Solved Examples

NCERT Solutions for Class 9 Maths Chapter 10: Heron's Formula — Free PDF

Complete solutions — finding areas of triangles when all three sides are known, and applying Heron's formula to quadrilaterals.

CBSEClass 9
The SparkEd Authors (IITian & Googler)15 March 202635 min read
NCERT Solutions Class 9 Maths Chapter 10 Herons Formula — SparkEd

Chapter Overview: Heron's Formula

Chapter 10 of the NCERT Class 9 Maths textbook introduces a remarkable formula that lets you calculate the area of any triangle when you know all three side lengths, without needing to find the height. This formula, named after Heron of Alexandria (a Greek mathematician who lived around 10-70 AD), is one of the most elegant results in elementary geometry.

The beauty of Heron's formula lies in its universality. While the familiar formula Area=12×base×height\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} requires you to know (or calculate) the perpendicular height, Heron's formula works directly with the three side lengths. This makes it especially valuable for scalene triangles, where computing the height can be complicated.

The chapter contains two exercises. Exercise 10.1 focuses on finding areas of triangles using Heron's formula, while Exercise 10.2 extends the technique to quadrilaterals by dividing them into triangles using a diagonal. Together, these exercises carry significant weight in CBSE board exams, typically contributing 3-5 marks. Mastering this chapter also strengthens your skills in simplifying square roots and working with large numerical computations, both of which are useful across many areas of mathematics.

Key Concepts and Definitions

Before solving problems, let us review the essential definitions and formulas from this chapter.

Semi-perimeter and Heron's Formula

Semi-perimeter: For a triangle with sides aa, bb, cc, the semi-perimeter is defined as:

s=a+b+c2s = \dfrac{a + b + c}{2}

It is exactly half the perimeter. Always compute ss as a separate step before substituting into Heron's formula — this earns marks and prevents errors.

Heron's Formula:

Area=s(s−a)(s−b)(s−c)\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}

This formula works for ALL types of triangles — scalene, isosceles, equilateral, right-angled, acute, or obtuse.

Special Cases from Heron's Formula

Equilateral triangle with side aa: s=3a2s = \dfrac{3a}{2}, and the area simplifies to:

Area=34a2\text{Area} = \dfrac{\sqrt{3}}{4}a^2

Isosceles triangle with equal sides aa and base bb: s=2a+b2s = \dfrac{2a+b}{2}, and the area simplifies to:

Area=b44a2−b2\text{Area} = \dfrac{b}{4}\sqrt{4a^2 - b^2}

Right triangle with legs aa, bb: Area =12ab= \dfrac{1}{2}ab (simpler than Heron's, but Heron's gives the same result).

Triangle Inequality and Pythagorean Triplets

Triangle Inequality: Before applying Heron's formula, verify that the three sides can form a valid triangle. The sum of any two sides must exceed the third:

a+b>c,b+c>a,a+c>ba + b > c, \quad b + c > a, \quad a + c > b

If any of these conditions fails, no triangle exists and Heron's formula will give a negative value inside the square root.

Important Pythagorean triplets (useful for verification):
- (3,4,5)(3, 4, 5) and multiples: (6,8,10)(6, 8, 10), (9,12,15)(9, 12, 15), (12,16,20)(12, 16, 20)
- (5,12,13)(5, 12, 13) and multiples: (10,24,26)(10, 24, 26)
- (8,15,17)(8, 15, 17)
- (7,24,25)(7, 24, 25)

Area of Quadrilaterals

Area of a quadrilateral: Divide it into two triangles using a diagonal. Find each area using Heron's formula and add them. If the diagonal length is not given, you may need to use the Pythagorean theorem or other geometric properties to find it.

Step-by-step method for Heron's formula:
1. Identify the three sides aa, bb, cc.
2. Compute the semi-perimeter: s=a+b+c2s = \dfrac{a+b+c}{2}.
3. Compute each factor: (s−a)(s-a), (s−b)(s-b), (s−c)(s-c).
4. Multiply: s×(s−a)×(s−b)×(s−c)s \times (s-a) \times (s-b) \times (s-c).
5. Take the square root of the product.
6. Simplify and include units.

