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Raavi Tiwari
@raavi-tiwari
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Joined: May 7, 2021
Last seen: Jun 14, 2021
Topics: 0 / Replies: 2064
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Answer to: What was the basic objective of the foreign policy of Independent India?

The basic objective of the foreign policy of Independent India was non-alignment, i.e. the American and Soviet alliances.

5 years ago
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Answer to: How was the Bhilai Steel Plant viewed?

The Bhilai Steel Plant was viewed as an important sign of the development of modem India after Independence.

5 years ago
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Answer to: What were the points of focus of the Second Five Year Plan?

Development of heavy industries. The building of large dams.

5 years ago
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Answer to: When did the new state of Andhra Pradesh come into existence?

The new state of Andhra Pradesh came into existence on 1 October 1953.

5 years ago
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Answer to: Who was Potti Sriramulu?

He was a veteran Gandhian who went on a hunger strike demanding the formation of Andhra state to protect the interests of Telugu speakers.

5 years ago
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Answer to: On what point did Nathuram Godse disagree with Gandhiji?

Nathuram Godse disagreed with Gandhiji’s conviction that Hindus and Muslims should live together in harmony.

5 years ago
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Answer to: Which step has been described as revolutionary?

All Indians above the age of 21 would be allowed to vote in state and national elections.

5 years ago
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Answer to: When was the Indian Constitution adopted?

The Indian Constitution was adopted on 26 January 1950.

5 years ago
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Answer to: In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

Given, ABC is an isosceles triangle. ∴ AB = AC ⇒ ∠ABD = ∠ECF In ΔABD and ΔECF, ∠ADB = ∠EFC (Each 90°) ∠BAD = ∠CEF (Already proved) ∴ ΔABD ...

5 years ago
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Answer to: CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively.

Given, CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. 2s.png ...

5 years ago
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Answer to: In the figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that: (i) ΔABC ~ ΔAMP (ii) CA/PA = BC/MP

Given, ABC and AMP are two right triangles, right angled at B and M respectively. (i) In ΔABC and ΔAMP, we have, ∠CAB = ∠MAP (common angles) ∠AB...

5 years ago
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Answer to: E is a point on the side AD produced of a parallelogram ABCD and BE intersect CD at F. Show that ΔABE ~ ΔCFB.

Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below, 2s.png In ΔABE and ΔC...

5 years ago
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Answer to: In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that: (i) ΔAEP ~ ΔCDP (ii) ΔABD ~ ΔCBE

Given, altitudes AD and CE of ΔABC intersect each other at the point P. (i) In ΔAEP and ΔCDP, ∠AEP = ∠CDP (90° each) ∠APE = ∠CPD (Vertically opp...

5 years ago
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Answer to: In the figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.

Given, ΔABE ≅ ΔACD. ∴ AB = AC [By CPCT] ……………….(i) And, AD = AE [By CPCT] ………………(ii) In ΔADE and ΔABC, dividing eq.(ii) by eq(i), AD/AB = AE/...

5 years ago
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Answer to: S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.

Given, S and T are point on sides PR and QR of ΔPQR 2s.png And ∠P = ∠RTS. In ΔRPQ and ΔRTS, ∠RTS = ∠QPS (Given) ∠R = ∠R (Common angle) ...

5 years ago
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Answer to: In the figure, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

In ΔPQR, ∠PQR = ∠PRQ ∴ PQ = PR ………………(i) Given, QR/QS = QT/PR Using equation (i), we get QR/QS = QT/QP ………….(ii) In ΔPQS and ΔTQR, by equ...

5 years ago
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Answer to: Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD.

2s.png In ΔDOC and ΔBOA, AB || CD, thus alternate interior angles will be equal, ∴ ∠CDO = ∠ABO Similarly, ∠DCO = ∠BAO Also, for the two ...

5 years ago
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