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# NCERT Solutions for Class 9 Maths exercise 12.2 Chapter 12 Heron’s Formula

NCERT Solutions for Class 9 Maths exercise 12.2 Chapter 12 Heron’s Formula are solved here for helping the 9 class students in doing their homework and preparation of the class test, CBSE exams, and completing assignments of the chapter 12 Heron’s Formula. Heron’s formula is related to the area of triangles when all sides are given. The NCERT solutions of the exercise 12.2  chapter 12 Heron’s Formula will help you in clearing doubts on unsolved questions of the class 9 maths NCERT textbook.

Area of Δ = √[s(s -a)(s-b)(s-c)]

Where s is semiperimeter and a,b and c are the sides of the triangle

s = (a + b+ c)/2

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## NCERT Solutions for Class 9 Maths Chapter 12 Heron’s Formula

Exercise 12.1- Heron’s Formula

Exercise 12.2- Heron’s Formula

### NCERT Solutions of class 9 maths

 Chapter 1- Number System Chapter 9-Areas of parallelogram and triangles Chapter 2-Polynomial Chapter 10-Circles Chapter 3- Coordinate Geometry Chapter 11-Construction Chapter 4- Linear equations in two variables Chapter 12-Heron’s Formula Chapter 5- Introduction to Euclid’s Geometry Chapter 13-Surface Areas and Volumes Chapter 6-Lines and Angles Chapter 14-Statistics Chapter 7-Triangles Chapter 15-Probability Chapter 8- Quadrilateral

## NCERT Solutions for Class 9 Maths exercise 12.2 Chapter 12 Heron’s Formula

Q1. A park in the shape of a quadrilateral ABCD has ∠C = 90°, AB = 9 m, BC = 12 m,CD = 5 m and AD = 8 m. How much area does it occupy.

Ans. We are given a quadrilateral ABCD , in which ∠C = 90°, AB = 9 m, BC = 12 m,CD = 5 m and AD = 8 m.

Since ΔBCD is  a right triangle therefore applying  the formula of area of triangle

ar Δ = 1/2(Base) ×Altitude

arΔ BCD = 1/2(DC) × (BC) = 1/2(5)(12) = 30 cm²

Determining the length of BD by applying the Pythagoras formula

BD = √(12² + 5²) = 13 cm

Now let’s find the area of the ΔABD by applying the Heron’s formula

$\fn_cm ar\Delta =\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}$

a = 8 cm, b = 9 cm, c =13 cm

s = (a +b+c)/2 = (8 +9+13)/2 =15 cm

$ar\Delta ABD=\sqrt{15\left ( 15-8 \right )\left ( 15-9 \right )\left ( 15-13 \right )}$

$=\sqrt{3\times 5\times 7\times 2\times 3\times 2}$

= 6√35 cm² =6 ×5.92 ≈35.5

Therefore area of ABCD = Area of BCD + Area of  ABD = 30 +35.5 = 65.5 cm²

### NCERT Solutions for Class 9 Maths exercise 12.2 Chapter 12 Heron’s Formula

Q2.Find the area of quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.

Ans. We are given a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm,CD = 4 cm and AD = 5 cm and AC = 5 cm

In ΔABC, AB = 3 cm, BC = 4 cm and AC = 5 cm

AC² = BC ² + AB² ⇒ 5² = 4² + 3²

Therefore ΔABC is a right triangle

Since ΔABC is  a right triangle therefore applying  the formula of area of triangle

ar Δ = 1/2(Base) ×Altitude

arΔ BCD = 1/2(AB) × (BC) = 1/2(3)(4) = 6 cm²

In ΔADC,we have CD = 4 cm, DA = 5 cm and AC = 5 cm.

Now let’s find the area of the ΔADC by applying the Heron’s formula

$\fn_cm ar\Delta =\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}$

a = 4 cm, b = 5 cm, c =5 cm

s = (a +b+c)/2 = (4 +5+5)/2 =7 cm

$\fn_cm ar\Delta ADC=\sqrt{7\left ( 7-4 \right )\left ( 7-5\right )\left ( 7-5 \right )}$

$\fn_cm =\sqrt{7\times 3\times 2\times 2}$

= 2√21 = 2 × 4.58 = 9.16 cm²

Therefore area of ABCD = Area of ABC + Area of  ADC = 6 +9.16 = 15.16 cm²

### NCERT Solutions for Class 9 Maths exercise 12.2 Chapter 12 Heron’s Formula

Q3.Radha made a picture of an aeroplane with the help of a coloured paper as shown in fig,12.5. Find the total area of the paper used.

