Ncert Solutions For Class 9 Maths Ex 12.1

Why Heron’s Formula…?

When the triangle is a right angled triangle:

 

Area = $\frac{1}{2}\times base\times altitude$

Example:

Now, Area of $\Delta ABC= \frac{1}{2}\times base\times altitude$

$Area\;of\;\Delta ABC= \frac{1}{2}\times 12\times 5 = 30\;cm^{2}$

When the triangle is an equilateral triangle:

Consider an equilateral triangle ABC with each side = 10cm

If we drop AM perpendicular to BC then BM = 5cm,

Now, AB² = AM² + BM²    (Pythagoras Theorem)

AM² = 10² – 5²

Therefore, AM = $5\sqrt{3}$ cm

Now, Area for triangle ABC = 2×(Area of triangle ABM)

And, $Area\;\Delta ABM =\frac{1}{2}\times Base\times Altitude$

Therefore, $Area\;\Delta ABM =\frac{1}{2}\times 5\times 5\sqrt{3}$

$Area\;\Delta ABM =\frac{25\sqrt{3}}{2}$cm²

Now, Area for triangle ABC = 2×($\frac{25\sqrt{3}}{2}$)cm²

Therefore, $Area\; of \;\Delta ABC= 25\sqrt{3}\; cm^{2}$

When the triangle is an isosceles triangle:         

Consider an isosceles triangle ABC with equal sides AB = AC = 13 cm and unequal side BC = 10cm

Drop AM perpendicular to BC. Therefore CM = 5cm

Now, AC² = AM² + CM²    (Pythagoras Theorem)

AM² = 13² – 5²

Therefore AM = 12cm

Now, Area for triangle ABC = 2×(Area of triangle ACM)

And, $Area\;\Delta ACM =\frac{1}{2}\times Base\times Altitude$

Therefore, $Area\;\Delta ACM =\frac{1}{2}\times 5\times 12$

$Area\;\Delta ABM$= 30 cm²

Now, Area for triangle ABC = 2× 30 cm²

Therefore, Area of triangle ABC = 60cm²

When the triangle is a scalene triangle:

Suppose that we know the lengths of sides of a scalene triangle and we have to find its area. Now, for applying the above formula we need to calculate its height. But in this case we have no idea about how to calculate the height of a scalene triangle and that is why it’s very difficult to find its area. Thus, Heron gave a formula to find the area of any triangle known as Heron’s Formula.

Area of a Triangle $ABC=\sqrt{s(s-a)(s-b)(s-c)}$

where a, b and c are sides of the triangle.

s = semi perimeter   (half of perimeter of triangle)

$s=\frac{a+b+c}{2}$

 

Finding square roots:

 (i) Find values of: $\sqrt{72895}$

 

(ii)Find values of $\sqrt{745}$

 

 

Q.1: The lengths of two sides of a triangle ABC are 10cm and 7cm and its perimeter is 28cm. Find the area of a triangle ABC.

Sol.

Given perimeter = 28cm

Therefore, s = $\frac{perimeter}{2}$ = 14cm

And a = 10cm, b =7cm and c = 11cm (Since, perimeter = a + b + c)

 On applying Heron’s Formula;

$Area\;of\;Triangle\;ABC=\sqrt{s(s-a)(s-b)(s-c)}$

Therefore, area of Triangle ABC

=$\sqrt{14(14-10)(14-7)(14-11)}$

(or)     =$\sqrt{14\times 4\times 7\times 3}$

= $14\sqrt{6}$

Therefore, area of triangle ABC = $14\sqrt{6}$cm²

 

Q.2: If perimeter of a triangular plot is 250m and the lengths of the sides of a triangular plot are in the ratio of 4 : 7 : 9. Find its area.

Sol.

Since, the lengths of the sides of a triangular plot are in the ratio of 4 : 7 : 9

Therefore, sides of triangular plot will be 4x, 7x and 9x

Given perimeter = 250m

Therefore, 4x + 7x + 9x = 250

Hence x = 12.5, that gives:

a = 4x = 50m, b = 7x = 87.5m and c = 9x = 112.5m

And s = $\frac{perimeter}{2}$= 125m

Now on applying Heron’s Formula;

$Area\;of\;plot\;ABC=\sqrt{s(s-a)(s-b)(s-c)}$

Therefore, area of plot ABC:

=$\sqrt{125(125-50)(125-87.5)(125-112.5)}$

(or)     =$\sqrt{125\times 75\times 37.5\times 12.5}$

= $937.5\sqrt{5}$

=2096.3m²

Therefore, area of triangular plot ABC = 2096.3m²

 

Q.3: There is a triangular park in a locality having sides 110m, 90m and 40m. A gardener Fauzia has to plant grass inside. In how much area does she need to plant grass? Also Fauzia has to put a fence all around that triangular park. Find the cost of fencing that triangular park with barbed wire if the cost of barbed wire is Rs 25 per meter and also a space of 2m is to be left for a gate on any one side of that park.

Sol.

Given a = 110m, b = 90m and c = 40m

Therefore, perimeter = 110 + 90 + 40 = 240m

And s = $\frac{perimeter}{2}$= 120m

Now applying Heron’s Formula;

$Area\;of\;park\;ABC=\sqrt{s(s-a)(s-b)(s-c)}$

Therefore, area of park ABC:

=$\sqrt{120(120-110)(120-90)(120-40)}$

(or)     =$\sqrt{120\times 10\times 30\times 80}$

= $1200\sqrt{2}$

= 1697.07m² (Approx.)

