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The Sides of a Triangular Field Are in the Ratio 5: 3: 4 and Its Perimeter is 180 M. Find: (I) Its Area. (Ii) the Altitude of the Triangle Corresponding to Its Largest Side. - Mathematics

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Question

The sides of a triangular field are in the ratio 5: 3: 4 and its perimeter is 180 m. Find:

(i) its area.

(ii) the altitude of the triangle corresponding to its largest side.

(iii) the cost of leveling the field at the rate of Rs. 10 per square meter.

Sum

Solution 1

(i) Given that the sides of a triangle are in the ratio 5: 3: 4.

Also, given that the perimeter of the triangle is 180.

Thus, we have, 5x + 4x + 3x = 180

⇒ 12x = 180

⇒ x = `180/12`

⇒ x = 15.

Thus, the sides are 5 × 15, 3 × 15 and 4 × 15

That is the sides are 75 m, 45 m and 60 m.

Since the sides are in the ratio, 5: 3: 4, it is a Pythagorean triplet,

Therefore, the triangle is a right-angled triangle,

Area of a right-angled triangle = `1/2` × base × altitude

⇒ `1/2`× 45 × 60

⇒ 45 × 30

= 1350 m2.

(ii) Consider the following figure.


In the above figure,

The largest side is AC = 75 m,

The altitude corresponding to AC is BD.

We need to find the value of BD.

Now consider the triangles ΔBCD and ΔBAD.
We have,

∠B = ∠B             .....[common]

BD = BD            .....[common]

∠D = ∠D = 90°

Thus, by Angle-Side-Angle criterion of congruence,

we have ΔBCD - ΔABD,

Similar triangles have similar proportionality.

Thus, we have,

`"CD"/"BD" = "BD"/"AD"`

⇒ BD2 = AD × CD             ....(1)

From subpart (i), the sides of the triangle are AC = 75 m, AB = 60 m and BC = 45 m

Let AD = x m ⇒ CD = (75 - x) m

Thus applying Pythagoras Theorem,
from right triangle ΔBCD, we have

452 = (75 - x)2 + BD2

⇒ BD2 = 452 - (75 - x)2

⇒ BD2 = 2025 - (5625 + x2 - 150x)

⇒ BD2 = 2025 - 5625 - x2 + 150x

⇒ BD2 = - 3600 - x2 + 150x          ....(2)

Now applying Pythagoras Theorem,
from right triangle ΔABD, we have

602 = x2 + BD2

⇒ BD2 = 602 - x2

⇒ BD2  = 3600 - x2                 ......(3)

From equations (2) and (3), we have,

- 3600 - x2 + 150x = 3600 -x2

⇒ 150x = 3600 + 3600

⇒ 150x = 7200

⇒ x = `7200/150`

⇒ x = 48

Thus, AD = 48 and CD = 75 - 48 = 27.

Substituting the values AD = 48 m
and CD = 27 m in equation (1), we have

BD2 = 48 x 27

⇒ BD2 = 1296

⇒ BD = 36 m

The altitude of the triangle corresponding to
its largest side is BD = 36 m

(iii) The area of the triangular field from subpart (i) is 1350 m2

The cost of levelling the field is Rs. 10 per square metre,

Thus the total cost of levelling the field = 1350 x 10 = Rs. 13,500.

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Solution 2

a = 5x = 5(15) = 75 m

b = 3x = 3(15) = 45 m

c = 4x = 4(15) = 60 m

P = 180 m

5x + 3x + 4x = 180

12x = 180

x = `180/12 = 15`

S = `180/2` = 90

ar(ABC) = `sqrt("s"("s - a")("s" - "b")("s" - "c"))`

`= sqrt(90(90 - 75)(90 - 45)(90- 60))`

`= sqrt(90 xx 15 xx 45 xx 30)`

`= sqrt(9 xx 10 xx 3 xx 5 xx 9 xx 5 xx 3 xx 10)`

`= 9 xx 5 xx 3 xx 10`

= 1350 m2

ar(Δ ABC) = 1350

`1/2 xx "b" xx "h" = 1350`

`1/2 xx 75 xx "h"`= 1350

h = `(1350 xx 2)/75`

h = 18 × 2

h = 36 m

∴ Cost of leveling = 1350 × 10 = 13500

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Chapter 20: Area and Perimeter of Plane Figures - Exercise 20 (A) [Page 248]

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Selina Concise Mathematics [English] Class 9 ICSE
Chapter 20 Area and Perimeter of Plane Figures
Exercise 20 (A) | Q 10 | Page 248
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