Exercise 10.1 — Area of Triangles (Solved)

Exercise 10.1 contains problems on finding the area of triangles using Heron's formula. Let us solve each problem with detailed working.

Problem 1: Equilateral Triangle Signal Board

Problem: A traffic signal board, indicating 'SCHOOL AHEAD', is an equilateral triangle with side aa. Find the area of the signal board using Heron's formula.

Solution:
All sides equal: a=b=c=aa = b = c = a

s=a+a+a2=3a2s = \dfrac{a+a+a}{2} = \dfrac{3a}{2}

Area=s(s−a)(s−a)(s−a)\text{Area} = \sqrt{s(s-a)(s-a)(s-a)}

=3a2⋅a2⋅a2⋅a2= \sqrt{\dfrac{3a}{2} \cdot \dfrac{a}{2} \cdot \dfrac{a}{2} \cdot \dfrac{a}{2}}

=3a416=34a2= \sqrt{\dfrac{3a^4}{16}} = \dfrac{\sqrt{3}}{4}a^2

This matches the standard formula for the area of an equilateral triangle. ✓

Problem 2: Triangle with sides 18, 24, 30 cm

Problem: Find the area of a triangle whose sides are 18 cm, 24 cm, and 30 cm. Also verify by using 12×base×height\dfrac{1}{2} \times \text{base} \times \text{height}.

Solution:
a=18a = 18, b=24b = 24, c=30c = 30

s=18+24+302=722=36s = \dfrac{18+24+30}{2} = \dfrac{72}{2} = 36

Compute each factor:
- s−a=36−18=18s - a = 36 - 18 = 18
- s−b=36−24=12s - b = 36 - 24 = 12
- s−c=36−30=6s - c = 36 - 30 = 6

Area=36×18×12×6=46656=216 cm2\text{Area} = \sqrt{36 \times 18 \times 12 \times 6} = \sqrt{46656} = 216 \text{ cm}^2

Verification: Check if it is a right triangle: 182+242=324+576=900=30218^2 + 24^2 = 324 + 576 = 900 = 30^2 ✓

So it is a right triangle with hypotenuse 30 cm and legs 18 cm and 24 cm.

Using base and height: Area=12×24×18=216 cm2\text{Area} = \dfrac{1}{2} \times 24 \times 18 = 216 \text{ cm}^2 ✓

Problem 3: Isosceles Triangle with Perimeter 30 cm

Problem: An isosceles triangle has perimeter 30 cm and each equal side is 12 cm. Find its area.

Solution:
Perimeter =30= 30 cm, equal sides =12= 12 cm each.
Third side =30−12−12=6= 30 - 12 - 12 = 6 cm.

a=12a = 12, b=12b = 12, c=6c = 6

s=302=15s = \dfrac{30}{2} = 15

Factors: s−a=3s - a = 3, s−b=3s - b = 3, s−c=9s - c = 9.

Area=15×3×3×9=1215=915≈34.84 cm2\text{Area} = \sqrt{15 \times 3 \times 3 \times 9} = \sqrt{1215} = 9\sqrt{15} \approx 34.84 \text{ cm}^2

Problem 4: Triangle with sides 122, 22, 120 cm

Problem: Find the area of a triangle with sides 122 cm, 22 cm, and 120 cm.

Solution:
a=122a = 122, b=22b = 22, c=120c = 120

s=122+22+1202=2642=132s = \dfrac{122+22+120}{2} = \dfrac{264}{2} = 132

Factors: s−a=10s - a = 10, s−b=110s - b = 110, s−c=12s - c = 12.

Area=132×10×110×12\text{Area} = \sqrt{132 \times 10 \times 110 \times 12}

Simplify step by step:
132×12=1584132 \times 12 = 1584 and 10×110=110010 \times 110 = 1100
1584×1100=17424001584 \times 1100 = 1742400
1742400=1320 cm2\sqrt{1742400} = 1320 \text{ cm}^2

Verification: 222+1202=484+14400=14884=122222^2 + 120^2 = 484 + 14400 = 14884 = 122^2 ✓ (right triangle)
Area =12×22×120=1320 cm2= \dfrac{1}{2} \times 22 \times 120 = 1320 \text{ cm}^2 ✓

Problem 5: Triangle with sides in ratio 3:5:7

Problem: The sides of a triangular plot are in the ratio 3:5:73:5:7 and its perimeter is 300 m. Find the area.

