Perimeter and Area: Formulas, Shortcuts and Solved Examples
This page has the perimeter and area formulas for every flat shape, unit conversions, shortcuts for how area changes when a side changes, and REET-level solved examples. REET 2024 had 5 questions from this topic, and all of them were straightforward formula-and-one-or-two-step calculations, so full marks here are genuinely within reach.
- 5 questions in REET 2024
- Class 3–8 level
- 7 min read
How REET asks this topic
REET 2024's 5 questions were all directly formula-based: working out a rectangle's perimeter and then its length from the cost of fencing, comparing the height of a parallelogram and a triangle on the same base with equal areas, the area of an equilateral triangle (√3/4 a²), and the area of a trapezium-shaped figure where one side had to be found using the Pythagoras theorem. In one question (reducing a square's side by a third), the correct answer, 4/9, wasn't among the options, so it was given as a bonus. In short: keep the formulas solid, and understand that when a side changes, area scales with the square of that ratio.
What perimeter and area mean
Perimeter is the total length of the boundary of a closed shape — like the length of fencing around a field. Its unit is a unit of length (cm, m). The perimeter of a circle is called its circumference.
Area is the measure of the flat region enclosed by a closed shape — like the number of tiles needed to cover a floor. Its unit is a square unit (cm², m²). Fencing, framing, edging = perimeter; painting, tiling, grass, carpet = area.
Shapes with the same perimeter can have different areas, and shapes with the same area can have different perimeters. Example: a 6 cm × 4 cm rectangle (perimeter 20, area 24) and a 5 cm square (perimeter 20, area 25). For a given perimeter, a square has the largest area among all rectangles.
All the key formulas in one place
| Shape | Perimeter | Area |
|---|---|---|
| Rectangle (length l, breadth b) | 2(l + b) | l × b; diagonal = √(l² + b²) |
| Square (side a) | 4a | a² = (diagonal)²/2; diagonal = a√2 |
| Triangle (base b, height h) | sum of the three sides | ½ × b × h |
| Equilateral triangle (side a) | 3a | (√3/4) a²; height = (√3/2) a |
| Right triangle | sum of the sides | ½ × base × perpendicular |
| Any triangle (sides a, b, c) | a + b + c = 2s | Heron's formula: √[s(s−a)(s−b)(s−c)] |
| Parallelogram | 2 × (sum of adjacent sides) | base × height |
| Rhombus (diagonals d₁, d₂) | 4 × side | ½ × d₁ × d₂ |
| Trapezium (parallel sides a, b; distance h) | sum of all four sides | ½ × (a + b) × h |
| Circle (radius r) | circumference = 2πr = πd | πr² |
| Semicircle | πr + 2r | ½ πr² |
π is taken as approximately 22/7 or 3.14; it's an irrational number, so 22/7 is only an approximation.
Unit conversion and shortcuts for changing dimensions
| Point | Rule |
|---|---|
| 1 m² | = 10,000 cm² (because 100 × 100) |
| 1 cm² | = 100 mm² |
| 1 are | = 100 m² |
| 1 hectare | = 10,000 m² |
| 1 km² | = 100 hectares = 10,00,000 m² |
| Every side scaled by k | Perimeter scales by k, area scales by k² (side doubled → area quadrupled) |
| Every side reduced by 1/3 (new side = 2/3) | New area = (2/3)² = 4/9 of the original; reduction = 5/9 |
| Radius doubled | Circumference doubles, area becomes four times |
| Same base, equal areas | Triangle's height = 2 × parallelogram's height (because ½bh₁ = bh₂) |
| Same base, between the same parallels | Area of a triangle = ½ × area of the parallelogram |
REET-level solved examples
Example 1 (cost of fencing): a rectangular plot's length is 10 m more than its breadth. Fencing at ₹15.50 per metre costs ₹1,550. Perimeter = 1550 ÷ 15.50 = 100 m. 2(l + b) = 100 ⇒ l + b = 50; l − b = 10 ⇒ l = 30 m, b = 20 m.
Example 2 (trapezium): PQ ∥ SR, PS ⊥ PQ, PQ = 11 cm, PS = 8 cm, QR = 10 cm, and the figure's area is 64 cm². Check: drop a perpendicular from Q to SR; with height 8 and slant side 10, the horizontal gap = √(10² − 8²) = 6 cm, so SR = 11 − 6 = 5 cm. Area = ½ × (11 + 5) × 8 = 64 cm².
Example 3 (Heron's formula): sides 13, 14, 15 cm. s = 21; area = √(21 × 8 × 7 × 6) = √7056 = 84 cm².
Example 4 (path): a 20 m × 15 m garden has a 2 m wide path running around the outside. Outer rectangle = 24 × 19 = 456 m²; path area = 456 − 300 = 156 m². If the path is inside instead, subtract the inner rectangle (l − 2w)(b − 2w).
