**
1. A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with
side a. Find the area of the signal board, using heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?
**

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We know that, an equilateral triangle has equal sides.

So, all sides are equal to a.

So, Perimeter of triangle = 180 cm (Given)

i.e., a + a + a = 180

i.e., 3a = 180

Thus, a = 60cm

So, \(s = \frac{a + a + a}{2}\) ...(Since, 2s = a + b + c)

i.e., \(s = \frac{3a}{2} = \frac{180}{2}\)

Therefore, s = 90cm

We also, know that, Area of an equilateral triangle = \(\sqrt{s(s - a)(s - a)(s - a)}\) ...(Since, Heron's formula = \(\sqrt{s(s - a)(s - b)(s - c)}\))

So, Area = \(\sqrt{90(90 - 60)(90 - 60)(90 - 60)}\)

i.e., = \(\sqrt{90 × 30 × 30 × 30}\)

i.e., = \(30 × 30\sqrt{3}\)

Therefore, Area of an equilateral triangle = \(900\sqrt{3} {cm}^2\))

**
2. The triangular side walls of a flyover have been used for advertisements. The sides of
the wails are 122 m, 22 m and 120 m (see figure). The advertisements yield earnings of Rs. 5000 per \({m}^2\) per year. A company hired one of its walls for 3 months. How much rent did it pay?
**

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Let a = 122 m, b = 22 m, c = 120 m

Also, we have,

\(b^2 + c^2 = (22)^2 + (120)^2 = 484 + 14400 = 14884 = (122)^2 = a^2\)

Thus, we observe that the side walls are in right triangular shape.

Thus, the area of the triangular side walls = \(\frac{1}{2} × a × c\)

i.e., = \(\frac{1}{2} × 22 × 120\)

i.e., = \(11 × 120 = 1320 m^2\)

We know that, yearly rent = Rs.5000 per \(m^2\)

Therefore, yearly rent = Rs.5000 × \(\frac{1}{2} per {m^}2\)

Now, the company has hired one of its walls for 3 months.

Thus, rent paid by the company for 3 months = 1320 × \(\frac{5000}{12}\) × 3

= 110 × 5000 × 3

= Rs. 1650000

Therefore, rent paid by the company for 3 months = Rs. 1650000.

**
3. There is a slide in a park. One of its side walls has been painted in same colour with a message “KEEP THE PARK GREEN AND CLEAN” (see figure). If the sides of the wall are 15
m, 11 m and 6 m, find the area painted in colour.
**

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The given figure formed a triangle whose sides are: a = 15m,b = 11m, c = 6m

\(s = \frac{15 + 11 + 6}{2}\) ...(Since, 2s = a + b + c)

i.e., \(s = \frac{32}{2}\)

Therefore, s = 16m

Therefore, area painted in colour = \(\sqrt{16(16 - 15)(16 - 11)(16 - 6)}\) ...(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

So, Area = \(\sqrt{16 × 1 × 5 × 10}\)

i.e., = \(\sqrt{2 × 2 × 2 × 2 × 5 × 5 × 2}\)

i.e., = \(\sqrt{400 × 2} {m}^2\)

Therefore, the area painted in colour is \(20\sqrt{2} m^2\))

**
4. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is
42 cm.
**

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Let the sides of a triangle a = 18 cm, b = 10 cm and c

We have, perimeter = 42 cm

So, a + b + c = 42

By substituting the values,

18 + 10 + c = 42

28 + c = 42cm

c = 42 - 28 cm

Thus, c = 14cm

Now, we know that, \(s = \frac{a + b + c}{2}\)

Therefore, \(s = \frac{18 + 10 + 14}{2} = \frac{42}{2} = 21cm\)

Now, Area of triangle = \(\sqrt{21(21 - 18)(21 - 10)(21 - 14)}\) ...(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{21 × 3 × 11 × 7}\)

= \(\sqrt{7 × 3 × 3 × 11 × 7}\)

= \(21\sqrt{11} {cm}^2\)

**
5. Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area.
**

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Suppose that the sides in cm, are 12x, 17x and 25x.