Solution:
Let sides be 3k,5k,7k3k, 5k, 7k.
3k+5k+7k=300  ⟹  15k=300  ⟹  k=203k + 5k + 7k = 300 \implies 15k = 300 \implies k = 20

Sides: 6060 m, 100100 m, 140140 m.

s=3002=150s = \dfrac{300}{2} = 150

Factors: s−60=90s - 60 = 90, s−100=50s - 100 = 50, s−140=10s - 140 = 10.

Area=150×90×50×10=6750000=15003≈2598.08 m2\text{Area} = \sqrt{150 \times 90 \times 50 \times 10} = \sqrt{6750000} = 1500\sqrt{3} \approx 2598.08 \text{ m}^2

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Exercise 10.2 — Area of Quadrilaterals (Solved)

To find the area of a quadrilateral using Heron's formula, divide it into two triangles by drawing a diagonal. Exercise 10.2 applies this technique to rhombi, trapeziums, and other quadrilaterals.

Problem 1: Rhombus-shaped Field

Problem: A rhombus-shaped field has green grass for 18 cows. If each side of the rhombus is 30 m and one diagonal is 48 m, find the area and the area each cow can graze.

Solution:
The diagonal divides the rhombus into two congruent triangles.

Each triangle has sides a=30a = 30, b=30b = 30, c=48c = 48 m.

s=30+30+482=54s = \dfrac{30+30+48}{2} = 54

Factors: s−30=24s - 30 = 24, s−30=24s - 30 = 24, s−48=6s - 48 = 6.

Area of one triangle=54×24×24×6=186624=432 m2\text{Area of one triangle} = \sqrt{54 \times 24 \times 24 \times 6} = \sqrt{186624} = 432 \text{ m}^2

Area of rhombus =2×432=864 m2= 2 \times 432 = 864 \text{ m}^2

Each cow grazes on 86418=48 m2\dfrac{864}{18} = 48 \text{ m}^2 of area.

Problem 2: Trapezium-shaped Field

Problem: A trapezium-shaped field has parallel sides 25 m and 10 m, and non-parallel sides 14 m and 13 m. Find its area.

Solution:
Draw DF∥BCDF \parallel BC meeting ABAB at FF. Then BFCDBFCD is a parallelogram.
DF=BC=14DF = BC = 14 m, AF=AB−FB=25−10=15AF = AB - FB = 25 - 10 = 15 m.

△ADF\triangle ADF has sides AD=13AD = 13, DF=14DF = 14, AF=15AF = 15 m.

s=13+14+152=21s = \dfrac{13+14+15}{2} = 21

Factors: s−13=8s - 13 = 8, s−14=7s - 14 = 7, s−15=6s - 15 = 6.

Area of △ADF=21×8×7×6=7056=84 m2\text{Area of } \triangle ADF = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84 \text{ m}^2

Height of trapezium: h=2×Areabase=2×8415=16815=11.2h = \dfrac{2 \times \text{Area}}{\text{base}} = \dfrac{2 \times 84}{15} = \dfrac{168}{15} = 11.2 m

Area of trapezium =12(25+10)×11.2=12×35×11.2=196 m2= \dfrac{1}{2}(25+10) \times 11.2 = \dfrac{1}{2} \times 35 \times 11.2 = 196 \text{ m}^2

Problem 3: Triangular Park Fencing

Problem: A triangular park ABCABC has sides 120120 m, 8080 m, and 5050 m. A gardener has to put a fence along the boundary and plant grass inside. Find the area and the cost of fencing at Rs 20 per metre.

Solution:
a=120a = 120, b=80b = 80, c=50c = 50

First check triangle inequality: 50+80=130>12050 + 80 = 130 > 120 ✓

s=120+80+502=125s = \dfrac{120+80+50}{2} = 125

Factors: s−120=5s - 120 = 5, s−80=45s - 80 = 45, s−50=75s - 50 = 75.

Area=125×5×45×75=2109375≈1452.37 m2\text{Area} = \sqrt{125 \times 5 \times 45 \times 75} = \sqrt{2109375} \approx 1452.37 \text{ m}^2

Perimeter =120+80+50=250= 120 + 80 + 50 = 250 m
Cost of fencing =250×20== 250 \times 20 = Rs 50005000

Worked Examples — Additional Practice

Here are additional worked examples covering common exam patterns and more challenging problems.

