The document Short Answers Type Questions- Heron’s Formula Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.

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**Question 1. The area of an equilateral triangle whose side is ‘a’ cm is **(√3/4)**a ^{2 }cm^{2}. Find its height.** Area of the triangle = (√3/4)a

Solution:

∵ Area of a triangle = (1/2)x base x height

∴ (1/2) x a x height =(√3/4)x a^{2}

⇒ height =(√3/4) a^{2} x (2/a) = (√3/2) a cm

**Question 2. Find the height of an equilateral triangle whose side is 2 cm. Solution: **Since height of an equilateral triangle is given by

height = (√3/2) x side

⇒ height =(√3/2) x 2 cm = 3 cm

** Question 3. Find the length of a diagonal of a square whose side is 2 cm. Solution: **The diagonal of a square = (√2) a cm

∴ Length of the diagonal = (√2) x 2 cm = 2√2cm.

**Question 4. The diagonal of a square is 9**√2** cm. What is the side? Solution:** Let side of the square = x cm.

∵ Its diagonal is given by √2 x side.

∴ √2 x x= 9 x √2

⇒

Thus, the required length of sides of the square is 9 cm.

**Question 1. The length of a rectangular plot of land is twice its breadth. If the perimeter of the plot be 180 metres, then find its area. Solution: **Let the breadth of the plot be ‘x’ metres.

∴ Its length = 2x metres

Since perimeter of a rectangular plot = 2[Length + Breadth]

∴ Perimeter of the given plot = 2[x + 2x]

= 2[3x]

= 6x metres

⇒ 6x = 180

⇒ x= 180/6= 30 metres

⇒ 2x = 2 x 30 = 60 metres.

∴ Length of the plot = 60 metres and Breadth of the plot = 30 meters.

∴ Area of the plot = Length x Breadth = 60 x 30 m^{2} = 1800 m^{2 }

**Question 2. The length of the sides containing the right angle in a right triangle differ by 7 cm. The area of the triangle is 60 cm ^{2}. Find the length of the hypotenuse. Solution: **Let the sides containing the right angle be ‘x’ cm and (x – 7) cm.

i.e. Base = x cm and height = (x – 7) cm

∴ Area = (1/2) x base x height

=(1/2) x x x (x – 7) cm^{2}

Now (1/2) x (x – 7) = 60

⇒ x(x – 7) = 120

⇒ x^{2} – 7x – 120 = 0

⇒ x^{2} – 15x + 8x – 120 = 0

⇒ x(x – 15) + 8(x – 15) = 0

⇒ (x + 8)(x – 15) = 0

⇒ x = – 8 or x = 15

Rejecting x = –8, we have x – 15 = 0

⇒ x = 15 cm x – 7 = 15 – 7 = 8 cm

Now, Hypotenuse = √289 = 17cm

Thus, the required length of the hypotenuse is 17 cm.

**Question 3. The lengths of sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 120 cm, find its area. Solution:** The sides are in the ratio of 3 : 4 : 5.

Let the sides be 3x, 4x and 5x.

∴ Perimeter = 3x + 4x + 5x = 12x

Now 12x = 120 [Perimeter = 120 cm]

⇒ x =(120/12) = 10

∴ Lengths are: a = 3x = 3 x 10 = 30 cm

b = 4x = 4 x 10 = 40 cm

c = 5x = 5 x 10 = 50 cm

Now, semi-perimeter (s) = (120/12)

cm = 60 cm

∵ (s – a) = 60 – 30 = 30 cm

(s – b) = 60 – 40 = 20 cm

(s – c) = 60 – 50 = 10 cm

Using Heron’s formula, we have

Area of the triangle =

= 2 x 30 x 10 cm^{2} = 600 cm^{2} Thus, the required area of the triangle = 600 cm^{2}.

**Question 4. Find the area of a quadrilateral ABCD in which AB = 8 cm, BC = 6 cm, CD = 8 cm, DA = 10 cm and AC = 10 cm. Solution:** In ΔABC, ∠B = 90°

∴ area of right (rt ΔABC) = (1/2) x 8 x 6 cm^{2} = 24 cm^{2}

In ΔACD,

a = AC = 10 cm b = AD = 10 cm c = CD = 8 cm

∴Area of ΔACD

= 2 x 4√21 = 8√21 cm^{2}

= 8 x 4,58 cm^{2} = 36.64 cm^{2}

Now, area of quadrilateral ABCD = ar (ΔABC) + ar (ΔACD)

= 24 cm^{2} + 8√21 cm^{2 }

= 24 cm^{2} + 36.64 cm^{2}

= 60.64 cm^{2}

**Question 5. How much paper of each shade is needed to make a kite given in the figure, in which ABCD is a square of diagonal 44 cm. Solution: **∵ The diagonals of a square bisect each other at right angles

∴ OB = OD = OA = OC = (44/2) = 22 cm

Now, ar rt (Δ –I) = (1/2)× OB × OA

= (1/2) × 22 × 22 cm^{2} = 242 cm^{2}

Similarly ar rt (Δ –II) = arrt(Δ–III) = ar rt (Δ –IV) = 242 cm^{2}

∵ Sides of ΔCEF are 20 cm, 20 cm and 14 cm

⇒ Area of ΔCEF

Now, area of yellow paper = ar (Δ – I) + ar (Δ – II)

= 242 cm^{2 }+ 242 cm^{2} = 484 cm^{2}

Area of red paper = ar (Δ – IV) = 242 cm^{2}

Area of green paper = ar (Δ – III) + ar ΔCEF

= 242 cm^{2} + 131.14 cm^{2}

= 373.14 cm^{2}

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