Class 9 Maths Chapter 10 Question Answers - Heron’s Formula

Q1. The area of an equilateral triangle whose side is ‘a’ cm is (√3/4)a² cm². Find its height.

Area of the triangle = (√3/4)a² cm²
∵ Area of a triangle = (1/2)x base x height

∴  (1/2) x a x height =(√3/4)x a²
⇒ height =(√3/4) a² x (2/a)  = (√3/2) a cm

Q2. Find the height of an equilateral triangle whose side is 2 cm.

Since height of an equilateral triangle is given by
height = (√3/2) x side
⇒ height =(√3/2) x 2 cm = √3 cm

Q3. Find the length of a diagonal of a square whose side is 2 cm.

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

Q4. The diagonal of a square is 9√2 cm. What is the side?

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.

Q5. 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.

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 m² = 1800 m² 

Q6. The length of the sides containing the right angle in a  right triangle differ by 7 cm. The area of the triangle is 60 cm². Find the length of the hypotenuse.

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) cm²
Now (1/2) x (x – 7) = 60
⇒ x(x – 7) = 120
⇒ x² – 7x – 120 = 0
⇒ x² – 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.

Q7. The lengths of sides of a triangle are in the ratio 3 : 4 : 5 and its perimeter is 120 cm, find its area. 

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 cm² = 600 cm² Thus, the required area of the triangle = 600 cm².

Q8. 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.

In ΔABC, ∠B = 90°

∴ area of right (rt ΔABC) = (1/2) x 8 x 6 cm² = 24 cm²
In ΔACD,
a = AC = 10 cm b = AD = 10 cm c = CD = 8 cm

∴Area of ΔACD 

= 2 x 4√21 = 8√21 cm²

= 8 x 4,58 cm² = 36.64 cm²

Now, area of quadrilateral ABCD = ar (ΔABC) + ar (ΔACD)
= 24 cm² + 8√21 cm² 
= 24 cm² + 36.64 cm²
= 60.64 cm²

Q9. 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.

∵ 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 cm² = 242 cm²
Similarly ar rt (Δ –II) = arrt(Δ–III) = ar rt (Δ –IV) = 242 cm²

∵ 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 cm² + 242 cm² = 484 cm²
Area of red paper = ar (Δ – IV) = 242 cm²
Area of green paper = ar (Δ – III) + ar ΔCEF
= 242 cm² + 131.14 cm²
= 373.14 cm²