Then, we know that 12x + 17x + 25x = 540 ...(Given, Perimeter of the triangle)

54x = 540

Thus, x = 10

So, the sides of the triangle are 12 × 10cm, 17 × 10cm, 25 × 10cm i.e., 120cm, 170cm and 250cm.

Now, we know that, \(s = \frac{a + b + c}{2}\)

Therefore, \(s = \frac{120 + 170 + 250}{2} = \frac{540}{2} = 270cm\)

Now, Area of triangle = \(\sqrt{270(270 - 120)(270 - 170)(270 - 250)}\) ...(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{270 × 150 × 100 × 20}\)

= \(\sqrt{27 × 10 × 3 × 15 × 10 × 100 × 20}\)

= \(100\sqrt{27 × 15 × 10 × 2} {cm}^2\)

= \(100\sqrt{9 × 3 × 3 × 5 × 10 × 2} {cm}^2\)

= \(100\sqrt{9 × 3 × 3 × 10 × 10} {cm}^2\)

= 100 × 10 × 9

Therefore, area of the given triangle is 9000\({cm}^2\)

**
6. An isosceles triangle has perimeter 30 cm arid each of the equal sides is 12 cm. Find
the area of the triangle.
**

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sol: Let \(\triangle{ABC}\) be an isosceles triangle, in which,

We have, AB = AC = 12cm ...(Given)

Now, AB + AC + BC = 30 ...(Given, perimeter = 30cm)

i.e.,
12 + 12 + BC = 30

i.e., BC = 30 - 24

So, BC = 6cm

Now, we know that, \(s = \frac{a + b + c}{2}\)

Therefore, \(s = \frac{30}{2} = 15cm\)

Now, Area of triangle = \(\sqrt{15(15 - 12)(15 - 6)(15 - 12)}\) ...(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{15 × 3 × 9 × 3}\)

= \(\sqrt{3 × 5 × 9 × 9}\)

= \(9\sqrt{15} {cm}^2\)

**
1. A park, in the shape of a quadrilateral ABCD, has \(\angle{C}\) = \(90^\circ\) , AB = 9m, BC = 12m, CD = 5m and AD = 8m. How much area does it occupy?
**

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In the quadrilateral ABCD,

we have, right angled triangle \(\triangle{BCD}\),

We have, \({BD}^2 = {BC}^2 + {CD}^2\) ...(By Pythagoras theorem)

i.e., = \({12}^2 + {5}^2 = 144 + 25 = 169\)

i.e., \({BD}^2 = {13}^2\)

i.e., BD = 13m

Also, In \(\triangle{ABD}\), we have, AB = 9m, BD = 13m, DA = 8m

Now, we know that, \(s = \frac{a + b + c}{2}\)

Therefore, \(s = \frac{9 + 13 + 8}{2} = \frac{30}{2} = 15m\)

Now, Area of triangle = \(\sqrt{15(15 - 9)(15 - 13)(15 - 8)}\) ...(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{15 × 6 × 2 × 7}\)

= \(\sqrt{3 × 5 × 3 × 2 × 2 × 7}\)

= \(3 × 2\sqrt{5 × 7} {m}^2\)

= \(6\sqrt{35} {m}^2\) = 6 × 5.9\({m}^2\) = 35.4\({m}^2\)(approx) ...(i)

Since, \(\triangle{BCD}\) is an right angled triangle,

Area of \(\triangle{BCD}\) = \(frac{1}{2}\) × BC × CD ...(Since, Area of triangle = \(frac{1}{2}\) × Base × Height)

= \(frac{1}{2} × 12 × 5 = 30{m}^2\) ...(ii)

Area of quadrilateral ABCD = Area of \(\triangle{ABD}\) + Area of \(\triangle{BCD}\)

Hence from (i) and (ii),

Area of quadrilateral ABCD = \(35.4{m}^2 + 30{m}^2 = 65.4{m}^2\)

