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Sia ? 4 years, 5 months ago
My Childhood” is an extract taken from the biography of APJ Abdul Kalam - ‘Wings of Fire’. He was a great scientist and also the 14th President of India. He was born in a middle - class Muslim family in 1931 in the island town of Rameswaram, Tamil Nadu. In his childhood he was greatly influenced by his parents, his teachers and his friends. His father, Jainulabdeen, was not highly educated but he was a generous and kind person. He was not rich but provided a secure childhood to Abdul and his siblings. Abdul inherited honesty and self-discipline from his father and faith in goodness and deep kindness from his mother.Kalam earned his first wages by working as a helping hand to his cousin, Samsuddin, who distributed newspapers in Rameswaram.In his childhood he had three close friends- Ramanadha Sastry,Aravindam and Sivaprakashan. Once when he was in fifth standard, a new teacher asked him not to sit in the front row along with the high caste Brahmin boys. Abdul found Ramanadha Sastry weeping as he went to the last row. This made a lasting impression on Abdul.Abdul was also greatly influenced by his science teacher, Sivasubramania Iyer. He learnt the lesson of breaking social barriers from him. Iyer invited him to his home for a meal. His wife refused to serve food to a Muslim boy in her pure kitchen. Iyer served him with his own hands and sat down beside him to eat his meal. He convinced his wife to serve meal with her own hands and thus was successful in changing her conservative attitude.For higher education, Abdul Kalam sought permission from his father to leave Rameswaram and study at the district headquarters in Ramanathapuram. His father said that Abdul had to go a long way in life just like a seagull bird which flies long distances. He calmed down APJ’s reluctant mother by quoting Kahlil Gibran’s poem ‘Your children’. He said that her children could not be dominated by her because they had their own thoughts. They did not belong to her but were a result of life’s desire for itself. He asked her to give them freedom to fulfil their thoughts.
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ICT skills are about understanding and applying a range of computer programmes, software and other applications. These include: word processing, spreadsheets, databases, power points and search engines.
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Sia ? 4 years, 5 months ago
In {tex}\triangle{/tex}ABC,
BP bisects {tex}\angle{/tex}ABC

{tex}\therefore{/tex} {tex}\angle{/tex}ABP = {tex}\angle{/tex}PBC = {tex}\frac{1}{2}{/tex}{tex}\angle{/tex}B ={tex}\frac{1}{2}{/tex}(2{tex}\angle{/tex}C) = {tex}\angle{/tex}C . . . . (1)
In PBC,
{tex}\therefore{/tex} {tex}\angle{/tex}PBC = {tex}\angle{/tex}PCB (= ∠C)
{tex}\therefore{/tex} PB = PC . . . [Sides opposite to equal angles] . . .(2)
In {tex}\triangle{/tex}APB and {tex}\triangle{/tex}DPC,
AB = CD . . . [Given]
PB = PC . . .[From (2)]
{tex}\angle{/tex}ABP = {tex}\angle{/tex}DCP (= {tex}\angle{/tex}C)
{tex}\therefore{/tex} {tex}\triangle{/tex}APB = {tex}\triangle{/tex}DPC . . . . [By SAS property]
{tex}\therefore{/tex} {tex}\angle{/tex}BAP = CDP (= {tex}\angle{/tex}A) . . .[c.p.c.t.] . . . (3)
and AP = DP . . . [c.p.c.t.] . . . .(4)
In {tex}\triangle{/tex}APD,
As AP = DC . . . [From (4)]
{tex}\therefore{/tex} {tex}\angle{/tex}PDA = {tex}\angle{/tex}PAD = {tex}\frac{A}{2}{/tex}
{tex}\therefore{/tex} {tex}\angle{/tex}DPA = {tex}\pi-\left(\frac{A}{2}+\frac{A}{2}\right)=\pi-A{/tex} . . . . (5)
From {tex}\angle{/tex}DPC
{tex}\angle{/tex}DPC = {tex}\pi{/tex} – (A + C)
{tex}\therefore{/tex} {tex}\angle{/tex}DPA = {tex}\pi{/tex} – {tex}\angle{/tex}DPC = {tex}\pi{/tex} – [{tex}\pi{/tex} – (A + C)] = A + C . . .(6)
From (5) and (6)
{tex}\pi{/tex} – A = A + C {tex}\therefore{/tex} 2A + C = {tex}\pi{/tex} . . . (7)
Again
A + B + C = {tex}\pi{/tex} . . .[Sum of three angles of a triangle = {tex}\pi{/tex}]
{tex}\therefore{/tex} A + 2C + C = {tex}\pi{/tex} . . .[As B = 2C]
{tex}\therefore{/tex} A + 3C = {tex}\pi{/tex} . . . (8)
Multiplying (7) by 3, we get
6A + 3C = 3{tex}\pi{/tex}
5A = 2{tex}\pi{/tex} . . .[By subtracting (8) from (9)]
{tex}\therefore{/tex} {tex}A=\frac{2 \pi}{5}=\frac{2}{5} \times 180^{\circ}=72^{\circ}{/tex}
{tex}\therefore{/tex} {tex}\angle{/tex}BAC = 72o
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Sia ? 4 years, 5 months ago
l = 22.5 m, b = 10 cm, h = 7.5 cm.
∴ Total surface area of brick = 2(lb + bh + hl)
= 2(22.5 × 10 + 10 × 7.5 + 7.5 × 22.5)
= 2(225 + 75 + 168.75)
= 2(468.75) = 937.5 cm2 = .09375 m2
The no.of brick that can be painted out ={tex}{9.375}\over.09375{/tex}
= 100 Brick.
Posted by Adarsh Kumar 4 years, 5 months ago
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