PQRS is a rhombus. If PR\(^2\) + QS\(^2\) = kPQ\(^2\), determine k.
1
2
3
4
Correct answer is D
PR\(^2\) + QS\(^2\) = kPQ\(^2\)
SQ\(^2\) = SR\(^2\) + RQ\(^2\)
PR\(^2\) + SQ\(^2\) = PQ\(^2\) + SR\(^2\) + 2RQ\(^2\)
= 2PQ\(^2\) + 2RQ\(^2\)
= 4PQ\(^2\)
∴ K = 4
A regular polygon of (2k + 1) sides has 140° as the size of each interior angle. Find k
4
4\(\frac{1}{2}\)
8
8\(\frac{1}{2}\)
Correct answer is A
A regular has all sides and all angles equal. If each interior angle is 140° each exterior angle must be
180° - 140° = 40°
The number of sides must be \(\frac{360^o}{40^o}\) = 9 sides
hence 2k + 1 = 9
2k = 9 - 1
8 = 2k
k = \(\frac{8}{2}\)
= 4
If -8, m, n, 19 are in arithmetic progression, find (m, n)
1, 10
2, 10
3, 13
4, 16
Correct answer is A
-8, m, n, 19 = m + 8
= 19 - n
m + n = 11
i.e. 1, 10
\(\frac{x}{y}\)
\(\frac{y}{x}\)
(\(\frac{x}{y}\))\(\frac{1}{n - 2}\)
(\(\frac{y}{x}\))\(\frac{1}{n - 2}\)
Correct answer is D
Sum of nth term of a G.P = Sn = \(\frac{ar^n - 1}{r - 1}\)
sum of the first two terms = \(\frac{ar^2 - 1}{r - 1}\)
x = a(r + 1)
sum of the last two terms = Sn - Sn - 2
= \(\frac{ar^n - 1}{r - 1}\) - \(\frac{(ar^{n - 1})}{r - 1}\)
= \(\frac{a(r^n - 1 - r^{n - 2} + 1)}{r - 1}\) (r2 - 1)
∴ \(\frac{ar^{n - 2}(r + 1)(r - 1)}{1}\)= arn - 2(r + 1) = y
= a(r + 1)r^n - 2
y = xrn - 2
= yrn - 2
\(\frac{y}{x}\) = r = (\(\frac{y}{x}\))\(\frac{1}{n - 2}\)
Simplify \(\frac{x(x + 1)^{\frac{1}{2}} - (x + 1)^{\frac{1}{2}}}{(x + 1)^{\frac{1}{2}}}\)
\(\frac{-1}{x + 1}\)
\(\frac{1}{x + 1}\)
\(\frac{1}{x}\)
\(\frac{1}{x - 1}\)
Correct answer is A
\(\frac{x}{(x + 1)}\) - \(\frac{\sqrt{(x + 1)}}{\sqrt(x + 1)}\)
= \(\frac{x}{x + 1}\) - 1
\(\frac{x - x - 1}{x + 1}\) = \(\frac{-1}{x + 1}\)