**
2. Find the area of a quadrilateral ABCD in which AB = 3cm, BC = 4cm, CD = 4cm, DA = 5cm and AC = 5cm.
**

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In the quadrilateral ABCD,

we have, \(\triangle{ABC}\),

We have, AB = 3cm, BC = 4cm, CA = 5cm

We have, \({AC}^2 = {AB}^2 + {BC}^2\) ...(By Pythagoras theorem)

i.e., = \({3}^2 + {4}^2 = 9 + 16 = 25\)

i.e., \({AC}^2 = {5}^2\)

i.e., AC = 5cm

Hence, \(\triangle{ABC}\) is a right triangle ...(i)

Now in \(\triangle{ABD}\), we have, AC = 5cm, CD = 4cm, DA = 5cm

We know that, \(s = \frac{a + b + c}{2}\)

Therefore, \(s = \frac{5 + 4 + 5}{2} = \frac{14}{2} = 7cm\)

Now, Area of triangle = \(\sqrt{7(7 - 5)(7 - 4)(7 - 5)}\) ...(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{7 × 2 × 3 × 2}\)

= \(2\sqrt{7 × 3} {cm}^2\)

= \(2\sqrt{21} {cm}^2\) = 2 × 4.6\({cm}^2\) = 9.2\({cm}^2\)(approx) ...(ii)

Since, \(\triangle{ABC}\) is an right angled triangle, ...(from(i))

Area of \(\triangle{ABC}\) = \(\frac{1}{2}\) × AB × BC ...(Since, Area of triangle = \(\frac{1}{2}\) × Base × Height)

= \(\frac{1}{2} × 3 × 4 = 6{cm}^2\) ...(iii)

Area of quadrilateral ABCD = Area of \(\triangle{ABC}\) + Area of \(\triangle{ACD}\)

Hence from (ii) and (iii),

Area of quadrilateral ABCD = \(9.2{cm}^2 + 6{m}^2 = 15.2{cm}^2\)

**
3. Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the
total area of the paper used.
**

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For part I :

It is a triangle with sides 5 cm, 5 cm and 1 cm.

Thus, We know that, \(s = \frac{a + b + c}{2}\)

Therefore, \(s = \frac{5 + 5 + 1}{2} = \frac{11}{2}cm\)

Now, Area of part I triangle = \(\sqrt{\frac{11}{2}(\frac{11}{2} - 5)(\frac{11}{2} - 5)(\frac{11}{2} - 1)}\) ...(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{\frac{11}{2} × \frac{1}{2} × \frac{1}{2} × \frac{9}{2}}\)

= \(\frac{3}{4}\sqrt{11} {cm}^2\)

= \(\frac{3}{4} × 3.31 {cm}^2\) = 3 × 0.829\({cm}^2\) = 2.487\({cm}^2\)(approx)

For part II :

It is a rectangle with sides 6.5 cm and 1 cm

Therefore, Area of part II = 6.5 × 1 ...(Since, Area of rectangle = Lenght × Breadth)

= 6.5\({cm}^2\)

For part III :

It is a trapezium ABCD.

\(\triangle{EBC}\) is an equilateral with side 1 cm.

Therefore, Area of \(\triangle{EBC}\) = \(\frac{1}{2}\) × EB × CF = \(\frac{\sqrt{3}}{4} × {1}^2\)

...(Since, Area of triangle = \(frac{1}{2}\) × Base × Height and Area of equilateral triangle = \(frac{\sqrt{3}}{4} × (side)^2\))

i.e.,
\(\frac{1}{2}\) × 1 × CF = \(\frac{\sqrt{3}}{4}\)

i.e., CF = \(\frac{\sqrt{3}}{2}cm\)

Now, Area of trapezium = \(\frac{1}{2}\) × Sum of parallel sides × Height

= \(\frac{1}{2}\) × (AB + CD) × CF

= \(\frac{1}{2}\) × (2 + 1) × \(\frac{\sqrt{3}}{2}\)

= \(\frac{3}{4}\) × \(\sqrt{3}\)

= 3 × 0.433

= 1.299 \({cm}^2\)

Therefore, Area of part III is 1.299 \({cm}^2\)

For part IV & V :

Both the parts IV and V are the same.

i.e.,it is a right triangle with sides 6 cm and 1.5 cm.