Example 1: Area of a Quadrilateral with Known Diagonal

Problem: Find the area of a quadrilateral ABCDABCD with AB=3AB = 3 cm, BC=4BC = 4 cm, CD=4CD = 4 cm, DA=5DA = 5 cm, and diagonal AC=5AC = 5 cm.

Solution:
Diagonal ACAC divides the quadrilateral into △ABC\triangle ABC and △ACD\triangle ACD.

Triangle ABC: sides 3,4,53, 4, 5.
Check: 32+42=9+16=25=523^2 + 4^2 = 9 + 16 = 25 = 5^2 — right triangle!
Area =12×3×4=6 cm2= \dfrac{1}{2} \times 3 \times 4 = 6 \text{ cm}^2

Triangle ACD: sides 5,4,55, 4, 5.
s=5+4+52=7s = \dfrac{5+4+5}{2} = 7
Area =7×2×3×2=84=221≈9.17 cm2= \sqrt{7 \times 2 \times 3 \times 2} = \sqrt{84} = 2\sqrt{21} \approx 9.17 \text{ cm}^2

Total area =6+221≈15.17 cm2= 6 + 2\sqrt{21} \approx 15.17 \text{ cm}^2

Example 2: Trapezium with Non-parallel Sides

Problem: A field is in the shape of a trapezium with parallel sides 11 m and 25 m, and non-parallel sides 15 m and 13 m. Find the area.

Solution:
Draw CE∥DACE \parallel DA, so ADCEADCE is a parallelogram.
BE=AB−AE=25−11=14BE = AB - AE = 25 - 11 = 14 m, CE=DA=15CE = DA = 15 m, BC=13BC = 13 m.

In △BCE\triangle BCE: sides =14,15,13= 14, 15, 13.

s=14+15+132=21s = \dfrac{14+15+13}{2} = 21

Area of △BCE=21×7×6×8=7056=84 m2\text{Area of } \triangle BCE = \sqrt{21 \times 7 \times 6 \times 8} = \sqrt{7056} = 84 \text{ m}^2

Height h=2×8414=12h = \dfrac{2 \times 84}{14} = 12 m

Area of trapezium =12(11+25)×12=12×36×12=216 m2= \dfrac{1}{2}(11 + 25) \times 12 = \dfrac{1}{2} \times 36 \times 12 = 216 \text{ m}^2

Example 3: Isosceles Triangle from Perimeter

Problem: The perimeter of an isosceles triangle is 42 cm and its base is 12 cm. Find the area.

Solution:
Equal sides =42−122=15= \dfrac{42 - 12}{2} = 15 cm each.

a=15a = 15, b=15b = 15, c=12c = 12.

s=422=21s = \dfrac{42}{2} = 21

Factors: s−15=6s - 15 = 6, s−15=6s - 15 = 6, s−12=9s - 12 = 9.

Area=21×6×6×9=6804=1821≈82.49 cm2\text{Area} = \sqrt{21 \times 6 \times 6 \times 9} = \sqrt{6804} = 18\sqrt{21} \approx 82.49 \text{ cm}^2

Example 4: Triangle with sides 5, 6, 7 cm

Problem: Find the area of a triangle whose sides are 5 cm, 6 cm, and 7 cm.

Solution:
a=5a = 5, b=6b = 6, c=7c = 7.

s=5+6+72=9s = \dfrac{5+6+7}{2} = 9

Factors: s−5=4s - 5 = 4, s−6=3s - 6 = 3, s−7=2s - 7 = 2.

Area=9×4×3×2=216=66≈14.70 cm2\text{Area} = \sqrt{9 \times 4 \times 3 \times 2} = \sqrt{216} = 6\sqrt{6} \approx 14.70 \text{ cm}^2

Example 5: Kite-shaped Field

Problem: A kite has two pairs of adjacent sides 20 m each and its longer diagonal is 32 m. Find the area using Heron's formula.

