 cm. Area?
- 2Diagonals 14 and 10. Area?
- 3Area 48, one diagonal 12. Other diagonal?
- 4The sticks of a kite are 20 cm and 15 cm. Area of the paper?
- 5Area 84, one diagonal 16. Other diagonal?
- 6A rhombus has side 5 cm and one diagonal 8 cm. Find the other diagonal and the area.
- 7A rhombus has side 25 cm and one diagonal 14 cm. Area?
- 8A rhombus has perimeter 68 cm and one diagonal 16 cm. Area?
Show answers (try them yourself first)
- 1½ × 6 × 4 = 12 sq cm.
- 2½ × 14 × 10 = 70.
- 348 × 2 ÷ 12 = 8.
- 4½ × 20 × 15 = 150 sq cm.
- 584 × 2 ÷ 16 = 10.5.
- 6Half-diagonal 4; other half-diagonal = √(25 − 16) = 3; other diagonal 6; area = ½ × 8 × 6 = 24.
- 7Half-diagonal 7; other half-diagonal = √(625 − 49) = √576 = 24; other diagonal 48; area = ½ × 14 × 48 = 336.
- 8Side = 68 ÷ 4 = 17; half-diagonal 8; other half-diagonal = √(289 − 64) = √225 = 15; other diagonal 30; area = ½ × 16 × 30 = 240.
Area of a trapezium
A field has two parallel sides, 9 m and 5 m, 4 m apart. We want its area.
Two equal trapeziums joined upside down form a parallelogram with base (a + b) and height h. So a trapezium is half of it.
In the figure: Parallel sides 9 and 14, height 6: area = ½ × (9 + 14) × 6 = 69.
Formula: Area = ½ × (sum of parallel sides) × height
Area of a trapezium
In the app you can move this figure with your finger.
Open in the app- Easy: Parallel sides 9 and 5, height 4: ½ × (9 + 5) × 4 = ½ × 14 × 4 = 28.
- Medium: Parallel sides 12 and 8, area 80. Height = 80 × 2 ÷ (12 + 8) = 160 ÷ 20 = 8.
- Exam-type: Garden: parallel sides 30 m and 20 m, 16 m apart. Area = ½ × 50 × 16 = 400 sq m.
Formulas
| Shape | Formula | What you need |
|---|---|---|
| Parallelogram | base × height | base, perpendicular height |
| Rhombus | ½ × d₁ × d₂ | two diagonals |
| Trapezium | ½ × (a + b) × h | two parallel sides, height |
A trapezium has only one pair of parallel sides: half their sum × height.
In a trapezium the height is the distance between the parallel sides, not the slanting side.
Take the average of the parallel sides, (a + b) ÷ 2, and multiply by the height. Memory clue: Trapezium: "average side × height"; the two parallel sides are its bases.
Spot the difference
| Point | Parallelogram | Trapezium |
|---|---|---|
| Parallel sides | two pairs | one pair |
| Formula | base × height | ½ × (a + b) × h |
| Example | 12 × 5 = 60 | ½ × (9 + 5) × 4 = 28 |
| Height | perpendicular on the base | perpendicular between the parallel sides |
Practice questions
- 1A trapezium has parallel sides 10 and 6 and height 5. Area?
- 2Parallel sides 18 and 12, height 8. Area?
- 3Parallel sides 7 and 5, height 3. Area?
- 4A trapezium has area 90 and parallel sides 11 and 7. Height?
- 5Area 150, height 10, one parallel side 18. The other parallel side?
- 6A plot has parallel sides 40 m and 24 m, 15 m apart. Turf costs ₹20 per sq m. Total cost?
- 7An isosceles trapezium has parallel sides 13 cm and 7 cm and each slanting side 5 cm. Find the height and the area.
- 8A right triangle with legs 3 and 4 is attached to one end of a 6 × 4 rectangle, forming a trapezium. Find the area in two ways.
Show answers (try them yourself first)
- 1½ × (10 + 6) × 5 = 40.
- 2½ × (18 + 12) × 8 = 120.
- 3½ × (7 + 5) × 3 = 18.
- 490 × 2 ÷ (11 + 7) = 180 ÷ 18 = 10.
- 5Sum of parallel sides = 150 × 2 ÷ 10 = 30; other side = 30 − 18 = 12.
- 6Area = ½ × 64 × 15 = 480 sq m; cost = 480 × 20 = ₹9600.
- 7(13 − 7) ÷ 2 = 3; height = √(25 − 9) = 4; area = ½ × (13 + 7) × 4 = 40.
- 8Rectangle + triangle = 24 + ½ × 3 × 4 = 24 + 6 = 30; formula: parallel sides 6 and 9, height 4: ½ × 15 × 4 = 30.
Area of a triangle (Heron’s formula)
A triangular plot has known sides but no known height. How do we find its area? Use Heron’s formula.
First find the semi-perimeter s = (sum of three sides) ÷ 2. Then multiply s by (s − each side) three times and take the square root.
In the figure: Sides 5, 12, 13: s = (5 + 12 + 13) ÷ 2 = 15; area = √(15 × 10 × 3 × 2) = 30.
Formula: Area = √[ s × (s − a) × (s − b) × (s − c) ] (s = (a + b + c) ÷ 2)
Area of a triangle (Heron’s formula)
In the app you can move this figure with your finger.