Area of part IV & V = \(\frac{1}{2}\) × 1.5 × 6 = \(\frac{9}{2}\) = 4.5 \({cm}^2\)

So, we get,

Total area of paper used = Area of (I + II + III + IV + V)

= (2.487 + 6.5 + 1.299 + 4.5 + 4.5)\({cm}^2\)

= 19.286\({cm}^2\)

= 19.3\({cm}^2\)(approx.)

**
4. A triangle and a parallelogram have the same base and the same area. If the sides of
the triangle are 26 cm, 28 cm and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
**

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Let ABC be a triangle with sides AB = 26 cm, BC = 28 cm, CA = 30 cm

Now, we know that, \(s = \frac{AB + BC + CA}{2}\)

Therefore, \(s = \frac{26 + 28 + 30}{2} = \frac{84}{2} = 42cm\)

Now, Area of triangle = \(\sqrt{42(42 - 26)(42 - 28)(42 - 30)}\) ...(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{42 × 16 × 14 × 12}\)

= \(\sqrt{7 × 6 × 4 × 4 × 2 × 7 × 2 × 6}\)

= \(7 × 6 × 4 × 2 {cm}^2\)

= \(336 {cm}^2\) ...(i)

We know that,

Area of parallelogram = Base × Height ...(ii)

We also have, Area of parallelogram = Area of \(\triangle{ABC}\) ...(Given)

Thus, from (i) and (ii),

Base × Height = 336

28 × Height = 336 ...(Given)

Height = 12cm

Therefore, the height of the parallelogram is 12cm.

**
5. A rhombus shaped field has green grass for 18 cows to graze. If each side of the
rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting?
**

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Let ABCD be a rhombus.

Area of the rhombus ABCD = 2 × area of \(\triangle{ABD}\) ... (i)

As we know that, in a rhombus, diagonals divides it in two equal parts.

In \(\triangle{ABD}\), we have, AB = 30 m, BD = 48 m, DA=30 m

Now, we know that, \(s = \frac{AB + BD + AD}{2}\)

Therefore, \(s = \frac{30 + 48 + 30}{2} = \frac{108}{2} = 54m\)

Now, Area of triangle = \(\sqrt{54(54 - 30)(54 - 48)(54 - 30)}\) ...(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{54 × 24 × 6 × 24}\)

= \(\sqrt{9 × 6 × 4 × 6 × 6 × 4 × 6}\)

= \(3 × 6 × 6 × 4 {m}^2\)

= \(432 {m}^2\)

Thus, from Equation (i),

Area of rhombus = 2 × 432m = 864 m

Also, Number of cows = 18

So, therefore, Area of grass field per cow = \(\frac{864}{18}\) = \(48 {m}^2\).

**
6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
**

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Each triangular piece in an umbrella is an isosceles triangle with sides 50 cm, 50 cm, 20 cm.

Now, we know that, \(s = \frac{a + b + c}{2}\)

Therefore, \(s = \frac{50 + 50 + 20}{2} = \frac{120}{2} = 60cm\)

Now, Area of triangle = \(\sqrt{60(60 - 50)(60 - 50)(60 - 20)}\) ...(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{60 × 10 × 10 × 40}\)

= \(\sqrt{3 × 2 × 10 × 10 × 10 × 10 × 2 × 2}\)

= \(2 × 100\sqrt{6} {cm}^2\)

= \(200\sqrt{6} {cm}^2\)

Thus, from Equation (i),

Area of each triangular piece = 2 × 432 cm = 864 cm

Since, there are 10 triangular pieces, out of which 5 & 5 are of different colours.

Hence, total area of cloth of each colour = \(5 × 200\sqrt{6} {cm}^2\) = \(1000\sqrt{6} {cm}^2\)

**
7. A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in figure. How much paper of each shade has been used in it?
**

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Since, the kite is in the shape of a square.

Each diagonal of square = 32 cm ...(Given)

We know that, the diagonals of a square bisect each other at right angle.

