English, asked by rai215215, 9 months ago

Like rootless weeds, the hair torn around their pallor.
The tall girl with her weigh down head,
The paper seeming boy, with rat's eyes.
Questions. (POEM)
a) Name the poet who wrote this poem ? (1marks)
b) Where do you think, are these children sitting ?(2 marks)
e) How do the faces and hair of these children look? (2 marks)
d) Why is the head of the tall girl weigh down? (2marks)
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Answers

Answered by TheValkyrie
3

Question:

Name the poet who wrote this poem

Answer:

This poem "An elementary school classroom in a slum" is written by the poet Stephen Slender.

Question:

Where do you think are these children sitting?

Answer:

The school is located in a slum and the children sits on the floor of the classroom. The classroom walls are painted in a dull colour which represent the bleak future of the children studying there. The children look outside and dream of a happier and comfortable life.

Question:

How do the faces and hair of these children look?

Answer:

The children are malnourished and most of them have stunted growth.  The poet compares them to rootless weeds. The poet describes a girl having a pallor (pale) face with hair torn aroud it. A boy is paper seeming which describes his weighlessness. Another boy is described as having stunted ad twisted bones inherited from his father.

Question:

Why is the head of the tall girl weighed down?

Answers:

The weighed down head of the tall girl represents the miseries and burdens she has to suffer in her life. She no longer finds joy and is overburdened with sorrows at such an early age. The girl is ill and exhausted and cannot find the energy to keep her head up.

More information about the poem:

This poem written by Stephen Splender talks about the miseries that children living in a slum faces. They do not get a proper education and are deprived of many of thheir basic right. They often suffer from malnutrition and does not show physical and emotional development.  They live in an environment of utter hopelessness. There is none for them to turn to and share their sorrows. The outside colourful world does not hold any meaning for these children. The poet urges thhe authorities to make the lives of these children happier by providing them good education, food and housing facilities. He urges them to ushackle the bond of poverty and ensure equal rights and opportunites for people living in a slum.

Answered by abdulrubfaheemi
0

Answer:

Answer:

Explanation:

\Large{\underline{\underline{\it{Given:}}}}

Given:

\sf{\dfrac{tan\:A}{sec\:A-1} -\dfrac{sin\:A}{1+cos\:A} =2\:cot\:A}

secA−1

tanA

1+cosA

sinA

=2cotA

\Large{\underline{\underline{\it{To\:Prove:}}}}

ToProve:

LHS = RHS

\Large{\underline{\underline{\it{Solution:}}}}

Solution:

→ Taking the LHS of the equation,

\sf{LHS=\dfrac{tan\:A}{sec\:A-1} -\dfrac{sin\:A}{1+cos\:A} }LHS=

secA−1

tanA

1+cosA

sinA

→ Applying identities we get

=\sf{\dfrac{\dfrac{sin\:A}{cos\:A} }{\dfrac{1}{cos\:A}-1 } -\dfrac{sin\:A}{1+cos\:A} }=

cosA

1

−1

cosA

sinA

1+cosA

sinA

→ Cross multiplying,

=\sf{\dfrac{\dfrac{sin\:A}{cos\:A} }{\dfrac{1-cos\:A}{cos\:A} } -\dfrac{sin\:A}{1+cos\:A} }=

cosA

1−cosA

cosA

sinA

1+cosA

sinA

→ Cancelling cos A on both numerator and denominator

=\sf{\dfrac{sin\:A}{1-cos\:A} -\dfrac{sin\:A}{1+cos\:A}}=

1−cosA

sinA

1+cosA

sinA

→ Again cross multiplying we get,

=\sf{\dfrac{sin\:A(1+cos\:A)-sin\:A(1-cos\:A)}{(1+cos\:A)(1-cos\:A)}}=

(1+cosA)(1−cosA)

sinA(1+cosA)−sinA(1−cosA)

→ Taking sin A as common,

\sf{=\dfrac{sin\:A[1+cos\:A-(1-cos\:A)]}{(1^{2}-cos^{2}\:A ) }}=

(1

2

−cos

2

A)

sinA[1+cosA−(1−cosA)]

\sf{=\dfrac{sin\:A[1+cos\:A-1+cos\:A]}{sin^{2}\:A } }=

sin

2

A

sinA[1+cosA−1+cosA]

→ Cancelling sin A on both numerator and denominator

\sf{=\dfrac{2\:cos\:A}{sin\:A} }=

sinA

2cosA

\sf=2\times \dfrac{cos\:A}{sin\:A} }

\sf{=2\:cot\:A}=2cotA

=\sf{RHS}=RHS

→ Hence proved.

\Large{\underline{\underline{\it{Identitites\:used:}}}}

Identititesused:

\sf{tan\:A=\dfrac{sin\:A}{cos\:A} }tanA=

cosA

sinA

\sf{sec\:A=\dfrac{1}{cos\:A} }secA=

cosA

1

\sf{(a+b)\times(a-b)=a^{2}-b^{2} }(a+b)×(a−b)=a

2

−b

2

\sf{(1-cos^{2}\:A)=sin^{2} \:A}(1−cos

2

A)=sin

2

A

\sf{\dfrac{cos\:A}{sin\:A}=cot\:A}

sinA

cosA

=cotA

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