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Priya Lalwani from Sophia Sr Sec School, Bikaner
Subject :IMO    Class : Class 9

Following are the steps of construction of a ∆ABC in which AB = 5 cm, ∠A = 30º and AC – BC = 2.5 cm. Arrange them and select the CORRECT option.
(i) Draw ∠BAX = 30º
(ii) Draw the perpendicular bisector of BD which cuts AX at C.
(iii) Draw AB = 5 cm
(iv) Join BD
(v) Join BC to obtain the required triangle ABC
(vi) From ray AX, cut off line segment AD = AC – BC = 2.5 cm

A(i) → (iii) → (iv) → (v) → (vi) → (ii)
B(iii) → (i) → (vi) → (iv) → (ii) → (v)
C(iii) → (i) → (ii) → (v) → (iv) → (vi)
D(iii) → (ii) → (iv) → (i) → (vi) → (v)


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bharat from oakridge international school, HYDERABAD
Subject :IMO    Class : Class 5

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adnan from Birla public school , Doha
Subject :IMO    Class : Class 1

Ans 1:

Debarpan Das from Delhi Public School , Durgapur
Class : Class 5

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Snigdha from Hindu senior secondary school, Chennai
Subject :IMO    Class : Class 4

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Snigdha from Hindu senior secondary school, Chennai
Subject :IMO    Class : Class 4

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Saarang roy from Indian school ghubra, Muscat
Subject :IMO    Class : Class 3

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Pranav Arun from garib school, Mumbai
Subject :IMO    Class : Class 7

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Vishal B from National Public School, Chennai
Class : Class 7

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Jaiman singh from The Mothers international school, New delhi
Subject :IMO    Class : Class 9

Ans 1:

Pratham Chaurasiya from Kendriya Vidyalaya Sarni, Pathakhera
Class : Class 9
Solve- In 1st figure, 6^3+4^3+3^3+2^3=315 ​ In 2nd figure, 11^3+8^3+7^3+6^3=2402 Similarly, in 3rd figure, 5^3+4^3+1^3+x^3=1190 = 190+x^3=1190 = x^3=1000 = x=10 So, correct answer is D) 10

Ans 2:

Malav from , Bharuch
Class : Class 10
how

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ishi gupta from DPS NOIDA, NOIDA
Subject :IMO    Class : Class 4

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Naitik from Delhi Public School Noida, Noida
Class : Class 4
please tell

Ans 2:

ishi gupta from DPS NOIDA, NOIDA
Class : Class 4

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ishi gupta from DPS NOIDA, NOIDA
Subject :IMO    Class : Class 4

Ans 1:

ishi gupta from DPS NOIDA, NOIDA
Class : Class 4

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