{"id":545640,"date":"2021-10-05T04:52:07","date_gmt":"2021-10-05T04:52:07","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=545640"},"modified":"2021-10-05T09:52:38","modified_gmt":"2021-10-05T09:52:38","slug":"rd-sharma-solutions-for-class-9-maths-chapter-12-herons-formula","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-9-maths-chapter-12-herons-formula\/","title":{"rendered":"RD Sharma Solutions for Class 9 Maths Chapter 12\u2013Heron’s Formula"},"content":{"rendered":"\n

Class 9: Maths Chapter 12 solutions. Complete Class 9 Maths Chapter 12 Notes.<\/p>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 12\u2013Heron’s Formula<\/h2>\n\n\n\n

RD Sharma 9th Maths Chapter 12, Class 9 Maths Chapter 12 solutions<\/p>\n\n\n\n

Exercise 12.1 Page No: 12.8<\/h4>\n\n\n\n

Question 1: Find the area of a triangle whose sides are respectively 150 cm, 120 cm and 200 cm.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know, Heron\u2019s Formula<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Here, a = 150 cm<\/p>\n\n\n\n

b = 120 cm<\/p>\n\n\n\n

c = 200 cm<\/p>\n\n\n\n

Step 1: Find s<\/p>\n\n\n\n

s = (a+b+c)\/2<\/p>\n\n\n\n

s = (150+200+120)\/2<\/p>\n\n\n\n

s = 235 cm<\/p>\n\n\n\n

Step 2: Find Area of a triangle<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 8966.56<\/p>\n\n\n\n

Area of triangle is 8966.56 sq.cm.<\/p>\n\n\n\n

Question 2: Find the area of a triangle whose sides are respectively 9 cm, 12 cm and 15 cm.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

We know, Heron\u2019s Formula<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Here, a = 9 cm<\/p>\n\n\n\n

b = 12 cm<\/p>\n\n\n\n

c = 15 cm<\/p>\n\n\n\n

Step 1: Find s<\/p>\n\n\n\n

s = (a+b+c)\/2<\/p>\n\n\n\n

s = (9 + 12 + 15)\/2<\/p>\n\n\n\n

s = 18 cm<\/p>\n\n\n\n

Step 2: Find Area of a triangle<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 54<\/p>\n\n\n\n

Area of triangle is 54 sq.cm.<\/p>\n\n\n\n

Question 3: Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given:<\/p>\n\n\n\n

a = 18 cm, b = 10 cm, and perimeter = 42 cm<\/p>\n\n\n\n

Let c be the third side of the triangle.<\/p>\n\n\n\n

Step 1: Find third side of the triangle, that is c<\/p>\n\n\n\n

We know, perimeter = 2s,<\/p>\n\n\n\n

2s = 42<\/p>\n\n\n\n

s = 21<\/p>\n\n\n\n

Again, s = (a+b+c)\/2<\/p>\n\n\n\n

Put the value of s, we get<\/p>\n\n\n\n

21 = (18+10+c)\/2<\/p>\n\n\n\n

42 = 28 + c<\/p>\n\n\n\n

c = 14 cm<\/p>\n\n\n\n

Step 2: Find area of triangle<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 4: In a triangle ABC, AB = 15cm, BC = 13cm and AC = 14cm. Find the area of triangle ABC and hence its altitude on AC.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

Let the sides of the given triangle be AB = a, BC = b, AC = c respectively.<\/p>\n\n\n\n

Here, a = 15 cm<\/p>\n\n\n\n

b = 13 cm<\/p>\n\n\n\n

c = 14 cm<\/p>\n\n\n\n

From Heron\u2019s Formula;<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 84<\/p>\n\n\n\n

Area = 84 cm2<\/sup><\/p>\n\n\n\n

Let, BE is a perpendicular on AC<\/p>\n\n\n\n

Now, area of triangle = \u00bd x Base x Height<\/p>\n\n\n\n

\u00bd \u00d7 BE \u00d7 AC = 84<\/p>\n\n\n\n

BE = 12cm<\/p>\n\n\n\n

Hence, altitude is 12 cm.<\/p>\n\n\n\n

Question 5: The perimeter of a triangular field is 540 m and its sides are in the ratio 25:17:12. Find the area of the triangle.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Let the sides of a given triangle be a = 25x, b = 17x, c = 12x respectively,<\/p>\n\n\n\n

