Geometry

The triangle with sides 3, 4, 5 is a right triangle. Using Apollonius theorem or coordinate geometry, the length of the median to side AB (c=5) is m_c = 1/2 * sqrt(2a^2 + 2b^2 - c^2) = 1/2 * sqrt(2*16 + 2*9 - 25) = 1/2 * sqrt(7) = sqrt(7)/2. The centroid G divides the median in a 2:1 ratio, so BG = 2/3 * m_c = 2/3 * sqrt(7)/2 = sqrt(7)/3. However, checking the options, sqrt(52)/3 is 2*sqrt(13)/3. Let's re-verify: median to AB is 1/2 * sqrt(2*16 + 2*9 - 25) = sqrt(7)/2. The distance BG is 2/3 of the median. None match perfectly. Re-evaluating: maybe median to BC? m_a = 1/2 * sqrt(2*16 + 2*25 - 9) = 1/2 * sqrt(32+50-9) = sqrt(73)/2. BG = 2/3 * sqrt(73)/2 = sqrt(73)/3. Option A is sqrt(73)/3. Wait, the question asks for BG, which is the segment from vertex B to centroid G. This is 2/3 of the median from B to AC. Median m_b = 1/2 * sqrt(2*a^2 + 2*c^2 - b^2) = 1/2 * sqrt(2*9 + 2*25 - 16) = 1/2 * sqrt(18+50-16) = 1/2 * sqrt(52) = sqrt(52)/2. BG = 2/3 * sqrt(52)/2 = sqrt(52)/3.

If two triangles are similar then the ratio of their perimeter is equal to the ratio of their corresponding sides.

$P _1=AB+BC+AC=60  cm$

$12\rightarrow \frac { 360 }{ 12 } 30$

Given: $AB = 3 cm$, $BC = 2 cm$ and $CA = 2.5 cm$ and $EF = 4 cm$ 

Ar. ($\triangle _{1}$) = $48 cm^2$
Ar. ($\triangle _{2}$) = $75 cm^2$
$a _{1} = 3.6 cm$
For similar triangles the ratio of areas is equal to the ratio of square of their altitudes.
Thus, $\frac{A(\triangle _{1})}{A(\triangle _{2})} = \frac{(a _1)^2}{(a _2)^2}$
$\frac{48}{75} = \frac{(3.6)^2}{(a _2)^2}$
$(a _2)^2 = \frac{12.96 \times 75}{48}$
$(a _2)^2 = 20.25$
$a _2 = 4.5$cm

Ar. ($\triangle ABC$) = $49 cm^2$
Ar. ($\triangle PQR$) = $25 cm^2$
$AB = 5.6 cm$
For similar triangles the ratio of areas is equal to the ratio of square of its sides.
Thus, $\dfrac{A(\triangle ABC)}{A(\triangle PQR)} = \dfrac{AB^2}{PQ^2}$
$\dfrac{49}{25} = \dfrac{(5.6)^2}{PQ^2}$
$PQ^2 = \dfrac{31.36 \times 25}{49}$
$PQ^2 = 16$
$PQ = 4$cm

$\triangle ABC = \triangle PQR$, then
$\dfrac {\text {Perimeter of}\triangle ABC}{\text {Perimeter of}\triangle PQR} = \dfrac {AB}{PQ} = \dfrac {BC}{QR} = \dfrac {AC}{PR}$
$\dfrac {36}{24} = \dfrac {AB}{10}; AB = \dfrac {36}{24}\times 10 = 15\ cm$.

Let $P$ be $(h,k)$
Now Chord of contact of tangent from $P$ to the hyperbola $x^2-y^2=a^2$ is,
$T =0\Rightarrow hx -ky = a^2$ (i)
And director circle of given hyperbola is, $x^2+y^2=a^2$
Thus equation of chord of contact to this circle from P is, $hx+ky = a^2$ (ii)
Now given line (i) and (ii) are perpendicular,
$\Rightarrow \cfrac{h}{k}\times \cfrac{-h}{k}=-1\Rightarrow h^2=k^2$
Hence locus of $P$ is given by, $x^2-y^2=0$ 

Let the altitude$=x:cm$
$\therefore$ The Base$=(x+4):cm$
and the Hypotenuse$=(x+8):cm$

Let one side be $xcm$. Then the other side will be $(x+5)cm$. Therefore, from Pythagoras theorem
${x}^{2}+{(x+5)}^{2}={25}^{2}$
$\Rightarrow { x }^{ 2 }+{ x }^{ 2 }+10x+25=625$
$\Rightarrow { x }^{ 2 }+5x-300=0\quad \quad \Rightarrow { x }^{ 2 }+20x-15x-300=0$
$\Rightarrow x(x+20)-15(x+20)=0$
$\Rightarrow (x-15)(x+20)=0\quad \Rightarrow \quad x=15\quad or\quad x=-20$
Rejecting $x=-20$, we have length of one side $=15cm$ and that of the other side $=(15+5)cm=20cm$

$0.35 \ m = 0.35 \times 100 \ cm = 35 \ cm.$
We have,
$(hypotenuse)^2\, =\, (side)^2\, +\, (side)^2$
$=\, (12)^2\, +\, (35)^2$
$= 144 + 1225$
$= 1369$
Hypotenuse $=\sqrt{1369} = 37 \ cm.$

$A\, =\, \displaystyle \frac {1}{2}\, lr\, =\, \displaystyle \frac {1}{2}\, \times\, 18.5\, \times\, 16\, =\, 148\, cm^2$

Arc length of the Circle  : Area of the Sector = Perimeter of the Circle :Area of the Circle

Let the radius of the circle be 'r'.

Hence, $20 : 60$=$ 2\pi r : \pi r^2$

$ 1:3 = 2: r$

$ r= 6 $ (Product of Means = Product of Extremes)

Therefore, $Diameter = 2r = 12cm$

Perimeter of the Sector =37cm