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**SEBA Class 9 Mathematics Chapter 12 Herons Formula**

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**Solutions****SEBA Class 9 Mathematics Chapter 12 Herons Formula****Herons Formula**

**Herons Formula****Chapter – 12**

Exercise 12.1 |

**Q.1. A traffic board, indicating ‘SCHOOL AHEAD’, is an equilat- eral 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?**

Ans: ‘a’ = a

‘b’ =a

‘c’ = a

∴ Area of the signal board

Perimeter = 180 cm

∴ Area of the signal board

Alternatively,

Area of the signal board

**Q.2. The triangular side walls of a flyover have been used for advertisement. The sides of the walls are 1200 m, 22m and 120 m (see figure). The advertisements yield an earning of Rs. 5000 per m ^ 2 per year. A company hired one of its walls for 3 months. How much rent did it pay?**

Ans: a = 122m

b = 22m

c = 120m

1 year = 12 months

∵ Rent for 12 months per m²= Rs. 5000

∴ Rent for 3 months of 1320 m²= Rs(1250x 1320) =Rs.1650000.

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

Ans: a = 15

b = 11 m

c = 6 m

∴ Area painted in colour

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

Ans: a = 18cm

b = 10 cm

Perimeter = 42 cm

⇒ a + b + c = 42 ⇒ 18 + 10 + c = 42

⇒ 28 + c = 42 ⇒ c = 42 – 28

c = 14 cm

∴ Area of the triangle

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

Ans: Let the sides of the triangle be 12k, 17k cm. Then,

Perimeter = 12k + 17k + 25k = 54k

According to the question,

54k = 540

∴ a = 12k = 12 x 10 = 120 cm

b = 17k = 17 x 10 = 170 cm

c = 25k = 25 x 10 = 250 cm

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

Ans: a = 12 cm

b = 12 cm

Perimeter = 30 cm

⇒ a+b+c=30 ⇒ 12 + 12 + c = 30

⇒ 24+c=30 ⇒ c = 30 – 24

⇒ c = 6 cm

∴ Area of the triangle

**Q.7. Find the area of an equilateral triangle whose perimeter is 24 cm.**

Ans: Perimeter of equilateral triangle 24 cm

∴ Each side of triangle

∴ Area of triangle

**Q.8. The length of hypotenuse of a right angled triangle is 13 and the length of another side is 5cm. Find its area.**

Ans: Let, ABC is a right angle triangle. ∠B = 90⁰, then BC = 5cm,

AC = 13cm.

∴ AB² = AC² – BC = 13² – 5² = 169 – 25 = 144

∴ AB = √144 = 12 cm

So, area of the triangle

Exercise 12.2 |

**Q.1. A park, in the shape of a quadrilateral ABCD, has ∠C = 90⁰, AB = 9m BC = 12m, CD = 5m and AD = 8m. How much area does it oc- cupy?**

Ans: Join BD

Area of right triangle BCD

In right triangle BCD,

BD² = BC² + CD² | By Pythagoras Theorem

= (12)² + (5)² = 144 + 25 = 169

⇒ BD = √169 = 13m

For ∆ABD

a = 13m,

b = 8m,

c = 9m

∴ Area of the AABD

∴ Area of the quadrilateral ABCD

= Area of ∆BCD + Area of ∆ABD

= 30m²+ 35.5m² = 65.5m² (approx.)

Hence the park occupies an area of 65.5 m² (approx.)

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

Ans: For ΔΑΒC

a = 4cm,

b = 5cm,

c = 3cm

∴ a² + c² = b²

∴ ∆ABC is right angled with ∠B = 90⁰

∴ Area of right triangle ABC

For ∆ACD

a = 4cm

b = 5cm

c = 5cm

∴ Area of the ∆ACD

∴ Area of the quadrilateral ABCD

= Area of ∆BCD + area of ∆ACD

= 6cm² + 9.2cm² = 15.2 cm². (Approx.)

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

Ans: For Triangular Area I

a = 5cm

b = 5cm

c = 1 cm

∴ Area I =

Area II = 6.5×1=6.5 cm²

For Area III

∴ Total area of the paper used

= Area I + Area II+Area III + Area IV + Area V

= 2.5 cm² +6.5 cm² + 1.3 cm² +4.5 cm² + 4.5 cm² = 19.3 cm²

**Q.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 parallelo- gram.**

Ans: For triangle

a = 26cm

b = 28cm

c = 30cm

= 6 x 4 x 7 x 2 = 336 cm²

Let the height of the parallelogram be h cm.

Then, area of the parallelogram

= Base x Height = 28 x h cm²

According to the question,

Hence the height of parallelogram is 12 cm.

**Q.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.**

Ans: For ∆ABC

a = 30 m,

b = 48m,

c = 30 m.

∴** **Area of the rhombus

= 2 Area of ∆ABC = 2 x 432 = 864 m²

Aliter: Draw BE⊥AC. Then E is the midpoint of AC.

In right triangle AEB,

AB²= AE²+ BE² | By Pythagoras Theorem

⇒ (30)² = (24)²+ BE²

⇒ 900 = 576 + BE²

⇒ BE²= 900 – 576

⇒ BE² = 324

⇒ BE = √(324) = 18m

∴ BD = 2 BE = 2×18 = 36m

∴ Area of rhombus ABCD = Area of right ∆AEB

= 864 m² ∴ Area of rhombus ABCD

= 4 area of ∆AEB

= 4 x 216 = 864 m²

∴ Area of grass for 18 cows = 864 m²

**Q.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?**

Ans: For one triangular piece

a = 20 cm,

b = 50 cm,

c = 50 cm.

∴ Area of one triangle =

∴ Area of five triangles of one colour

Hence 1000 √6 cm² of each colour is required for the umbrella.

**Q.7. A kite in the shape of a square with a diagonal 32 cm and as isosceles triangle of base 8 cm and side 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?**

Ans: For one tile

a = 9 cm,

b = 28 cm,

c = 35cm.

∴ Area of one tile

Cost of polishing the tiles at the rate of 30 per cm²

**Q.8. A field is in the shape of a trapezium whose parallel sides are 25 m and 10m. **

The non-parallel sides are 14 m and 13 m. Find the area of the field.

Ans: Let the given field be in the shape of a trapezium ABCD in which AB = 25 m CD = 10m BC = 13 m and AD = 14m.

From D, draw DE || BC meeting

AB at E. Also, draw DF⊥AB

∴ DE = BC = 13m

AE = AB – EB = AB – DC

= 25 – 10 = 15m

For ∆AED

a = 14m

b = 13m

c = 15m

∴ Area of the ∆AED

∴ Area of parallelogram EBCD = Base x Height

∴ Area of the field

= Area of ∆AED + Area of parallelogram EBCD

= 84 m² + 112 m² = 196 m²