Heron's Formula — Practice

Question 1

Find the area of a triangle with sides 13 cm, 14 cm, 15 cm.

Accepted Answer

Step 1 — Find s.
  s = (13 + 14 + 15)/2 = 42/2 = 21 cm.

Step 2 — Find s−a, s−b, s−c.
  s − a = 21 − 13 = 8.
  s − b = 21 − 14 = 7.
  s − c = 21 − 15 = 6.

Step 3 — Apply Heron's Formula.
  Area = √(21 × 8 × 7 × 6)
       = √7056.

Step 4 — Simplify (look for perfect-square factors).
  21 × 8 × 7 × 6 = (21 × 6)(7 × 8) = 126 × 56 — alternative: (3·7) × (2³) × 7 × (2·3) = 2⁴ × 3² × 7² = (4·3·7)² = 84².
  So √7056 = 84.

✦ Answer: Area = 84 cm².

Question 2

Find the area of a triangle with sides 6 cm, 8 cm, 10 cm.

Accepted Answer

Step 1 — Recognise the Pythagorean triple.
  6² + 8² = 36 + 64 = 100 = 10². ✓ So this is a RIGHT triangle, right-angled at the corner between sides 6 and 8.

Step 2 — Use the shortcut.
  Area = (1/2) × leg × leg = (1/2) × 6 × 8 = 24 cm².

Step 3 — Verify with Heron's (optional).
  s = (6 + 8 + 10)/2 = 12. Area = √(12 × 6 × 4 × 2) = √576 = 24. ✓

✦ Answer: Area = 24 cm².

Lesson: always check for Pythagorean triples first — much faster than Heron.

Question 3

Find the area of an equilateral triangle with side 8 cm.

Accepted Answer

Step 1 — Use the equilateral shortcut.
  Area = (√3 / 4) × a², where a = side.

Step 2 — Substitute a = 8.
  Area = (√3 / 4) × 64 = 16√3 cm² ≈ 27.71 cm².

✦ Answer: 16√3 cm² (≈ 27.71 cm²).

Question 4

A triangle has sides 11 m, 15 m, 20 m. Find the semi-perimeter.

Accepted Answer

Step 1 — Sum the sides.
  11 + 15 + 20 = 46 m.

Step 2 — Half it.
  s = 23 m.

✦ Answer: s = 23 m.

Question 5

An isosceles triangle has equal sides 5 cm and base 6 cm. Find its area using Heron's formula.

Accepted Answer

Step 1 — Compute s.
  s = (5 + 5 + 6)/2 = 8 cm.

Step 2 — Find s−a, s−b, s−c.
  s − 5 = 3, s − 5 = 3, s − 6 = 2.

Step 3 — Apply formula.
  Area = √(8 × 3 × 3 × 2) = √144 = 12 cm².

Step 4 — Verify by altitude method.
  Drop perpendicular from apex to base midpoint. Half-base = 3. Height² = 5² − 3² = 16 → height = 4.
  Area = (1/2)(6)(4) = 12 cm². ✓

✦ Answer: 12 cm².

Question 6

A triangular field has sides 50 m, 78 m and 112 m. Find its area.

Accepted Answer

Step 1 — Compute s.
  s = (50 + 78 + 112)/2 = 120 m.

Step 2 — Compute factors.
  s − 50 = 70.
  s − 78 = 42.
  s − 112 = 8.

Step 3 — Apply Heron.
  Area = √(120 × 70 × 42 × 8).
  = √(120 × 70 × 42 × 8).
  Factor: 120 = 8·15, 70 = 10·7, 42 = 6·7, 8 = 8.
  Product = 8·15 × 10·7 × 6·7 × 8 = 8² × 15 × 10 × 6 × 7².
  = 64 × 49 × (15·10·6)
  = 64 × 49 × 900
  = 64 × 49 × 900.
  √ that = 8 × 7 × 30 = 1680.

✦ Answer: Area = 1680 m².

Question 7

An equilateral triangle has area 49√3 cm². Find its side and perimeter.

Accepted Answer

Step 1 — Use the equilateral formula.
  Area = (√3/4) × a².
  49√3 = (√3/4) × a².

Step 2 — Solve for a².
  Divide both sides by √3: 49 = a²/4 → a² = 196 → a = 14 cm.

Step 3 — Compute perimeter.
  P = 3a = 42 cm.

✦ Answer: Side = 14 cm, Perimeter = 42 cm.

Question 8

A triangle has sides in the ratio 5 : 12 : 13 and perimeter 60 cm. Find its area.

Accepted Answer

Step 1 — Find the actual sides.
  Let sides = 5x, 12x, 13x. Sum = 30x = 60 → x = 2.
  Sides: 10, 24, 26.

Step 2 — Recognise the Pythagorean triple.
  10² + 24² = 100 + 576 = 676 = 26². ✓ Right triangle.

Step 3 — Use the shortcut.
  Area = (1/2) × 10 × 24 = 120 cm².

✦ Answer: 120 cm².

Via Heron: s = 30, factors 20, 6, 4. Area = √(30·20·6·4) = √14400 = 120. ✓

Question 9

Find the area of quadrilateral ABCD where AB = 9, BC = 12, CD = 5, DA = 8 cm and diagonal AC = 15 cm.

Accepted Answer

Step 1 — Divide into two triangles using diagonal AC.

Step 2 — Triangle ABC (sides 9, 12, 15).
  Check: 9² + 12² = 81 + 144 = 225 = 15². Right triangle!
  Area₁ = (1/2)(9)(12) = 54 cm².

Step 3 — Triangle ACD (sides 15, 5, 8).
  Wait — these don't form a valid triangle: 5 + 8 = 13 < 15. Triangle inequality fails.

Step 4 — Re-read: the problem gives AC = 15 and AD = 8, CD = 5. Indeed 5 + 8 = 13 < 15. So no such quadrilateral exists with these measurements.

(In a real exam, the figures would satisfy the triangle inequality. Let's assume CD = 14 instead — then ACD has sides 15, 8, 14; valid.)

Step 5 — Heron for △ACD with sides 15, 14, 8 (revised).
  s = (15 + 14 + 8)/2 = 18.5. Hmm non-integer — likely the original problem intended different numbers.

✦ Answer: As stated, the quadrilateral doesn't exist; the problem has a typo. The TECHNIQUE is correct — split by diagonal, find each triangle by Heron, sum.

Lesson: always check triangle inequality first when given any triangle's sides.

Question 10

A rhombus has perimeter 40 cm and one diagonal 16 cm. Find the area using Heron's formula on one half.

Accepted Answer

Step 1 — Find the side of the rhombus.
  Perimeter = 4 × side = 40 → side = 10 cm.

Step 2 — A diagonal splits the rhombus into 2 congruent triangles.
  Each triangle has sides 10, 10, 16.

Step 3 — Apply Heron to one triangle.
  s = (10 + 10 + 16)/2 = 18.
  s − 10 = 8, s − 10 = 8, s − 16 = 2.
  Area = √(18 × 8 × 8 × 2) = √(2304) = 48 cm².

Step 4 — Total rhombus area = 2 × 48 = 96 cm².

✦ Answer: 96 cm².

Cross-check: in a rhombus the other diagonal can be found from the side-diagonal relationship. Half-diag = √(side² − (half of 16)²) = √(100 − 64) = 6 → full = 12.
Rhombus area = (1/2) × d1 × d2 = (1/2)(16)(12) = 96. ✓ Both methods agree.

Practice — Heron's Formula

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