9
300-YEAR-OLD MATHEMATICAL RIDDLE SOLVED BY GERMAN TEEN
problem had no solutions,
G. Rewrite the newspaper report in the active voice.
us by our teacher that the
A mathematical problem
for centuries before it was
posed by Sir Isaac Newton worked out by the Indian-
I thought to myself, well,
more than 300 years ago has bom teen while working on
there's no harm in trying."
been solved for the first time a school project.
he said.
by 16-year-old Shouryya
A research award has been
Ray
The problem was posed
given to Ray for his efforts
The exact calculation of the and he has been labelled
by Newton, relating to the
path of a projectile under a genius by the German
movement of projectiles
gravity and subject to air media, but "curiosity and
through the air, in the
resistance was worked out schoolboy naivety" was
seventeenth century. Only
by Shouryya Ray
what the young lad put it partial solutions had been
down to.
offered by mathematicians
Mathematicians had been
until now.
stumped by this problem -When it was explained to
The w
Miss
Miss
Answers
Answer:
Perimeter of trapezium = 104 m
Length of Non-parallel sides = 18 m and 22 m
Altitude = 16m
Area of Trapezium = 0.5 * (sum of parallel sides)*altitude ------ (1)
Sum of parallel sides = Perimeter - (sum of non-parallel sides) = 104 m - (18+22)m = 104-40 m = 64 m
From (1), Area of Trapezium = 0.5*64*16 = 512 m^2
Ans: Area of trapezium = 512 m^2
Explanation:
Question : Prove that√5 is irrational.
Answer :
Let us assume that √5 is a rational number.
Sp it t can be expressed in the form p/q where p,q are co-prime integers and q≠0
⇒√5=p/q
On squaring both the sides we get,
⇒5=p²/q²
⇒5q²=p² —————–(i)
p²/5= q²
So 5 divides p
p is a multiple of 5
⇒p=5m
⇒p²=25m² ————-(ii)
From equations (i) and (ii), we get,
5q²=25m²
⇒q²=5m²
⇒q² is a multiple of 5
⇒q is a multiple of 5
Hence, p,q have a common factor 5. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number
√5 is an irrational number
Hence proved