Ans.  Figure (I) is a isoscell triangular and (II) is rectangular section , the sides of triangle (I) are  5 cm,5cm and 1 cm

Fig 12.5 is following

Now let’s find the area of  (I)  by applying the Heron’s formula

$\fn_cm ar\Delta =\sqrt{s\left ( s-a \right )\left ( s-b \right )\left ( s-c \right )}$

a = 5 cm, b = 5 cm, c =1 cm

s = (a +b+c)/2 = (5 +5+1)/2 =5.5 cm

$\fn_cm ar(I)=\sqrt{5.5\left ( 5.5-5 \right )\left ( 5.5-5 \right )\left ( 5.5-1 \right )}$

= √ (5.5×0.5 × 0.5× 4.5)= 1/100 √(55×5×5×45 =1/100√(11 ×5 ×5×5×9 ×5) =75√11/100=75×3.32/100 =249/100 =2.49 cm²

Figure (II) is a rectangle of the sides 1 cn and  6.5 cm ,area of figure (II) = 1 ×6.5 = 6.5 cm²

Figure (III)  is a trapizium with parallel sides 1 cm and 2 cm, which contains a parallelogram with each sides 1 cm  and and anequilateral triangle  with each sides 1 cm

Area of figure (III) =1/2(sum of parallel sides)× height

Hight of  the trapizium = Height of triangle = h

Area of triangular section = a²√3/4 = 1²√3/4= √3/4 cm ²

Area of Δ = 1/2(Base)× Height = 1/2 (1)h =h/2

h/2 = √3/4 ⇒ h = √3/2  cm

Therefore ar (III) = (1/2)(1+2) ×√3/2 = 3√3/4 = (3 ×1.732)/4 ≈1.3cm²

Figure (IV) and (V) are right triangles

Area of figure (IV) = Area of figure (V) = (1/2 )×6×1.5 = 4.5 cm²

Area of aeroplane = 2.49 + 6.5 +1.3 +4.5 +4.5 =19.29≈19.3 cm²

### NCERT Solutions for Class 9 Maths exercise 12.2 Chapter 12 Heron’s Formula

Q4. Triangle and 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.

Ans. We are given the sides of a ΔDCE, DC = 28 cm, EC = 26 cm, DE = 30 cm

Also given EC = 28 cm common base of triangle DCE and quadrilateral ABCD and

area of ΔDCE = ar ABCD(parallelogram)

√[s(s -a)(s-b)(s-c)] = (1/2) Base ×Height

s = (a +b+ c)/2 = (26 + 28 +30)/2 = 84/2 = 42 cm

√[42(42 -26)(42-28)(42-30)] =  28 ×h

28 ×h = √[42×16×14×12]

28h = √[3×7×2×16×2×7×2×2×3]

28 h = 3 ×4×2×2×7 =336

h = 12

Hence height of parallelogram is 12 cm

### NCERT Solutions for Class 9 Maths exercise 12.2 Chapter 12 Heron’s Formula

Q5.A rhombus field has green grass for 18 cows to graze.If each sides of rhombus is 30 m and longer diagonal is 48 m.How much area of grass field will each cow be getting ?

Ans. We are given a rhombus ABCD ,whose each side is 30 cm and longer diagonal BD = 48 m

In ΔBCD,we have BC =30 m, CD = 30 m and BD = 48 m

ar Δ  = √[s(s -a)(s-b)(s-c)]

s = (a +b+ c)/2 = (30 + 30 +48)/2 =10 8/2 = 54 m

ar ΔBCD  = √[54(54 -30)(54-30)(54-48)]

= √[54×24×24×6] =√[6×9×6×4×6×4×6] = 6×6×4×3 =432

ar ABCD = 2ar ΔBCD  ( Diagonal bisect the parallelogram in two equal parts)

= 2 × 432 = 864 m

The number of cows grazing in the field = 18

Therefore each cow be getting the area to graze in the field = 864/18 =48 m²

### NCERT Solutions for Class 9 Maths exercise 12.2 Chapter 12 Heron’s Formula

Q6. An umbrella is made by stitching 10 triangular pieces of cloths 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.

