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**Exercise 12.2****1. A park, in the shape of a quadrilateral ABCD, has âˆ C = 90Âº, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?**

**Answer**

âˆ C = 90Âº, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m

BD is joined.

In Î”BCD,

By applying Pythagoras theorem,

BD^{2} = BC^{2 }+ CD^{2 }^{ }

â‡’ BD^{2} = 12^{2 }+ 5^{2 }

â‡’ BD^{2} = 169

â‡’ BD = 13 m

Area of Î”BCD = 1/2 Ã— 12 Ã— 5 = 30 m^{2}

Now,

Semi perimeter of Î”ABD(s) = (8 + 9 + 13)/2 m = 30/2 m = 15 m

Using heron's formula,

Area of quadrilateral ABCD = Area of Î”BCD + Area of Î”ABD = 30 m^{2 }+^{ }35.5m^{2} = 65.5m^{2 }

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

AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm

In Î”ABC,

By applying Pythagoras theorem,

AC^{2} = AB^{2 }+ BC^{2 }^{ }

â‡’ 5^{2} = 3^{2 }+ 4^{2 }

â‡’ 25 = 25

Thus, Î”ABC is a right angled at B.

Area of Î”BCD = 1/2 Ã— 3 Ã— 4 = 6 cm^{2}

Now,

Semi perimeter of Î”ACD(s) = (5 + 5 + 4)/2 cm = 14/2 cm = 7 m

Using heron's formula,

Area of quadrilateral ABCD = Area of Î”ABC + Area of Î”ABD = 6 cm^{2 }+^{ }9.17 cm^{2} = 15.17 cm^{2 } **3. Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.**

**Answer**

Length of the sides of the triangle section I = 5cm, 1cm and 5cm

Perimeter of the triangle = 5 + 5 + 1 = 11cm

Semi perimeter = 11/2 cm = 5.5cm

Using heron's formula,

Length of the sides of the rectangle of section I = 6.5cm and 1cm

Area of section II = 6.5 Ã— 1 cm^{2} =^{ }6.5 cm^{2}

Section III is an isosceles trapezium which is divided into 3 equilateral of side 1cm each.

Area of the trapezium = 3 Ã— âˆš3/4 Ã— 1^{2 }cm^{2 }= 1.3 cm^{2 }(approx)

Section IV and V are 2 congruent right angled triangles with base 6cm and height 1.5cm

Area of region IV and V = 2 Ã— 1/2 Ã— 6 Ã— 1.5cm^{2 }= 9cm^{2}

Total area of the paper used = (2.488 + 6.5 + 1.3 + 9)cm^{2 }=^{ }19.3 cm^{2}**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.****Answer**

Given,

Area of the parallelogram and triangle are equal.

Length of the sides of the triangle are 26 cm, 28 cm and 30 cm.

Perimeter of the triangle = 26 + 28 + 30 = 84 cm

Semi perimeter of the triangle = 84/2 cm = 42 cm

Using heron's formula,

Area of parallelogram = Area of triangle

28cm Ã— h = 336 cm^{2}

â‡’^{ }h = 336/28 cm

â‡’^{ }h = 12 cm

The height of the parallelogram is 12 cm.**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? ****Answer**

Diagonal AC divides the rhombus ABCD into two congruent triangles of equal area.

Semi perimeter of Î”ABC = (30 + 30 + 48)/2 m = 54 m

Using heron's formula,

Area of field = 2 Ã— area of the Î”ABC = (2 Ã— 432)m^{2 }= 864 m^{2}

Thus,

Area of grass field which each cow will be getting = 864/18^{ }m^{2 }=^{ }48 m^{2}**6. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig.12.16), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?**

**Answer**

Semi perimeter of each triangular piece = (50 + 50 + 20)/2 cm = 120/2 cm = 60cm

Using heron's formula,**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 Fig. 12.17. How much paper of each shade has been used in it?**

**Answer**

We know that,

As the diagonals of a square bisect each other at right angle.

Area of given kite = 1/2 (diagonal)^{2}

= 1/2 Ã— 32 Ã— 32 = 512 cm^{2}

Area of shade I = Area of shade II

â‡’ 512/2 cm^{2 }= 256^{ }cm^{2}

So, area of paper required in each shade = 256 cm^{2}

For the III section,

Length of the sides of triangle = 6cm, 6cm and 8cm

Semi perimeter of triangle = (6 + 6 + 8)/2 cm = 10cm

Using heron's formula,

**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 Fig. 12.18). Find the cost of polishing the tiles at the rate of 50p per cm ^{2} .**

**Answer**

Semi perimeter of the each triangular shape = (28 + 9 + 35)/2 cm = 36 cm

Using heron's formula,

Total area of 16 tiles = 16 Ã— 88.2 cm^{2 }= 1411.2 cm^{2}Cost of polishing tiles = 50p per cm^{2}

Total cost of polishing the tiles = Rs. (1411.2 Ã— 0.5) = Rs. 705.6**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.****Answer**

Let ABCD be the given trapezium with parallel sides AB = 25m and CD = 10mand the non-parallel sides AD = 13m and BC = 14m.

CM âŠ¥ AB and CE || AD.

In Î”BCE,

BC = 14m, CE = AD = 13 m and

BE = AB - AE = 25 - 10 = 15m

Semi perimeter of the Î”BCE = (15 + 13 + 14)/2 m = 21 m

Using heron's formula,

also, area of the Î”BCE = 1/2 Ã— BE Ã— CM = 84 m^{2 }

â‡’ 1/2 Ã— 15 Ã— CM = 84 m^{2 }

â‡’ CM = 168/15 m^{2 }

â‡’ CM = 56/5 m^{2 }

Area of the parallelogram AECD = Base Ã— Altitude = AE Ã— CM = 10 Ã— 84/5 = 112 m^{2 }

Area of the trapezium ABCD = Area of AECD + Area of Î”BCE = (112+ 84) m^{2 }= 196 m^{2 }