Given, Perimeter of triangle = 540 cm<\/p>\n\n\n\n

2s = a + b + c<\/p>\n\n\n\n

a + b + c = 540 cm<\/p>\n\n\n\n

25x + 17x + 12x = 540 cm<\/p>\n\n\n\n

54x = 540 cm<\/p>\n\n\n\n

x = 10 cm<\/p>\n\n\n\n

So, the sides of a triangle are<\/p>\n\n\n\n

a = 250 cm<\/p>\n\n\n\n

b = 170 cm<\/p>\n\n\n\n

c = 120 cm<\/p>\n\n\n\n

Semi perimeter, s = (a+b+c)\/2<\/p>\n\n\n\n

= 540\/2<\/p>\n\n\n\n

= 270<\/p>\n\n\n\n

s = 270 cm<\/p>\n\n\n\n

From Heron\u2019s Formula;<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 9000<\/p>\n\n\n\n

Hence, the area of the triangle is 9000 cm2 <\/sup>.<\/p>\n\n\n\n

\n\n\n\n

Exercise 12.2 Page No: 12.19<\/h4>\n\n\n\n

Question 1: Find the area of the quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Area of the quadrilateral ABCD = Area of \u25b3ABC + Area of \u25b3ADC \u2026.(1)<\/p>\n\n\n\n

\u25b3ABC is a right-angled triangle which B.<\/p>\n\n\n\n

Area of \u25b3ABC = 1\/2 x Base x Height<\/p>\n\n\n\n

= 1\/2\u00d7AB\u00d7BC<\/p>\n\n\n\n

= 1\/2\u00d73\u00d74<\/p>\n\n\n\n

= 6<\/p>\n\n\n\n

Area of \u25b3ABC = 6 cm2<\/sup> \u2026\u2026(2)<\/p>\n\n\n\n

Now, In \u25b3CAD,<\/p>\n\n\n\n

Sides are given, apply Heron\u2019s Formula.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Perimeter = 2s = AC + CD + DA<\/p>\n\n\n\n

2s = 5 cm + 4 cm + 5 cm<\/p>\n\n\n\n

2s = 14 cm<\/p>\n\n\n\n

s = 7 cm<\/p>\n\n\n\n

Area of the \u25b3CAD = 9.16 cm2<\/sup> \u2026(3)<\/p>\n\n\n\n

Using equation (2) and (3) in (1), we get<\/p>\n\n\n\n

Area of quadrilateral ABCD = (6 + 9.16) cm2<\/sup><\/p>\n\n\n\n

= 15.16 cm2<\/sup>.<\/p>\n\n\n\n

Question 2: The sides of a quadrilateral field, taken in order are 26 m, 27 m, 7 m, 24 m respectively. The angle contained by the last two sides is a right angle. Find its area.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Here,<\/p>\n\n\n\n

AB = 26 m, BC = 27 m, CD = 7 m, DA = 24 m<\/p>\n\n\n\n

AC is the diagonal joined at A to C point.<\/p>\n\n\n\n

Now, in \u25b3ADC,<\/p>\n\n\n\n

From Pythagoras theorem;<\/p>\n\n\n\n

AC2<\/sup> = AD2<\/sup> + CD2<\/sup><\/p>\n\n\n\n

AC2 <\/sup>= 142 <\/sup>+ 72<\/sup><\/p>\n\n\n\n

AC = 25<\/p>\n\n\n\n

Now, area of \u25b3ABC<\/p>\n\n\n\n

All the sides are known, Apply Heron\u2019s Formula.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Perimeter of \u25b3ABC= 2s = AB + BC + CA<\/p>\n\n\n\n

2s = 26 m + 27 m + 25 m<\/p>\n\n\n\n

s = 39 m<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 291.84<\/p>\n\n\n\n

Area of a triangle ABC = 291.84 m2<\/sup><\/p>\n\n\n\n

Now, for area of \u25b3ADC, (Right angle triangle)<\/p>\n\n\n\n

Area = 1\/2 x Base X Height<\/p>\n\n\n\n

= 1\/2 x 7 x 24<\/p>\n\n\n\n

= 84<\/p>\n\n\n\n

Thus, the area of a \u25b3ADC is 84 m2<\/sup><\/p>\n\n\n\n

Therefore, Area of rectangular field ABCD = Area of \u25b3ABC + Area of \u25b3ADC<\/p>\n\n\n\n

