Q-30. Find the value of y for which the distance between the point p (2,-3)
adn Q(10,y) is 10 units.
Q-31. Show that
tan 8 + tane = Seco - Sec20
Q-32. If the 9th term of A.P. is zero, prove that 29th term is twice its 19th
term.
Q-33. Ritu can row downstream 20 km in 2 hours and upsteram 4 km in 2
hours. Find her speed of rowing in still water and speed of the current.
.Q-34. In an equilateral Triangle ABC, D is a point on side BC such that BD
= 1/3 BC.Prove that 9 AD2 = 7 AB2
Answers
Answer 30:-
the formula to calculate the distance between two points (a,b) and (c,d) is given by
dist d = √(a-c)² + (b-d)²
applying it for this ques we have d=10 units
10 = √(2 - 10)² + (-3-y)²
squaring both sides
100 = (-8)² + (9 + y² + 6y)
100 = 64 +9 + y² + 6y
y² + 6y - 27 = 0
y² + 9y - 3y - 27 = 0
y(y + 9) -3(y + 9) = 0
(y-3)(y+9) = 0
y=3 or y=-9
Answer 32:-
Let a and d respectively be the first term and common difference of the AP.
Given a9 = 0
So, a + (9-1)d = 0
a+8d=0
a= -8d
Now, 29th term = a+28d
=-8d+28d
= 20d = 2 x 10d
= 2(-8d + 18d)
=2(a+18d)
= 2 x 19th term
Thus, the 29th term of the AP is twice the 19th term.
Answer 33:-
The speed of the rowing on still water denoted by M kmph.
The speed of the stream denoted by N kmph.
The upstream speed of Ritu’s rowing is (M – N) kmph
The downstream speed of Ritu’s rowing is (M + N) kmph
2(M+N)=20 _______________________ (1)
2(M-N)=4 ________________________ (2)
Solving the equation, we get
2(M+N)=20;
2(M-N)=4;
M-N=2;
M=2+N;
Substituting M=2+N in equation (1)
2(M+N)=20
2(2+N+N)=20 ;
N=4
Substituting N = 4 in equation (1),
2(M+N)=20
2(M+4)=20
2M+8=20
2M=20-8
2M=12
M=6
the value of M = 6.
Therefore, the speed at which Ritu rows is 6 kmph and the speed of the current is 4 kmph.
Answer 34:-
ABC is an equilateral triangle , where D point on side BC in such a way that BD = BC/3 . Let E is the point on side BC in such a way that AE⊥BC .
Now, ∆ABE and ∆AEC
∠AEB = ∠ACE = 90°
AE is common side of both triangles ,
AB = AC [ all sides of equilateral triangle are equal ]
From R - H - S congruence rule ,
∆ABE ≡ ∆ACE
∴ BE = EC = BC/2
Now, from Pythagoras theorem ,
∆ADE is right angle triangle ∴ AD² = AE² + DE² ------(1)
∆ABE is also a right angle triangle ∴ AB² = BE² + AE² ------(2)
From equation (1) and (2)
AB² - AD² = BE² - DE²
= (BC/2)² - (BE - BD)²
= BC²/4 - {(BC/2) - (BC/3)}²
= BC²/4 - (BC/6)²
= BC²/4 - BC²/36 = 8BC²/36 = 2BC²/9
∵AB = BC = CA
So, AB² = AD² + 2AB²/9
9AB² - 2AB² = 9AD²
Hence, 9AD² = 7AB²
Figure in image...
Answer:
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Step-by-step explanation: