Question

NCERT Solutions for class 9 maths​

Answer 1

☞︎︎︎ Given :

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• Side of equilateral △ = a units

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⇒ a = b = c

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↣ Now,

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• Formula

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

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↣ Here,

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• a = a

• b = a

• c = a

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

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

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• Area of triangle

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☆ Herons Formula

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

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

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

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

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

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

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

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• (derived Formula)

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

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➢ Perimeter of board = 180cm

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⇒ a + a + a = 180cm

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⇒ 3a = 180cm

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• Transposing The Terms

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

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

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

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

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

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

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

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⇒ area = 900√3cm²

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Area = 901.73 cm²

Answer 2

It is given that:

X

Equating the corresponding elements of the two matrices, we have:

a+4c=−7

b+4d=2

2a+5c=−8

2b+5d=4

3a+6c=−9

3b+6d=6

Now, a+4c=−7⇒a=−7−4c

∴2a+5c=−8⇒−14−8c+5c=−8

⇒−3c=6⇒c=−2

∴a=−7−4(−2)=−7+8=1

Now, b+4d=2⇒b=2−4d

∴2b+5d=4⇒4−8d+5d=4

⇒−3d=0

⇒d=0

∴b=2−4(0)=2

Thus, a=1,b=2,c=−2,d=0

Hence, the required matrix X is [

1

2

−2

0

]