= 291.84 m2<\/sup> + 84 m2<\/sup><\/p>\n\n\n\n

= 375.8 m2<\/sup><\/p>\n\n\n\n

Question 3: The sides of a quadrilateral, taken in order as 5, 12, 14, 15 meters respectively, and the angle contained by first two sides is a right angle. Find its area.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Here, AB = 5 m, BC = 12 m, CD =14 m and DA = 15 m<\/p>\n\n\n\n

Join the diagonal AC.<\/p>\n\n\n\n

Now, area of \u25b3ABC = 1\/2 \u00d7AB\u00d7BC<\/p>\n\n\n\n

= 1\/2\u00d75\u00d712 = 30<\/p>\n\n\n\n

Area of \u25b3ABC is 30 m2<\/sup><\/p>\n\n\n\n

In \u25b3ABC, (right triangle).<\/p>\n\n\n\n

From Pythagoras theorem,<\/p>\n\n\n\n

AC2<\/sup> = AB2<\/sup> + BC2<\/sup><\/p>\n\n\n\n

AC2<\/sup> = 52 <\/sup>+ 122<\/sup><\/p>\n\n\n\n

AC2<\/sup> = 25 + 144 = 169<\/p>\n\n\n\n

or AC = 13<\/p>\n\n\n\n

Now in \u25b3ADC,<\/p>\n\n\n\n

All sides are know, Apply Heron\u2019s Formula:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Perimeter of \u25b3ADC = 2s = AD + DC + AC<\/p>\n\n\n\n

2s = 15 m +14 m +13 m<\/p>\n\n\n\n

s = 21 m<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 84<\/p>\n\n\n\n

Area of \u25b3ADC = 84 m2<\/sup><\/p>\n\n\n\n

Area of quadrilateral ABCD = Area of \u25b3ABC + Area of \u25b3ADC<\/p>\n\n\n\n

= (30 + 84) m2<\/sup><\/p>\n\n\n\n

= 114 m2<\/sup><\/p>\n\n\n\n

Question 4: A park in the shape of a quadrilateral ABCD, has \u2220 C = 900<\/sup>, AB = 9 m, BC = 12 m, CD = 5 m, AD = 8 m. How much area does it occupy?<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Here, AB = 9 m, BC = 12 m, CD = 5 m, DA = 8 m.<\/p>\n\n\n\n

And BD is a diagonal of ABCD.<\/p>\n\n\n\n

In right \u25b3BCD,<\/strong><\/p>\n\n\n\n

From Pythagoras theorem;<\/p>\n\n\n\n

BD2<\/sup> = BC2<\/sup> + CD2<\/sup><\/p>\n\n\n\n

BD2 <\/sup>= 122 <\/sup>+ 52<\/sup> = 144 + 25 = 169<\/p>\n\n\n\n

BD = 13 m<\/p>\n\n\n\n

Area of \u25b3BCD = 1\/2\u00d7BC\u00d7CD<\/p>\n\n\n\n

= 1\/2\u00d712\u00d75<\/p>\n\n\n\n

= 30<\/p>\n\n\n\n

Area of \u25b3BCD = 30 m2<\/sup><\/p>\n\n\n\n

Now, In \u25b3ABD,<\/strong><\/p>\n\n\n\n

All sides are known, Apply Heron\u2019s Formula:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Perimeter of \u25b3ABD = 2s = 9 m + 8m + 13m<\/p>\n\n\n\n

s = 15 m<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 35.49<\/p>\n\n\n\n

Area of the \u25b3ABD = 35.49 m2<\/sup><\/p>\n\n\n\n

Area of quadrilateral ABCD = Area of \u25b3ABD + Area of \u25b3BCD<\/p>\n\n\n\n

= (35.496 + 30) m2<\/sup><\/p>\n\n\n\n

= 65.5m2<\/sup>.<\/p>\n\n\n\n

Question 5: Two parallel sides of a trapezium are 60 m and 77 m and the other sides are 25 m and 26 m. Find the area of the trapezium.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Given: AB = 77 m , CD = 60 m, BC = 26 m and AD = 25m<\/p>\n\n\n\n