Open in the app- Easy: Sides 5, 12, 13: s = 30 ÷ 2 = 15 ; 15 × 10 × 3 × 2 = 900 ; √900 = 30. (Check: right triangle ½ × 5 × 12 = 30 ✔)
- Medium: Sides 13, 14, 15: s = 42 ÷ 2 = 21 ; 21 × 8 × 7 × 6 = 7056 ; √7056 = 84.
- Exam-type: Sides 8, 15, 17: s = 40 ÷ 2 = 20; 20 × 12 × 5 × 3 = 3600; √3600 = 60. (Check: right triangle ½ × 8 × 15 = 60 ✔)
s is the semi-perimeter, not the full perimeter.
Using the full perimeter instead of s. s = perimeter ÷ 2.
Sides like 5, 12, 13 or 8, 15, 17 form a right triangle; just use ½ × the two shorter sides. Memory clue: Heron: first s = half the perimeter, then s − a, s − b, s − c; multiply these four numbers and take the square root.
Spot the difference
| Point | Base-height formula | Heron’s formula |
|---|---|---|
| Needed | base and height | three sides |
| Formula | ½ × base × height | √[s(s − a)(s − b)(s − c)] |
| Example | ½ × 12 × 5 = 30 | 5, 12, 13 ⇒ 30 |
| When to use | the height is known | the height is not known |
Practice questions
- 1Area of the triangle with sides 3, 4, 5?
- 2Sides 6, 8, 10. Area?
- 3Sides 10, 10, 12. Area?
- 4Sides 9, 12, 15. Area?
- 5Sides 17, 25, 28. Area?
- 6An isosceles triangle has sides 13, 13, 10. Area?
- 7A triangular plot has sides 15 m, 20 m, 25 m. Levelling costs ₹10 per sq m. Total cost?
- 8Find the area of the triangle with sides 20, 21, 29 and the height on the side 21.
Show answers (try them yourself first)
- 1s = 6; 6 × 3 × 2 × 1 = 36; √36 = 6.
- 2s = 12; 12 × 6 × 4 × 2 = 576; √576 = 24.
- 3s = 16; 16 × 6 × 6 × 4 = 2304; √2304 = 48.
- 4s = 18; 18 × 9 × 6 × 3 = 2916; √2916 = 54.
- 5s = 35; 35 × 18 × 10 × 7 = 44100; √44100 = 210.
- 6s = 18; 18 × 5 × 5 × 8 = 3600; √3600 = 60.
- 7s = 30; 30 × 15 × 10 × 5 = 22500; area = 150 sq m; cost = 150 × 10 = ₹1500.
- 8s = 35; 35 × 15 × 14 × 6 = 44100; area = 210; height = 2 × 210 ÷ 21 = 20.
Area of a circle
Cut a round pizza into thin slices and place them head to tail: you get almost a rectangle of length = half circumference (πr) and width = radius (r).
Area of a circle = π × r × r. We take π = 22/7 (or 3.14).
In the figure: A circle of radius 7 unrolled into a rectangle: length 22 (half the circumference), width 7: 22 × 7 = 154.
Formula: Area = π r² (π = 22/7)
Area of a circle
In the app you can move this figure with your finger.
Open in the app- Easy: r = 7 cm: 22/7 × 7 × 7 = 154 sq cm.
- Medium: Area 616: r² = 616 × 7 ÷ 22 = 196 ; r = 14.
- Exam-type: Diameter 21 cm: r = 10.5 ; area = 22/7 × 10.5 × 10.5 = 346.5 sq cm.
Irregular shapes: On squared paper count full squares; count squares that are half or more covered as 1; ignore those covered less than half. The total gives the approximate area.
π = 22/7: convenient when the radius is a multiple of 7.
The area has r × r, not 2 × r. 2πr is the circumference.
If the diameter is given, halve it first to get the radius. Memory clue: Circle: r appears twice in πr²: "radius × radius × π".
Spot the difference
| Point | Circumference | Area |
|---|---|---|
| Formula | 2πr | πr² |
| Unit | cm | sq cm |
| For r = 7 | 44 cm | 154 sq cm |
| If r doubles | circumference doubles | area becomes four times |
Practice questions
- 1A parallelogram has base 9 cm and height 6 cm. Area?
- 2Rhombus with diagonals 10 and 8. Area?
- 3Area of a circle with r = 14 cm?
- 4A trapezium has parallel sides 14 and 10, height 6. Area?
- 5Area of a triangle with sides 9, 10, 17?
- 6A round ground has area 2464 sq m. Find its radius.
- 7A rhombus has side 13 cm and one diagonal 24 cm. Area?
- 8A trapezium-shaped field has parallel sides 20 m and 12 m, 9 m apart. Ploughing costs ₹50 per sq m. Total cost?
Show answers (try them yourself first)
- 19 × 6 = 54 sq cm.
- 2½ × 10 × 8 = 40.
- 322/7 × 14 × 14 = 616 sq cm.
- 4½ × (14 + 10) × 6 = 72.
- 5s = 36 ÷ 2 = 18 ; 18 × 9 × 8 × 1 = 1296 ; √1296 = 36.
- 6r² = 2464 × 7 ÷ 22 = 784 ; r = 28 m.
- 7Half-diagonal 12 ; other half-diagonal = √(13² − 12²) = √25 = 5 ; other diagonal 10 ; area = ½ × 24 × 10 = 120.
- 8Area = ½ × (20 + 12) × 9 = 144 sq m ; cost = 144 × 50 = ₹7200.