Example 5 (circle): for a circle with a circumference of 44 cm, 2 × 22/7 × r = 44 ⇒ r = 7 cm; area = 22/7 × 49 = 154 cm².
How to teach children
Activity (Class 3–5): string and tile
Have children run a piece of string around the edge of a book or table and measure it — that's perimeter. Then lay down 1 cm square paper pieces (or squares on graph paper) and count them — that's area. The difference between 'boundary' and 'spread' becomes something they feel with their hands.
Activity (Class 6–8): cut-and-rearrange to derive formulas
Cut off the triangular piece from a paper parallelogram and stick it on the other side — it becomes a rectangle, so area = base × height. Join two congruent triangles to form a parallelogram, so a triangle = ½ × base × height. The formula emerges rather than being memorised.
Teaching aids
A geoboard and rubber bands, graph paper, square-centimetre tiles, string, dot paper; a project to measure the school playground or the classroom floor.
Common misconceptions
Mixing up the units for perimeter and area (writing area in 'cm'); assuming area always increases when perimeter increases; mistaking a slant side for the height; thinking 1 m² = 100 cm².
Assessment idea
'Make as many different rectangles as you can with a 24 sq cm area (whole-number sides), and write down their perimeters' — this checks understanding, reasoning and pattern recognition all at once.
Memory trick / exam tip
In questions where a side changes, don't rebuild the formula — use the ratio: new side/old side = k, so new area = k² × old area. 'Reduced by a third' means k = 2/3 (area 4/9), 'a third of' means k = 1/3 (area 1/9) — read the wording carefully; REET 2024's bonus question turned on exactly this distinction.
Questions asked in REET 2024 (with solutions)
Question 103Bonus — dropped in the final key
If each side of the square is reduced to one-third, then the area will be
- A1/3
- B1/2
- C1/9
- D1/6
Explanation
BSER ने अंतिम उत्तर कुंजी में इस प्रश्न को BONUS (निरस्त) कर दिया है। कारण यह प्रतीत होता है कि अंग्रेज़ी और हिंदी रूप में अर्थ अलग है: अंग्रेज़ी 'reduced to one-third' के अनुसार भुजा a/3 होने पर क्षेत्रफल 1/9 गुना हो जाता है (विकल्प C), जबकि हिंदी 'एक तिहाई कम कर दी जाए' के अनुसार भुजा 2a/3 होने पर क्षेत्रफल 4/9 गुना होता है, जो किसी विकल्प में नहीं है। साथ ही प्रश्न में 'मूल क्षेत्रफल का कितना भाग' स्पष्ट नहीं लिखा गया है।
Question 106
The length of a rectangular plot is 10 meters more than its breadth. If the cost of fencing around the plot is ₹ 1,550 at the rate of ₹ 15.50 per meter, then the length of the plot will be
- A20 मीटर
- B30 मीटर
- C40 मीटर
- D10 मीटर
Explanation
परिमाप = कुल खर्च ÷ दर = 1550 ÷ 15.50 = 100 मीटर। यदि चौड़ाई b है तो लम्बाई b + 10, अतः 2(b + 10 + b) = 100 ⇒ 2b + 10 = 50 ⇒ b = 20 मीटर। इसलिए लम्बाई = 30 मीटर; विकल्प (A) 20 मीटर चौड़ाई है।
Question 96
A parallelogram and a triangle are equal in area and lie on the same base. If the height of the parallelogram is 2 cm, then the height of the triangle is :
- A1 सेमी
- B2 सेमी
- C3 सेमी
- D4 सेमी
Explanation
समांतर चतुर्भुज का क्षेत्रफल = आधार × ऊँचाई = b × 2, और त्रिभुज का क्षेत्रफल = ½ × b × h। दोनों बराबर हैं: 2b = ½ bh, अतः h = 4 सेमी। समान आधार पर समान क्षेत्रफल के लिए त्रिभुज की ऊँचाई समांतर चतुर्भुज की दोगुनी होनी चाहिए।
Question 99
The area of an equilateral triangle of side 3 cm will be :
- A9√3/4 सेमी²
- B6√2/4 सेमी²
- C3√3/4 सेमी²
- D6√2/5 सेमी²
Explanation
समबाहु त्रिभुज का क्षेत्रफल = (√3/4) × a²। a = 3 सेमी रखने पर क्षेत्रफल = (√3/4) × 9 = 9√3/4 सेमी²। विकल्प (C) 3√3/4 में भुजा का वर्ग न लेने की गलती है।
Question 98
In given figure PQ||SR and PS⊥PQ. If PQ = 11 cm, PS = 8 cm and QR = 10 cm then area of figure PQRS will be
Figure: Trapezium PQRS with PQ=11 cm (bottom), PS=8 cm (left, perpendicular to PQ), QR=10 cm (slanted right side), SR parallel to PQ. Diagram added and verified: the horizontal offset works out to 6 cm, giving SR=5 cm and an area of 64 square cm, matching the stored answer.