AE and CF are diagonals.<\/p>\n\n\n\n

DE and CF are two perpendiculars on AB.<\/p>\n\n\n\n

Therefore, we get, DC = EF = 60 m<\/p>\n\n\n\n

Let\u2019s say, AE = x<\/p>\n\n\n\n

Then BF = 77 \u2013 (60 + x)<\/p>\n\n\n\n

BF = 17 \u2013 x \u2026(1)<\/p>\n\n\n\n

In right \u25b3ADE<\/strong>,<\/p>\n\n\n\n

From Pythagoras theorem,<\/p>\n\n\n\n

DE2<\/sup> = AD2 <\/sup>\u2212 AE2<\/sup><\/p>\n\n\n\n

DE2 <\/sup>= 252 <\/sup>\u2212 x2 <\/sup>\u2026.(2)<\/p>\n\n\n\n

In right \u25b3BCF<\/strong><\/p>\n\n\n\n

From Pythagoras theorem,<\/p>\n\n\n\n

CF2 <\/sup>= BC2 <\/sup>\u2212 BF2<\/sup><\/p>\n\n\n\n

CF2 <\/sup>= 262 <\/sup>\u2212 (17\u2212x) 2<\/sup>[Uisng (1)]<\/p>\n\n\n\n

Here, DE = CF<\/p>\n\n\n\n

So, DE2<\/sup> = CF2<\/sup><\/p>\n\n\n\n

(2) \u21d2 252<\/sup> \u2212 x2 <\/sup>= 262 <\/sup>\u2212 (17\u2212x)2<\/sup><\/p>\n\n\n\n

625 \u2212 x2 <\/sup>= 676 \u2013 (289 \u221234x + x2<\/sup>)<\/p>\n\n\n\n

625 \u2212 x2 <\/sup>= 676 \u2013 289 +34x \u2013 x2<\/sup><\/p>\n\n\n\n

238 = 34x<\/p>\n\n\n\n

x =7<\/p>\n\n\n\n

(2) \u21d2 DE2<\/sup> = 252<\/sup> \u2013 (7)2<\/sup><\/p>\n\n\n\n

DE2<\/sup> = 625\u221249<\/p>\n\n\n\n

DE = 24<\/p>\n\n\n\n

Area of trapezium = 1\/2\u00d7(60+77)\u00d724 = 1644<\/p>\n\n\n\n

Area of trapezium is 1644 m2<\/sup> (Answer)<\/p>\n\n\n\n

Question 6: Find the area of a rhombus whose perimeter is 80 m and one of whose diagonal is 24 m.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n\n\n

Perimeter of a rhombus = 80 m (given)<\/p>\n\n\n\n

We know, Perimeter of a rhombus = 4\u00d7side<\/p>\n\n\n\n

Let a be the side of a rhombus.<\/p>\n\n\n\n

4\u00d7a = 80<\/p>\n\n\n\n

or a = 20<\/p>\n\n\n\n

One of the diagonal, AC = 24 m (given)<\/p>\n\n\n\n

Therefore OA = 1\/2\u00d7AC<\/p>\n\n\n\n

OA = 12<\/p>\n\n\n\n

In \u25b3AOB,<\/p>\n\n\n\n

Using Pythagoras theorem:<\/p>\n\n\n\n

OB2<\/sup> = AB2 <\/sup>\u2212 OA2<\/sup> = 202 <\/sup>\u2212122 <\/sup>= 400 \u2013 144 = 256<\/p>\n\n\n\n

or OB = 16<\/p>\n\n\n\n

Since diagonal of rhombus bisect each other at 90 degrees.<\/p>\n\n\n\n

And OB = OD<\/p>\n\n\n\n

Therefore, BD = 2 OB = 2 x 16 = 32 m<\/p>\n\n\n\n

Area of rhombus = 1\/2\u00d7BD\u00d7AC = 1\/2\u00d732\u00d724 = 384<\/p>\n\n\n\n

Area of rhombus = 384 m2<\/sup>.<\/p>\n\n\n\n

Question 7: A rhombus sheet, whose perimeter is 32 m and whose diagonal is 10 m long, is painted on both the sides at the rate of Rs 5 per m2<\/sup>. Find the cost of painting.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Perimeter of a rhombus = 32 m<\/p>\n\n\n\n

We know, Perimeter of a rhombus = 4\u00d7side<\/p>\n\n\n\n

\u21d2 4\u00d7side = 32<\/p>\n\n\n\n

side = a = 8 m<\/p>\n\n\n\n

Each side of rhombus is 8 m<\/p>\n\n\n\n

AC = 10 m (Given)<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Then, OA = 1\/2\u00d7AC<\/p>\n\n\n\n

OA = 1\/2\u00d710<\/p>\n\n\n\n

OA = 5 m<\/p>\n\n\n\n

In right triangle AOB,<\/p>\n\n\n\n

From Pythagoras theorem;<\/p>\n\n\n\n

OB2<\/sup> = AB2<\/sup>\u2013OA2<\/sup> = 82 <\/sup>\u2013 52 <\/sup>= 64 \u2013 25 = 39<\/p>\n\n\n\n

