Question

Ncert Solutions for class 9 maths Herons Formula

ex1

q1​

Answer 1

Step-by-step explanation:

☞︎︎︎ Given :

$\:$

Side of equilateral △ = a units

$\:$

⇒ a = b = c

$\:$

↣ Now,

$\:$

Formula

$\:$

$\large \rm \bigstar \: \: s = \dfrac{a + b + c}{2}$

$\:$

↣ Here,

$\:$

a = a

b = a

c = a

$\:$

$\large \rm \implies \: \: s = \dfrac{a +a + a}{2}$

$\:$

$\large \rm \implies \: \: s = \dfrac{3a}{2}$

$\:$

Area of triangle

$\:$

☆ Herons Formula

$\:$

$\large \boxed{\mathtt{ = \sqrt{s(s - a)(s - b)(s - c)} }}$

$\:$

$\large \rm⇒ \: \sqrt{ \dfrac{3a}{2} \left( \dfrac{3a}{2} - a\right)\left( \dfrac{3a}{2} - a\right)\left( \dfrac{3a}{2} - a\right)}$

$\:$

$\large \rm⇒ \: \sqrt{ \dfrac{3a}{2} \left( \dfrac{3a - 2a}{2} \right)\left( \dfrac{3a - 2a}{2} \right)\left( \dfrac{3a - 2a}{2} \right)}$

$\:$

$\large \rm⇒ \: \sqrt{ \dfrac{3a}{2} \left( \dfrac{a}{2} \right)\left( \dfrac{a}{2} \right)\left( \dfrac{ a}{2} \right)}$

$\:$

$\large \rm⇒ \: \sqrt{ \dfrac{3a}{2} \times \dfrac{a}{2} \times \dfrac{a}{2} \times \dfrac{ a}{2} }$

$\:$

$\large \rm⇒ \: \sqrt{ \dfrac{3{a }^{4} }{16} }$

$\:$

$\: \: \: \: \: \large \rm \therefore\: { \dfrac{ \sqrt{ 3} \: {a }^{2} }{4} }$

$\:$

(derived Formula)

$\:$

$: \implies \large \rm area = { \dfrac{ \sqrt{ 3} \: {a }^{2} }{4} }$

$\:$

➢ Perimeter of board = 180cm

$\:$

⇒ a + a + a = 180cm

$\:$

⇒ 3a = 180cm

$\:$

Transposing The Terms

$\:$

$\large \rm \: ⇒ \: a = \dfrac{180}{3}$

$\:$

$\large \rm \: ⇒ \: a = \dfrac{ \cancel{180} \: \: 60}{ \cancel3}$

$\:$

$\large \rm \therefore \: a = 60cm$

$\:$

$\: \: \: \: : \mapsto \: \: \: \large \rm area = { \dfrac{ \sqrt{ 3} \: {a }^{2} }{4} }$

$\:$

$\: \: : \implies \large \rm area = { \dfrac{ \sqrt{ 3} \: {(60) }^{2} }{4} }$

$\:$

$\: \: : \implies \large \rm area = { \dfrac{ \sqrt{ 3} \: \times 60 \times 60 }{4} }$

$\:$

$: \implies \large \rm area = { \dfrac{ \sqrt{ 3} \: \times 60 \times \cancel{60} 15}{ \cancel4} }$

$\:$

⇒ area = 900√3cm²

$\:$

Area = 901.73 cm²

Answer 2

Answer:

For a quadrilateral, when one of its diagonal value and the sides are given, the area can be calculated by splitting the given quadrilateral into two triangles and use the Heron's formula. Example :A park, in the shape of a quadrilateral ABCD, has ∠C=90∘, AB = 9 cm, BC = 12 cm, CD = 5 cm and AD = 8 cm.