Jharkhand Board JAC Class 9 Maths Important Questions Chapter 12 Heron’s Formula Important Questions and Answers.

## JAC Board Class 9th Maths Important Questions Chapter 12 Heron’s Formula

Question 1.

The area of a triangle is 30 cm^{2}. Find the base if the altitude exceeds the base by 7 cm.

Solution :

Let base BC = x cm

Then altitude = (x + 7) cm

Area of ΔABC = \(\frac {1}{2}\) × base × height

⇒ 30 = \(\frac {1}{2}\)(x)(x + 7)

⇒ 60 = x^{2} + 7x

⇒ x^{2} + 7x – 60 = 0

⇒ x^{2} + 12x – 5x – 60 = 0

⇒ x(x + 12) – 5(x + 12) = 0

⇒ (x – 5)(x + 12) = 0

⇒ x = 5 or x = -12+

⇒ x = 5 [∵ x ≠ -12]

∴ Base (x) = 5 cm and

Altitude = x + 7 = 5 + 7 = 12 cm.

Question 2.

The cost of turfing a triangular field at the rate of ₹ 45 per 100 m^{2} is ₹ 900. Find the height, if double the base of the triangle is 5 times the height.

Solution :

Let the height of triangular field be h metres.

It is given that 2 × (base) = 5 × (Height)

∴ Base = \(\frac {5}{2}\) h

Area = \(\frac {1}{2}\) × Base × Height

Area = \(\frac {1}{2}\) × \(\frac {5}{2}\)h × h = \(\frac {5}{4}\)h^{2}m^{2} …………(i)

∴ Cost of turfing the field is ₹ 45 per 100m^{2}

∴ Area = Total cost / Rate per sq.m

= \(\frac{900}{45 / 100}=\frac{90000}{45}\)

= 2000 m^{2} …………(ii)

From (i) and (ii), we get

\(\frac {5}{4}\)h^{2} = 2000

⇒ 5h^{2} = 8000

⇒ h^{2} = 1600

⇒ h = 40 m

∴ Height of the triangular field is 40 m.

Question 3.

From a point in the interior of an equilateral triangle, perpendicular drawn to the three sides are 8 cm, 10 cm and 11 cm respectively. Find the area of the triangle.

Solution :

Let each side of the equilateral ΔABC = x cm,

From an interior point O, OD, OE and OF perpendiculars be drawn to BC, AC and AB respectively.

It is given that OD = 11 cm, OE = 8 cm and OF = 10 cm.

Join OA, OB and OC.

Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB

= \(\frac {1}{2}\) × 11 + \(\frac {1}{2}\) × 8 + \(\frac {1}{2}\) × 10 = \(\frac {29}{2}\) × cm^{2}

But, area of an equilateral triangle, whose each side is x = \(\frac{\sqrt{3}}{4}\) x^{2}cm^{2}

Therefore,

\(\frac{\sqrt{3}}{4}\) x^{2} = \(\frac {29}{2}\)x

∴ x = \(\frac{4 \times 29}{2 \times \sqrt{3}}=\frac{58}{\sqrt{3}}\) cm

∴ Area of ΔABC

= \(\frac{29}{2} \times \frac{58}{\sqrt{3}}\) cm^{2} = \(\frac{\sqrt{841}}{1.73}\)cm^{2}

∴ Area of ΔABC = 486.1 cm^{2}

Question 4.

The difference between the sides at right angles in a right-angled triangle is 14 cm. The area of the triangle is 120 cm^{2}. Calculate the perimeter of the triangle.

Solution :

Let the sides containing the right angle be x cm and (x – 14) cm.

Then, its area = [\(\frac {1}{2}\).x. (x – 14)] cm^{2}

But, area = 120 cm^{2} [Given]

∴ \(\frac {1}{2}\)x (x – 14) = 120

⇒ x^{2} – 14x – 240 = 0

⇒ x^{2} – 24x + 10x – 240 = 0

⇒ x(x – 24) + 10 (x – 24) = 0

⇒ (x – 24) (x + 10) = 0

⇒ x = 24 [Neglecting x = -10]

∴ One side = 24 cm,

other side = (24 – 14) cm = 10 cm

Hypotenuse

= \(\sqrt{(24)^2+(10)^2}\) cm

= \(\sqrt{576+100}\) cm

= \(\sqrt{676}\) cm = 26 cm.

∴ Perimeter of the triangle

= (24 + 10 + 26) cm = 60 cm.

Question 5.

Find the percentage increase in the area of a triangle if its each side is doubled.

Solution :

Let a, b, c be the sides of the given triangle and s be its semi-perimeter

∴ s = \(\frac {1}{2}\)(a + b + c) …………(i)

The sides of the new triangle are 2a, 2b and 2c. Let s’ be its semi-perimeter.

∴ s’ = (2a + 2b + 2c)

= a + b + c = 2s [Using (i)]

Let Δ = Area of given triangle

Δ = \(\sqrt{s(s-a)(s-b)(s-c)}\) …….(ii)

And, Δ’ = Area of new triangle

Δ’ = \(\sqrt{s^{\prime}\left(s^{\prime}-2 a\right)\left(s^{\prime}-2 b\right)\left(s^{\prime}-2 c\right)}\)

= \(\sqrt{2 s(2 s-2 a)(2 s-2 b)(2 s-2 c)}\)

= \(\sqrt{16 s(s-a)(s-b)(s-c)}\)

Δ’ = 4Δ

∴ Increase in the area of the triangle

= Δ’ – Δ = 4Δ – Δ = 3Δ

∴ % increase in area = (\(\frac {3Δ}{Δ}\) × 100)%

= 300%

Multiple Choice Questions

Question 1.

The area of the field ABGFEA is:

(a) 7225 m^{2}

(b) 7230 m^{2}

(c) 7235 m^{2}

(d) 7240 m^{2}

Solution :

(a) 7225 m^{2}

Question 2.

Area of shaded portion as shown in the figure is:

(a) 12 m^{2}

(b) 13 m^{2}

(c) 14 m^{2}

(d) 15 m^{2}

Solution :

(a) 12 m^{2}

Question 3.

The lengths of four sides and a diagonal of the given quadrilateral are indicated in the diagram. If A denotes the area of quadrilateral in cm^{2}, then A is

(a) 12\(\sqrt{6}\) sq. unit

(b) 6 sq. unit

(c) 6\(\sqrt{6}\) sq. unit

(d) \(\sqrt{6}\) sq. unit

Solution :

(a) 12\(\sqrt{6}\) sq. unit

Question 4.

If the sides of a triangle are doubled, then its area:

(a) Remains the same

(b) Becomes doubled

(c) Becomes three times

(d) Becomes four times

Solution :

(d) Becomes four times

Question 5.

Inside a triangular garden there is a flower bed in the form of a similar triangle. Around the flower bed runs a uniform path of such a width that the side of the garden are double of the corresponding sides of the flower bed. The areas of the path and the flower bed are in the ratio:

(a) 1 : 1

(b) 1 : 2

(c) 1 : 3

(d) 3 : 1

Solution :

(d) 3 : 1