OB = \u221a39 m<\/p>\n\n\n\n

And, BD = 2 x OB<\/p>\n\n\n\n

BD = 2\u221a39 m<\/p>\n\n\n\n

Area of the sheet = 1\/2\u00d7BD\u00d7AC = 1\/2 x (2\u221a39 \u00d7 10 ) = 10\u221a39<\/p>\n\n\n\n

Area of the sheet is 10\u221a39 m2<\/sup><\/p>\n\n\n\n

Therefore, cost of printing on both sides of the sheet, at the rate of Rs. 5 per m2<\/sup><\/p>\n\n\n\n

= Rs. 2 x (10\u221a39 x 5) = Rs. 625.<\/p>\n\n\n\n

\n\n\n\n

Exercise VSAQs Page No: 12.23<\/h4>\n\n\n\n

Question 1: Find the area of a triangle whose base and altitude are 5 cm and 4 cm respectively.<\/strong><\/p>\n\n\n\n

Solution<\/strong>:<\/p>\n\n\n\n

Given: Base of a triangle = 5 cm and altitude = 4 cm<\/p>\n\n\n\n

Area of triangle = 1\/2 x base x altitude<\/p>\n\n\n\n

= 1\/2 x 5 x 4<\/p>\n\n\n\n

= 10<\/p>\n\n\n\n

Area of triangle is 10 cm2<\/sup>.<\/p>\n\n\n\n

Question 2: Find the area of a triangle whose sides are 3 cm, 4 cm and 5 cm respectively.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Given: Sides of a triangle are 3 cm, 4 cm and 5 cm respectively<\/p>\n\n\n\n

Apply Heron\u2019s Formula:<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

S = (3+4+5)\/2 = 6<\/p>\n\n\n\n

Semi perimeter is 6 cm<\/p>\n\n\n\n

Now,<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

= 6<\/p>\n\n\n\n

Area of given triangle is 6 cm2<\/sup>.<\/p>\n\n\n\n

Question 3: Find the area of an isosceles triangle having the base x cm and one side y cm.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

In right triangle APC,<\/p>\n\n\n\n

Using Pythagoras theorem,<\/p>\n\n\n\n

AC2<\/sup> = AP2<\/sup> + PC2<\/sup><\/p>\n\n\n\n

y2<\/sup> = h2<\/sup> + (x\/2)2<\/sup><\/p>\n\n\n\n

or h2<\/sup> = y2<\/sup> \u2013 (x\/2)2<\/sup><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Question 4: Find the area of an equilateral triangle having each side 4 cm.<\/strong><\/p>\n\n\n\n

Solution:<\/strong> Each side of an equilateral triangle = a = 4 cm<\/p>\n\n\n\n

Formula for Area of an equilateral triangle = ( \u221a3\/4 ) \u00d7 a\u00b2<\/p>\n\n\n\n

= ( \u221a3\/4 ) \u00d7 4\u00b2<\/p>\n\n\n\n

= 4\u221a3<\/p>\n\n\n\n

Area of an equilateral triangle is 4\u221a3 cm2<\/sup>.<\/p>\n\n\n\n

Question 5: Find the area of an equilateral triangle having each side x cm.<\/strong><\/p>\n\n\n\n

Solution:<\/strong><\/p>\n\n\n\n

Each side of an equilateral triangle = a = x cm<\/p>\n\n\n\n

Formula for Area of an equilateral triangle = ( \u221a3\/4 ) \u00d7 a\u00b2<\/p>\n\n\n\n

= ( \u221a3\/4 ) \u00d7 x\u00b2<\/p>\n\n\n\n

= x2<\/sup> \u221a3\/4<\/p>\n\n\n\n

Area of an equilateral triangle is \u221a3x2<\/sup>\/4 cm2<\/sup>.<\/p>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 12: Download PDF<\/strong><\/h2>\n\n\n\n

RD Sharma Solutions for Class 9 Maths Chapter 12\u2013Heron’s Formula<\/p>\n\n\n\n

Download PDF<\/strong>: RD Sharma Solutions for Class 9 Maths Chapter 12\u2013Heron’s Formula PDF<\/a><\/p>\n\n\n\n

Chapterwise RD Sharma Solutions for Class 9 Maths :<\/strong><\/h2>\n\n\n\n