Solution:
The longer diagonal divides the kite into two congruent triangles, each with sides 20,20,3220, 20, 32.

s=20+20+322=36s = \dfrac{20+20+32}{2} = 36

Factors: s−20=16s - 20 = 16, s−20=16s - 20 = 16, s−32=4s - 32 = 4.

Area of each triangle=36×16×16×4=36864=192 m2\text{Area of each triangle} = \sqrt{36 \times 16 \times 16 \times 4} = \sqrt{36864} = 192 \text{ m}^2

Total area =2×192=384= 2 \times 192 = 384 m2^2

Common Mistakes to Avoid

Mistake 1: Forgetting to compute the semi-perimeter.
Students sometimes substitute the full perimeter instead of s=a+b+c2s = \dfrac{a+b+c}{2} into Heron's formula. Always halve the perimeter first.

Mistake 2: Errors in simplifying square roots.
When you get 46656\sqrt{46656}, factor it: 46656=216246656 = 216^2. Practice prime factorisation to simplify square roots efficiently. A useful approach: a×b=a×b\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}, so break the product into perfect-square factors.

Mistake 3: Not checking the triangle inequality.
Before applying Heron's formula, verify that the three sides can actually form a triangle: the sum of any two sides must be greater than the third side. If the triangle inequality fails, the triangle does not exist.

Mistake 4: Rounding too early.
Keep exact values (like 9159\sqrt{15}) as long as possible. Only convert to decimals in the final answer if asked. Premature rounding leads to inaccurate results and may cost half a mark.

Mistake 5: Forgetting units.
Always include the unit (cm2\text{cm}^2, m2\text{m}^2, etc.) in your final answer. Area is always in square units. Writing "216216" instead of "216 cm2216 \text{ cm}^2" loses marks.

Key Formulas Summary

FormulaExpression
Semi-perimeters=a+b+c2s = \dfrac{a+b+c}{2}
Heron's formulaArea=s(s−a)(s−b)(s−c)\text{Area} = \sqrt{s(s-a)(s-b)(s-c)}
Equilateral triangleArea=34a2\text{Area} = \dfrac{\sqrt{3}}{4}a^2
Isosceles triangle (equal sides aa, base bb)Area=b44a2−b2\text{Area} = \dfrac{b}{4}\sqrt{4a^2 - b^2}
Area using base & heightArea=12×b×h\text{Area} = \dfrac{1}{2} \times b \times h
Trapezium areaArea=12(a+b)×h\text{Area} = \dfrac{1}{2}(a+b) \times h
Rhombus area (using diagonals)Area=12×d1×d2\text{Area} = \dfrac{1}{2} \times d_1 \times d_2
QuadrilateralDivide into two triangles using a diagonal

Practice Questions with Answers

Try these questions on your own first, then check the solutions below.

Q1: Triangle with sides 13, 14, 15 cm

Question: Find the area of a triangle with sides 1313 cm, 1414 cm, and 1515 cm.

Answer:
s=13+14+152=21s = \dfrac{13+14+15}{2} = 21

Area =21×8×7×6=7056=84= \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84 cm2^2.

Verification: This is one of the classic triangles. Its height on side 1414 is h=2×8414=12h = \dfrac{2 \times 84}{14} = 12 cm.

Q2: Right Triangle Check

Question: The sides of a triangle are 2020 cm, 2121 cm, and 2929 cm. Is it a right triangle? Find its area.

Answer:
Check: 202+212=400+441=841=29220^2 + 21^2 = 400 + 441 = 841 = 29^2 ✓ — yes, it is a right triangle.

Area =12×20×21=210= \dfrac{1}{2} \times 20 \times 21 = 210 cm2^2.

Using Heron's: s=35s = 35, Area =35×15×14×6=44100=210= \sqrt{35 \times 15 \times 14 \times 6} = \sqrt{44100} = 210 cm2^2 ✓.

Q3: Cost of Planting Grass

Question: A triangular garden has sides 4040 m, 2424 m, and 3232 m. Find the area and the cost of planting grass at Rs 55 per m2^2.

Answer:
Check: 242+322=576+1024=1600=40224^2 + 32^2 = 576 + 1024 = 1600 = 40^2 ✓ (right triangle)

Area =12×24×32=384= \dfrac{1}{2} \times 24 \times 32 = 384 m2^2.

