Samacheer Kalvi 10th Maths Unit Exercise 4 Solutions

In $\Delta BEF$, $CD \parallel EF$, so $\Delta BDC \sim \Delta BEF$:

$\dfrac{x}{4} = \dfrac{BD}{BE} = \dfrac{5}{5+y}$ ... (1)

In $\Delta EAB$, $CD \parallel AB$, so $\Delta EDC \sim \Delta EBA$:

$\dfrac{x}{6} = \dfrac{ED}{EB} = \dfrac{y}{5+y}$ ... (2)

Adding (1) and (2): $\dfrac{x}{4} + \dfrac{x}{6} = \dfrac{5+y}{5+y} = 1$

$\dfrac{5x}{12} = 1 \Rightarrow x = \dfrac{12}{5} = 2.4$ cm

Put $x = 2.4$ in (2): $\dfrac{2.4}{6} = \dfrac{y}{5+y}$

$0.4(5+y) = y \Rightarrow 2 = 0.6y \Rightarrow y = \dfrac{10}{3}$ cm

In $\Delta AOB$, $OD$ is the bisector of $\angle AOB$ and $D$ lies on $AB$.

By the angle bisector theorem, $\dfrac{AD}{DB} = \dfrac{OA}{OB}$ ... (1)

In $\Delta BOC$, $OE$ bisects $\angle BOC$ and $E$ lies on $BC$:

$\dfrac{BE}{EC} = \dfrac{OB}{OC}$ ... (2)

In $\Delta COA$, $OF$ bisects $\angle COA$ and $F$ lies on $CA$:

$\dfrac{CF}{FA} = \dfrac{OC}{OA}$ ... (3)

Multiplying (1), (2) and (3):

$\dfrac{AD}{DB} \times \dfrac{BE}{EC} \times \dfrac{CF}{FA} = \dfrac{OA}{OB} \times \dfrac{OB}{OC} \times \dfrac{OC}{OA} = 1$

Therefore $AD \times BE \times CF = DB \times EC \times FA$.

$AB = AC$ and $AD = AE$.

So $DB = AB - AD = AC - AE = EC$.

Then $\dfrac{AD}{DB} = \dfrac{AE}{EC}$, so by the converse of the Basic Proportionality Theorem, $DE \parallel BC$.

Since $DE \parallel BC$, $\angle ADE = \angle ABC$ (corresponding angles).

So $\angle BDE = 180^\circ - \angle ADE = 180^\circ - \angle ABC$ ... (1)

Since $AB = AC$, $\angle ABC = \angle ACB$, i.e. $\angle ABC = \angle ECB$ ... (2)

From (1) and (2): $\angle BDE + \angle ECB = 180^\circ$.

In quadrilateral $BDEC$ the opposite angles $\angle BDE$ and $\angle ECB$ are supplementary.

Hence $BDEC$ is a cyclic quadrilateral, so $B$, $C$, $E$, $D$ lie on the same circle.

Distance of first train in 2 hours (west) $= 20 \times 2 = 40$ km.

Distance of second train in 2 hours (north) $= 30 \times 2 = 60$ km.

West and north directions are perpendicular, so the paths form a right angle at the station.

By Pythagoras theorem, distance $= \sqrt{40^2 + 60^2}$

$= \sqrt{1600 + 3600} = \sqrt{5200}$

$= 20\sqrt{13} \approx 72.11$ km

$D$ is the midpoint of $BC$, so $BD = DC = \dfrac{a}{2}$, and $EC = ED + DC = x + \dfrac{a}{2}$.

In right triangle $AED$: $p^2 = h^2 + x^2$, so $h^2 = p^2 - x^2$ ... (1)

In right triangle $AEC$: $b^2 = h^2 + EC^2$

$b^2 = h^2 + \left(x + \dfrac{a}{2}\right)^2$

Using (1): $b^2 = p^2 - x^2 + x^2 + ax + \dfrac{a^2}{4}$

$b^2 = p^2 + ax + \dfrac{a^2}{4}$

Let the Emu be at a distance $x$ ft from the pillar and let its shadow be $s$ ft long.

The tip of the pillar, the top of the Emu and the tip of the shadow lie on one line, so the two right triangles formed are similar.

$\dfrac{\text{height of pillar}}{\text{distance from pillar to shadow tip}} = \dfrac{\text{height of Emu}}{\text{shadow}}$

$\dfrac{30}{x+s} = \dfrac{8}{s}$

$30s = 8x + 8s \Rightarrow 22s = 8x$

$s = \dfrac{4x}{11}$

Let $TPT'$ be the tangent at $P$ to the first circle; $A$, $B$ lie on both circles.

In the first circle, by the tangent–chord angle (alternate segment) theorem with chord $PA$:

$\angle TPA = \angle ABP$ ... (1)

$A$, $B$, $D$, $C$ lie on the second circle, so $ABDC$ is a cyclic quadrilateral:

$\angle ABD + \angle ACD = 180^\circ$ ... (2)

$P$, $B$, $D$ are collinear, so $\angle ABP + \angle ABD = 180^\circ$ ... (3)

From (2) and (3): $\angle ABP = \angle ACD$.

Using (1): $\angle TPA = \angle ACD = \angle PCD$.

These are alternate angles for the lines $TPT'$ and $CD$ with transversal $PC$.

Hence $CD \parallel$ tangent at $P$.

The points $D$, $E$, $F$ on the sides (or extensions) of $\Delta ABC$ are collinear, so Menelaus' theorem applies.

Menelaus' theorem: $\dfrac{AD}{DB}\times\dfrac{BE}{EC}\times\dfrac{CF}{FA}=1$.

Substitute the given ratios: $\dfrac{5}{3}\times\dfrac{3}{2}\times\dfrac{CF}{FA}=1$.

$\dfrac{5}{2}\times\dfrac{CF}{FA}=1 \Rightarrow \dfrac{CF}{FA}=\dfrac{2}{5}$.

Here $F$ is a point on the side $AC$, so $AF+FC=AC$.

Let $CF=2k$ and $FA=5k$; then $2k+5k=21$, i.e. $7k=21$, so $k=3$.

Therefore $CF=2k=2\times 3=6$ units.