Question

NCERT Solutions for class 9 maths Herons Formula.


Q1)Students of a school staged a rally for cleanliness campaign. They walked through the lanes in two groups. One group walked through the lanes AB, BC and CA; while the other through AC, CD and DA . Then they cleaned the area enclosed within their lanes. If AB = 9 m, BC = 40 m, CD = 15 m, DA = 28 m and ∠B = 90º. Find the total area cleaned by the students (neglecting the width of the lanes).


Q2) Sanya has a piece of land which is in the shape of a rhombus
produce different crops. She divided the land in two equal parts. If the perimeter of. She wants her one daughter and one son to work on the land and the land is 400 m and one of the diagonals is 160 m, how much area each of them will
for their crops?​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

☞︎︎︎ Solution :

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↬ Q1)

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↣ Since -

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• AB = 9m

• BC = 40m

• ∠ B = 90°

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➝ We have -

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$\large ➢ \: \: \mathtt{AC = \sqrt{ {9}^{2} + {40}^{2} } \: m}$

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$\large \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{= \sqrt{ {81}^{} + {1600}^{} } \: m}$

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$\large \: \: \: \: \: \: \: \: \: \: \: \: \mathtt{ = \sqrt{{1600 }} \: m}$

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$\large \boxed{ \therefore \mathtt{ \: AC = 41m}}$

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• ABC = right angled triangle

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➦ Finding Area of △ ABC

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$\large ➝ \: \boxed{ \mathtt{area = \dfrac{1}{2} \times base \times height}}$

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$\large \: \: \: \: \: \: \: : ⇒ \: \: \: \: \: \mathtt{ \dfrac{1}{2} \times 40 \times 9}$

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$\large \: \: \: \: \: \: \: : ⇒ \: \: \: \: \: \mathtt{ \dfrac{1}{ \cancel2} \times \cancel{ 40} \: \: 20 \times 9}$

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$\large :⇒ \mathtt{ \sqrt{42(42 - 41)(42 - 15)(42 - 28)} }$

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$\large \: \: \: \: \: \: \: : ⇒ \: \: \: \boxed{ \mathtt{area = 180 {m}^{2} }}$

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• ACD = Scalene Triangle

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↬ Here,

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$\large ↣ \: \: \mathtt{s = \dfrac{41 + 15 + 28}{2} }$

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$\large ↣ \: \: \mathtt{s = \dfrac{ \cancel{84} \: \: 42}{ \cancel2} }$

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$\: \: \: \: \: \: \: \: \: \: \: \: \: \large \therefore \: \: \underline{ \underline \mathtt{s = 42m}}$

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☞ Finding Area by Herons Formula :

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

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$\large : ⇒ \mathtt{ \sqrt{42(42 - 41)(42 - 15)(42 - 28)} }$

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$\: \: \: \: \: \: \large: ⇒ \: \mathtt{ \sqrt{42 \times 1 \times 27 \times 14} \: m {}^{2} }$

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$\large \therefore \underline{\boxed{ \large \mathtt{ area= 126 {m}^{2} }}}$

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☛ Group 1 cleaned more area by 180 - 126 = 54m²

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☛ Total Area = 306m²

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↬ Q2)

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⇀ Given :

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• Perimeter = 400m

• Diagonal = 160m

• So side = 400 ÷ 4 = 100m

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$\large ↣ \: \: \mathtt{s = \dfrac{100 + 100 + 160}{2} }$

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☆ Herons Formula :

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$\large \mathtt{ :⇒ \: \sqrt{180(180 - 100)(180 - 100)(180 - 160} }$

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$\large \: \: \: \: \mathtt{ : ⇒ \: \: \sqrt{180 \times 80 \times 80 \times 20} } \\ \\ \\\large \: \: \: \: \mathtt{ : ⇒ \: \: \sqrt{23040000}}$

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$\boxed{\large \mathtt{ \therefore \: \: \underline{area = 4800 {m}^{2} }}}$

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• Alternative Method :

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$\large \mathtt{ \: \: \: \: \: \: \: \: \: : \: \: bd = 160m } \\ \\ \large \mathtt{ \: \: \: \: \: \: \: \: \: : \: \: de= 160m }$

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$\large \: \: \: \mathtt{ : ⇒ \: de = 160 \div 2 = \underline{80 {m}^{2}} }$

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$\large \mathtt{ \: \: \: \: \: \: \: \: \: : \: \: {de}^{2} + {ce}^{2} = {dc}^{2} } \\ \\ \large \mathtt{ \: \: \: \: \: \: \: \: \: : \: \: ce= \sqrt{ {dc}^{2} - {de}^{2} } }$

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$\large \mathtt{ \: \: \: \: \: \: \: : ⇒ \: \: ce= \sqrt{ {100}^{2} - {80}^{2} } }$

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$\large \mathtt{ \: \: \: \: \: \: \: : ⇒ \: \: ce= \sqrt{ {10000} - {6400} } } \\ \\ \large \mathtt{ \: \: \: \: \: \: \: : ⇒ \: \: ce= \sqrt{ {3600} } }$

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$\large \mathtt{ \therefore \: \: \underline{ce = 60m}}$

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• Finding Area :

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☞︎︎︎ Formula -

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$\large \mathtt{ \mapsto \boxed{ \mathtt{area = \dfrac{1}{2} \times base \times height }}}$

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$\large \mathtt{ \: \: \: : ⇒ \: \: \: \dfrac{1}{2} \times 160 \times 60 } \\ \\ \\ \large \mathtt{ \: \: \: : ⇒ \: \: \: \dfrac{1}{ \cancel2} \times \cancel{160} \: \: 80 \times 60 }$

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$\large \mathtt{ \: \: \: \: : ⇒ \: \: \: 80 \times 60}$

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$\underline{ \boxed{ \large \mathtt{ \therefore \: area = 4800 {m}^{2} }}}$