Ans. It is given to us that an umbrella is made of 10 triangular pieces of the sizze 20 cm,50 cm and 50 cm

ar Δ  = √[s(s -a)(s-b)(s-c)]

s = (a +b+ c)/2 = (20 + 50 +50)/2 =120/2 = 60 m

ar Δ  = √[60(60 -20)(60-50)(60-50)]

=√[60×40×10×10]=√(6×10×4×10×10×10) = 2×10×10√6 =200√6

Area of each triangular  piece  is = 200√6 cm²

There are 10 triangular  pieces  in which 5 pieces are of the same colour a 5 are of another colour

Therefore cloth of each colour is required for the umbrella = 5× 200√6 =1000√6 cm²

### NCERT Solutions for Class 9 Maths exercise 12.2 Chapter 12 Heron’s Formula

Q7.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?

Ans. We are given a  kite of the shape of a square with 32 cm diagonal

(I )shade is a an isosceles triangle of the height 32/2 = 16 cm and base 32 cm

∴Area of I shade = (1/2) ×32 ×16 = 256 cm²

Area of II shade = Area of I shade (diagonal of the square divides the square in two equal parts)

Area of II shade = 256 cm²

III shade is an isisceles triangle of the base 8 cm and sides 6 cm,6 cm

ar Δ  = √[s(s -a)(s-b)(s-c)]

s = (a +b+ c)/2 = (6 + 6+8)/2 =20/2 = 10 cm²

ar (III)  = √[10(10 -6)(10-6)(10-8)]

= √[5×2×4×4×2] =8√5 cm²

### NCERT Solutions for Class 9 Maths exercise 12.2 Chapter 12 Heron’s Formula

Q8.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 the fig.12.18) Find the cost of polishing the tiles at the rate of 50p per cm2 .

Ans. The given triangular tiles are 16, with sides 9 cm,28 cm and 35 cm

Area of a tile is computed by the Heron’s formula

ar Δ  = √[s(s -a)(s-b)(s-c)]

s = (a +b+ c)/2 = (9 + 28+35)/2 =72/2 = 36 cm²

=√[36(36 -9)(36-28)(36-35)]

=√[36×27×8×1]= =√[36×9×3×4×2]=6×3×2√6 = 36√6 =36×2.449≈88.2cm²

Therefore area of 16 tiles = 16× 88.2 =1411.2 cm²

The rate of polishing = 50 p/cm² = Rs 0.50 /cm²

Hence cost of polishing the tiles are = 0.50 × 1411.2 = Rs 705.6

### NCERT Solutions for Class 9 Maths exercise 12.2 Chapter 12 Heron’s Formula

Q9.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.

Ans. We are given a trapizium ABCD in which parallel sides are 25 m and 10 m and non-parallel sides are 14 m and 13 m

Drawing CE parallel to AD and  CF ⊥ BE

We get a parallelogram ADCE and ΔBCE

Considering the ΔBCE in which BE = AB -AE = AB – DC= 25 – 10 = 15 m

Applyig the Heron’s formula for the area of ΔBCE

ar Δ  = √[s(s -a)(s-b)(s-c)]

s = (a +b+ c)/2 = (13 + 14+15)/2 =42/2 = 21 cm

ar Δ BCE = √[21(21 -13)(21-14)(21-15)]

= √[21×8×7×6]=√[7×3×4×2×7×2×3] =7×2×3×2 = 84 m²

We also can get area of Δ BCE  = (1/2) ×BE ×CF = (1/2)×15 CF

(1/2)×15 CF = 84

CF = (2×84)/15 =11.2 m

Since ABCD is a trapizium

Area of a trapizium = (1/2) [10 + 25]×11.2 =(1/2)×35×11.2 = 196 m²

Hence area of the field is 196 cm²

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