Therefore, area of triangular plot ABC = 1697.07m² (Approx.)

Since, perimeter = 240m

Therefore for fencing length of wire needed = 240 – 2 = 238m

So the total cost of fencing the triangular park = 238×25 = 5950Rs

 

Q.4: The side walls of a flyover are triangular in shape and are used for advertisements. Dimensions of side walls are: 120m, 140m and 50m. The cost of posting advertisements is 4500Rs per m² per year. How much rent an automobile company paid if the company hired one of its walls for 4 months?

Sol.

Given a = 120m, b = 140m and c = 50m

Therefore, perimeter of each triangular wall = 120 + 140 + 50 = 310m

And s = $\frac{perimeter}{2}$= 155m

Now applying Heron’s Formula;

$Area\;of\;triangle\;ABC=\sqrt{s(s-a)(s-b)(s-c)}$

Therefore, area of each triangular wall:

=$\sqrt{155(155-120)(155-140)(155-50)}$

(or)     =$\sqrt{155\times 35\times 15\times 105}$

= $525\sqrt{31}$

Therefore, area of each triangular wall = $525\sqrt{31}$m²

Since, rent of posting advertisements is 4500 per m² per year (12 months)

Therefore rent for $525\sqrt{31}$m² area = $525\sqrt{31}$ × 4500

= $2362500\sqrt{31}$Rs per year(12 months).

Now, For 12 months’ rent =  $2362500\sqrt{31}$ Therefore for 4months = $\frac{4\times 2362500\sqrt{31}}{12}=787500\sqrt{31}$

Hence, rent paid by company for hiring one of its walls for 4 months = $787500\sqrt{31}$Rs.

 

Q.5: If each side of a triangle is increased by six times then find the total percentage of increase in its area.

Sol.

Let a, b and c be the sides of the original triangle. Then,

Perimeter of original triangle = a + b + c and s₁ = $\frac{a + b + c}{2}$

And sides of the new triangle will be 6a, 6b and 6c. Then,

Perimeter of new triangle = 6a + 6b + 6c and s₂ = $\frac{6a + 6b + 6c}{2}$ = 3(a + b + c) = 6s₁   (Since, 2s₁ = a + b + c)

Let A₁ and A₂ be the areas of the original and new triangle respectively. Then,

        Area of triangle A₁=$\sqrt{s_{1}(s_{1} – a)(s_{1} – b)(s_{1} – c)}$

 

And  Area of triangle A₂=$\sqrt{s_{2}(s_{2} – a)(s_{2} – b)(s_{2} – c)}$

Since, s₂ = 6s₁

Therefore, Area of triangle A₂=$\sqrt{6s_{1}(6s_{1} – a)(6s_{1} – b)(6s_{1} – c)}$

Area of triangle A₂=$\sqrt{12s_{1}(s_{1} – a)(s_{1} – b)(s_{1} – c)}$

Therefore, A₂ = 12A₁

Therefore, increase in the area of triangle = A₂ – A₁

= 12A₁ – A₁ = 11A₁

Therefore, percentage increase in area = $\frac{11A_{1}}{A_{1}}\times 100$ = 1100%

Hence, area of the new triangle is increased by 1100%

 

Q.6: If perimeter of a triangular plot is 490 and the lengths of the sides of a triangular plot are in the ratio of 22 : 16 : 32. Find its area.

Sol.

Since, the lengths of the sides of a triangular plot are in the ratio of 22 : 16 : 32

Therefore, sides of triangular plot will be 22x, 16x and 32x

Given perimeter = 490m

Therefore, 22x + 16x + 32x = 490

Hence x =7 , that gives:

a =22x = 154m, b = 16x = 112m and c = 32x = 224m

Therefore perimeter = a + b + c = 490m

And s = $\frac{perimeter}{2}$= 245m

Now on applying Heron’s Formula;

$Area\;of\;plot\;ABC=\sqrt{s(s-a)(s-b)(s-c)}$

Therefore, area of plot ABC:

=$\sqrt{245(245-154)(245-112)(245-224)}$

(or)     =$\sqrt{245\times 91\times 133\times 21}$

 = 7891.13m² (Approx.)

Therefore, area of triangular plot ABC = 7891.13 m² (Approx.)

 

Q.7: The length of equal sides of an isosceles triangle ABC is 8cm and its perimeter is 28cm. Find the area of a triangle ABC.

Sol.

Given perimeter = 28cm

Therefore, s = $\frac{perimeter}{2}$ = 14cm

And a = 8cm, b =8cm and c = 12m (Since, perimeter = a + b + c)

On applying Heron’s Formula;

$Area\;of\;Triangle\;ABC=\sqrt{s(s-a)(s-b)(s-c)}$

Therefore, area of Triangle ABC

=$\sqrt{14(14-8)(14-8)(14-12)}$

(or)     =$\sqrt{14\times 6\times 6\times 2}$

= $12\sqrt{7}$

=31.75cm² (Approx.)

Therefore, area of triangle ABC = 31.75cm²(Approx.)