**HERON’S FORMULA**

**(Questions from CBSE Papers)****Points to Remember**

*1*. Area of a triangle = ½ base × height

As in figure,

Area = ½ BC × AM square units.

*2*. When the triangle is right angled.

Area of a triangle = ½ base × height

=

**½ × Product of sides containing the triangle.**

*3*. When the triangle is equilateral i.e. all sides are of same length (say = a).

Take a mid-point on BC (say M) and join it to A.

Δ ABM is a right angles triangle.

Using Pythagoras theorem.

AB

^{2}= AM

^{2}+ BM

^{2}

a

^{2}= AM

^{2}+ (a/2)

^{2}

⇒ AM

^{2}= a

^{2}- a

^{2}/4 = 3a

^{2}/4

⇒ AM = (√3)a/2

Area of Δ ABC = ½ × BC × AM = ½ × a × (√3)a/2

⇒ Area of Δ ABC =

**(√3/4)a**

^{2}**. For an isosceles triangle, two sides are equal. As in figure in Δ ABC, AB = AC = a and BC = b.**

*4*Take a mid-point M on BC and draw a ⊥ AM from BC.

BM = MC = b/2

and AM = h

In rt-angled Δ ABM, using Pythagoras theorem:

AB

^{2}= AM

^{2}+ BM

^{2}

⇒ AM

^{2}= AB

^{2}- BM

^{2}

^{}

⇒ h

^{2}= a

^{2}- (b/2)

^{2}= (4a

^{2}- b

^{2})/4

Area of Δ ABC = ½ × BC × AM =

**½ × b × (4a**

^{2}- b^{2})/4**. For scalene triangles, since we do not know the height we cannot find area as we did in above cases.**

*4***. Heron's Formula or Hero's Formula is the method to find area of any type of triangle, provided the length of the three sides are known. If a, b, c are the lengths of the triangle,**

*5*s is the semi-perimeter of the triangle.

*. To compute the area of a quadrilateral whose sides and one diagonal are given. With the help of diagonal, partition the quadrilateral region into two triangular regions. Then apply Heron's formula to find the area of each triangle. The sum of areas of the two triangles gives us the area of the quadrilateral.*

**6****Questions from CBSE Papers**

**Q1(CBSE 2010): The semi perimeter of a triangle having the length of its sides as 20 cm, 15 cm and 9 cm is:**

(a) 44 cm

(b) 21 cm

(c) 22 cm

(d) none

Answer: (c) 22 cm

**Q2(CBSE 2010/2011): Heron's formula is**:

Answer: D

**Q3: An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find area of the triangle.**

Answer: An isosceles triangle has two sides equal a = b = 12 cm.

Given Perimeter = 30 cm

⇒ third side of the Δ is = c = 30 - (12 + 12) = 6 cm

Semi-perimeter s = 30 /2 = 15cm

Area of Δ = [(15×(15-12)×(15-12)×(15-6)]

^{1/2}

⇒ = [15 × 3 × 3 × 9]

^{1/2}

⇒ = 9√(15) cm

^{2}

**Q4: Find the area of an equilateral triangle with side 10 cm.**

Answer: Area of equilateral Δ = (√3/4)a

^{2}= (√3/4) × 10

^{2}= (√3/4) × 100 = 25√3 cm

^{2}

Using Heron's formula, s = 30/2 = 15

Area = [(15×(15-10)×(15-10)×(15-10)]

^{1/2}= [15 × 5 × 5 × 5 ]

^{1/2}= 25√3 cm

^{2}

**Q5: A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 15 cm, 14 cm and 13 cm, and the parallelogram stands on the base 15 cm, find the height of the parallelogram.**

Answer: Given, Area of Triangle = Area of the Parallelogram

and both have same base = 15cm.

Sides of the triangle are: a = 15cm, b = 14 cm and c = 13 cm

Semi-perimeter s = (15+14+13)/2 = 42/2 = 21 cm

Area of Δ = [(21×(21-15)×(21-14)×(21-13)]

^{1/2}

= (21 × 6 × 7 × 8)

^{1/2}

= (3 × 7 × 2 × 3 × 7 × 2 × 4)

^{1/2}= 3 × 7 × 2 × 2 = 84cm

^{2}

∴ Area of parallelogram = base × height = 84

⇒ height = 84 / 15 = 5.6 cm

**Q6: A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?**

Answer: Each side of the equilateral triangle = a

Semi-perimeter (s) = (a+a+a)/3 = 3a/2

Using Heron's formula, area of Δ = [(3a/2)×(3a/2 -a)×(3a/2 -a)×(3a/2 -a)]

^{1/2}

⇒ = [(3a/2)×(a/2)×(a/2)×(a/2)]

^{1/2}= a

^{2}/4(√3)

⇒ = a

^{2}√3/4

Given perimeter of Δ = 180m = 3a

⇒ a = 60m

Therefore area of Δ = (60)

^{2}√3/4 =

**900√3 cm**

^{2}**Q7: The sides of a Δ are 7 cm, 24 cm and 25 cm. Its area is :**

(a) 168 cm

^{2}

(b) 84 cm

^{2}

(c) 87.5 cm

^{2}

(d) 300 cm

^{2}

Answer: (b) 84 cm

^{2}

s = (7 + 24 + 25)/2 = 28

Area = [(28)×(28-7)×(28-24)×(28-25)]

^{1/2}= [(28)×(21)×(4)×(3)]

^{1/2}

= [(7×4)×(7 ×3)×(4)×(3)]

^{1/2}= 7 × 4 × 3 = 84 cm

^{2}

**Q8: A square and an equilateral triangle have equal perimeters. If the diagonal of the square is 12√2cm, then area of the triangle is**:

(a) 24√2 cm

^{2}

(b) 24√3 cm

^{2}

(c) 48√3 cm

^{2}

(d) 64√3 cm

^{2}

Answer: (d) 64√3 cm

^{2}

Let side of square = a

Using Pythagoras theorem, 2a

^{2}= 144 × 2 ⇒ a = 12cm

Perimeter of equilateral Δ = 4 × 12 = 48

Side of equilateral Δ (b) = 48/3 = 16cm

Area of equilateral Δ = (√3/4)b

^{2}= (√3/4)(16)

^{2}= 64√3 cm

^{2}

**Miscellaneous**

**Q: How much area of a triangle will increase if its each side is doubled. Calculate the percentage increase.**

Answer: Let a, b, c be the sides of the triangle.

Semi-perimeter (s

_{1}) = ½ (a + b + c)

The area of the Δ = A

_{1}= [(s

_{1})×(s

_{1}- a)×(s

_{1}- b)×(s

_{1}- a)]

^{1/2}

When the sides are doubled i.e. 2a, 2b and 2c,

Semi-perimeter (s

_{2}) = ½ (2a + 2b + 2c) = 2s

_{1}

The new area of Δ = A

_{2}= [(2s

_{1})×(2s

_{1}- 2a)×(2s

_{1}- 2b)×(2s

_{1}- 2c)]

^{1/2}

⇒ A

_{2}= [(2s

_{1})×2(s

_{1}- a)×2(s

_{1}- b)×2(s

_{1}- c)]

^{1/2}= 4 [(s

_{1})×(s

_{1}- a)×(s

_{1}- b)×(s

_{1}- c)]

^{1/2}

⇒ A

_{2}= 4A

_{1}

New area of Δ will be four times the area of original triangle.

% Increase = (A

_{2}- A

_{1})/(A

_{1}) × 100 = (3A

_{1})/(A

_{1}) × 100 = 300%

.

More MCQ can be created

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