Cost =384×5== 384 \times 5 = Rs 19201920.

Q4: Quadrilateral Area

Question: A quadrilateral ABCDABCD has sides AB=9AB = 9 cm, BC=40BC = 40 cm, CD=28CD = 28 cm, DA=15DA = 15 cm, and diagonal AC=41AC = 41 cm. Find its area.

Answer:
△ABC\triangle ABC: sides 9,40,419, 40, 41. Check: 92+402=81+1600=1681=4129^2 + 40^2 = 81 + 1600 = 1681 = 41^2 ✓.
Area =12×9×40=180= \dfrac{1}{2} \times 9 \times 40 = 180 cm2^2.

△ACD\triangle ACD: sides 41,28,1541, 28, 15. s=42s = 42.
Area =42×1×14×27=15876=126= \sqrt{42 \times 1 \times 14 \times 27} = \sqrt{15876} = 126 cm2^2.

Total area =180+126=306= 180 + 126 = 306 cm2^2.

Q5: Equilateral Triangle with Heron's Formula

Question: Find the area of an equilateral triangle with side 1010 cm using Heron's formula.

Answer:
s=302=15s = \dfrac{30}{2} = 15.

Area =15×5×5×5=1875=253≈43.30= \sqrt{15 \times 5 \times 5 \times 5} = \sqrt{1875} = 25\sqrt{3} \approx 43.30 cm2^2.

Alternatively, 34×100=253\dfrac{\sqrt{3}}{4} \times 100 = 25\sqrt{3} ✓.

Exam Tips for Heron's Formula

Tip 1 — Always compute the semi-perimeter first. Write s=a+b+c2s = \dfrac{a+b+c}{2} as a separate step before substituting into Heron's formula. This earns a mark and reduces errors.

Tip 2 — Simplify inside the square root. Factor the product s(s−a)(s−b)(s−c)s(s-a)(s-b)(s-c) into perfect squares when possible. For example, 36×18×12×6=(6)2×(6×3)×(6×2)×6=6×6×6=216\sqrt{36 \times 18 \times 12 \times 6} = \sqrt{(6)^2 \times (6 \times 3) \times (6 \times 2) \times 6} = 6 \times 6 \times 6 = 216.

Tip 3 — For quadrilaterals, draw a diagonal. Split the quadrilateral into two triangles, find each area using Heron's formula, and add them.

Tip 4 — Check with the right-triangle shortcut. If a2+b2=c2a^2 + b^2 = c^2, the triangle is right-angled and area =12×a×b= \dfrac{1}{2} \times a \times b. Use this to verify your Heron's formula answer.

Tip 5 — For trapezium problems, find the height first. Use Heron's formula on a triangle formed by the non-parallel sides and the difference of parallel sides, then extract the height from the triangle's area.

Tip 6 — Memorise common Pythagorean triplets. (3,4,5)(3,4,5), (5,12,13)(5,12,13), (8,15,17)(8,15,17), (7,24,25)(7,24,25) and their multiples. Recognising a right triangle saves time.

Tip 7 — Show prime factorisation for large square roots. When simplifying square roots of large numbers, break them into prime factors systematically.

Tip 8 — Keep answers in surd form. Unless the question asks for a decimal approximation, leave answers like 9159\sqrt{15} in exact form. This shows mathematical precision.

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Key Takeaways

  • Heron's formula Area=s(s−a)(s−b)(s−c)\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} finds the area of any triangle from its three sides alone — no height needed.
    - The semi-perimeter s=a+b+c2s = \dfrac{a+b+c}{2} must always be calculated first as a separate step.
    - Heron's formula works for all triangle types: scalene, isosceles, equilateral, right-angled, acute, and obtuse.
    - For quadrilaterals, split into two triangles using a diagonal, apply Heron's formula to each, and add the areas.
    - For trapeziums, use Heron's formula on a constructed triangle to find the height, then apply Area=12(a+b)×h\text{Area} = \dfrac{1}{2}(a+b) \times h.
    - Always verify your answer by checking if the triangle is right-angled (using Pythagorean triplets).
    - Keep answers in exact surd form unless decimal approximation is specifically requested.
    - The formula is named after Heron of Alexandria (1st century AD), though similar results may have been known to Archimedes earlier.

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