For Part I :

Area of Part I = \(frac{1}{2}\) × Base × Height = \(frac{1}{2}\) × 32 × 16

= 16 × 16 = 256\({cm}^2\)

For Part II :

Area of Part II = \(frac{1}{2}\) × Base × Height = \(frac{1}{2}\) × 32 × 16

= 16 × 16 = 256\({cm}^2\)

For Part III :

It is a triangle with sides 6 cm, 6 cm and 8 cm.

Now, we know that, \(s = \frac{a + b + c}{2}\)

Therefore, \(s = \frac{6 + 6 + 8}{2} = \frac{20}{2} = 10cm\)

Now, Area of triangle = \(\sqrt{10(10 - 6)(10 - 6)(10 - 8)}\) ...(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{10 × 4 × 4 × 2}\)

= \(\sqrt{5 × 2 × 4 × 4 × 2}\)

= \(2 × 4\sqrt{5} {cm}^2\)

= \(8\sqrt{5} {cm}^2\)

= 8 × 2.24 cm = 17.92 \({cm}^2\)

Hence, the area of colour used for Paper I = 256 \({cm}^2\)

The area of colour used for Paper II = 256 \({cm}^2\)

The area of colour used for Paper III = 17.92 \({cm}^2\)

**
8. A floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm (see figure). Find the cost polishing the tiles at the rate of Rs.50 paise per \({cm}^2\).
**

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Given, the sides of a triangular tile are 9 cm, 28 cm and 35 cm.

For each triangular tile, we have, \(s = \frac{a + b + c}{2}\)

Therefore, \(s = \frac{9 + 28 + 35}{2} = \frac{72}{2} = 36cm\)

Now, Area of triangle = \(\sqrt{36(36 - 9)(36 - 28)(36 - 35)}\) ...(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{36 × 27 × 8 × 1}\)

= \(\sqrt{6 × 6 × 9 × 3 × 4 × 2 × 1}\)

= \(6 × 3 × 2\sqrt{6} {cm}^2\)

= \(36\sqrt{6} {cm}^2\)

Therefore, Total area of 16 such titles = 16 × \(36\sqrt{6}\) = 16 × 36 × 2.45 \({cm}^2\) = 1411.20 \({cm}^2\)

Total cost of ploshing the titles at the rate of Rs.50 paise per \({cm}^2\) =

= Rs. \(\frac{50}{100} × 1411.20\) = Rs.705.60

**
9. A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non - parallel sides are 14 m and 13 m. Find the area of the field.
**

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Here, ABCD is a trapezium and AB || DC.

Here, ABCD is a trapezium and CE || DA and \(CF\perp{AB}\).

Now, AB = 25 m, BC = 14 m, CD = 10 m, DA = 13 m, AE = 10 m and CE = 13m

So, EB = 25 - 10 = 15 m

Now, for \(\triangle{EBC}\),

we have, \(s = \frac{a + b + c}{2}\)

Therefore, \(s = \frac{15 + 14 + 13}{2} = \frac{42}{2} = 21 m\)

Now, Area of triangle = \(\sqrt{21(21 - 15)(21 - 14)(21 - 13)}\) ...(Since, Heron's formula [area = \(\sqrt{s(s - a)(s - b)(s - c)}\)])

= \(\sqrt{21 × 6 × 7 × 8}\)

= \(\sqrt{7 × 3 × 3 × 2 × 7 × 4 × 2}\)

= \(7 × 3 × 4 {m}^2\)

= \(84 {cm}^2\)

Also, Area of triangle \(\triangle{EBC}\) = \(frac{1}{2}\) × Base × Height = \(\frac{1}{2}\) × 15 × CF

Therefore, \(frac{1}{2}\) × 15 × CF = 84

i.e., CF = \(\frac{84 × 2}{15} = \frac{168}{15} = 11.2m\)

Now, area of the trapezium ABCD = \(\frac{1}{2}\) × (Sum of parallel sides) × (Distance between parallel sides)

= \(\frac{1}{2}\) × (AB + CD) × CF

= \(\frac{1}{2}\) × (25 + 10) × 11.2

= \(\frac{1}{2}\) × 35 × 11.2

= 35 × 5.65

Therefore, the area of the field is 196 \({m}^2\).