- A32 सेमी²
- B64 सेमी²
- C72 सेमी²
- D44 सेमी²
Explanation
PQRS एक समलम्ब (trapezium) है जिसकी ऊँचाई PS = 8 सेमी है। R से PQ पर लम्ब डालें तो पाइथागोरस से क्षैतिज दूरी = √(10² − 8²) = 6 सेमी, अतः SR = 11 − 6 = 5 सेमी। क्षेत्रफल = ½ × (PQ + SR) × ऊँचाई = ½ × (11 + 5) × 8 = 64 सेमी²।
Practice questions
Question 1
A rectangle has length 12 cm and breadth 7 cm. Its perimeter is
- A19 सेमी
- B38 सेमी
- C84 सेमी
- D42 सेमी
Explanation
Perimeter = 2(l + b) = 2(12 + 7) = 38 cm. 84 cm² would be its area, not its perimeter.
Question 2
The diagonal of a square is 10 cm. Its area is
- A100 सेमी²
- B50 सेमी²
- C25 सेमी²
- D40 सेमी²
Explanation
A square's area = (diagonal)²/2 = 100/2 = 50 cm².
Question 3
A circle has radius 7 cm. Its circumference is (π = 22/7)
- A22 सेमी
- B44 सेमी
- C154 सेमी
- D49 सेमी
Explanation
Circumference = 2πr = 2 × 22/7 × 7 = 44 cm. 154 cm² is its area.
Question 4
If the radius of a circle is tripled, its area becomes
- A3 गुना
- B6 गुना
- C9 गुना
- D27 गुना
Explanation
Area scales with the square of the radius ratio: 3² = 9 times. The circumference would only become 3 times as large.
Question 5
The diagonals of a rhombus are 16 cm and 12 cm. Its area is
- A192 सेमी²
- B96 सेमी²
- C48 सेमी²
- D56 सेमी²
Explanation
A rhombus's area = ½ × d₁ × d₂ = ½ × 16 × 12 = 96 cm².
Question 6
The area of a triangle with sides 5 cm, 12 cm and 13 cm is
- A30 सेमी²
- B60 सेमी²
- C65 सेमी²
- D32.5 सेमी²
Explanation
5² + 12² = 13², so this is a right triangle. Area = ½ × 5 × 12 = 30 cm² (Heron's formula also gives this: s = 15, √(15×10×3×2) = 30).
Question 7
A 1 m wide path runs inside along all sides of a 30 m × 20 m rectangular field. The area of the path is
- A96 मी²
- B100 मी²
- C104 मी²
- D196 मी²
Explanation
Inner rectangle = (30 − 2)(20 − 2) = 28 × 18 = 504 m². Path area = 600 − 504 = 96 m².
Question 8
5 hectares is equal to
- A5,000 मी²
- B50,000 मी²
- C500 मी²
- D5,00,000 मी²
Explanation
1 hectare = 10,000 m², so 5 hectares = 50,000 m².
Question 9
A Class 6 student says, "A rectangle with a greater perimeter will always have a greater area." The best way to correct this misconception is
- Aसूत्र दोबारा लिखवाना
- Bग्राफ़ पेपर पर 1×12 और 4×4 जैसे आयत बनवाकर परिमाप और क्षेत्रफल की तुलना करवाना
- Cविद्यार्थी को गलत कहकर आगे बढ़ना
- Dकेवल पाठ्यपुस्तक के प्रश्न हल करवाना
Explanation
A 1×12 rectangle has perimeter 26 and area 12, while a 4×4 square has perimeter 16 and area 16. Building these and comparing them lets the child see the counter-evidence for themselves and revise their own belief.
Question 10
A triangle and a parallelogram have the same base and equal areas. If the triangle's height is 12 cm, the parallelogram's height is
- A24 सेमी
- B12 सेमी
- C6 सेमी
- D3 सेमी
Explanation
½ × b × 12 = b × h ⇒ h = 6 cm. On the same base with equal area, a triangle's height is always twice the parallelogram's height.
Frequently asked questions
What's the difference between perimeter and area?+
Perimeter is the length of a shape's boundary (unit: cm, m), while area is the measure of the surface enclosed by the shape (unit: cm², m²).
How do you find the area of an equilateral triangle?+
The formula is (√3/4) × a². For a side of 3 cm, the area = 9√3/4 cm² — this exact question came up in REET 2024.
What is the area formula for a trapezium?+
½ × (sum of the parallel sides) × the perpendicular distance between them.
How many square metres are in 1 hectare?+
1 hectare = 10,000 square metres, and 1 are = 100 square metres.
How many questions come from perimeter and area in REET?+
REET 2024's papers had 5 questions from this topic (with one given as a bonus). It's a regular topic in both Level 1 and Level 2.
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Official Sources
Change Log
- 17 September 2026 — English page published — mirrors the fact-checked Hindi page, independently bilingual-accuracy checked