Samacheer Kalvi 10th Maths Exercise 4.4 Solutions

The tangent is perpendicular to the radius at the point of contact $Q$.

So $\triangle OQP$ is right angled at $Q$.

$OP^2 = OQ^2 + PQ^2$

$25^2 = r^2 + 24^2$

$r^2 = 625 - 576 = 49$

$r = 7$ cm

Sides containing the right angle: $LM = 6$ cm, $LN = 8$ cm.

Hypotenuse $MN = \sqrt{6^2+8^2} = \sqrt{100} = 10$ cm.

Let the incircle of radius $r$ touch $LM$ at $P$ and $LN$ at $Q$.

$OP \perp LM$, $OQ \perp LN$ and $OP = OQ = r$, so $OPLQ$ is a square, giving $LP = LQ = r$.

Tangents from $M$: $MP = 6 - r$; tangents from $N$: $NQ = 8 - r$.

$MN = MP + NQ$ gives $10 = (6-r)+(8-r)$

$10 = 14 - 2r \Rightarrow r = 2$ cm

Tangents from an external point are equal.

From $A$: $AD = AF = x$; from $B$: $BD = BE = y$; from $C$: $CE = CF = z$.

$AB = x + y = 12$ … (1)

$BC = y + z = 8$ … (2)

$CA = z + x = 10$ … (3)

Adding, $2(x+y+z) = 30$, so $x+y+z = 15$ … (4)

$(4) - (2)$: $x = 15 - 8 = 7$

$(4) - (3)$: $y = 15 - 10 = 5$

$(4) - (1)$: $z = 15 - 12 = 3$

$OP = 5$ cm (radius), $OT = 13$ cm, $TP$ is a tangent so $\angle OPT = 90^\circ$.

$PT = \sqrt{OT^2 - OP^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm.

$OE = 5$ cm (radius), so $ET = OT - OE = 13 - 5 = 8$ cm.

Tangents from $A$: $AP = AE$. Let $AE = x$, then $AT = PT - AP = 12 - x$.

$AB$ is a tangent at $E$, so $\angle AET = 90^\circ$.

In $\triangle AET$: $AE^2 + ET^2 = AT^2$

$x^2 + 64 = (12-x)^2 = 144 - 24x + x^2$

$24x = 80 \Rightarrow x = \dfrac{10}{3}$

By symmetry $BE = AE$, so $AB = 2x = \dfrac{20}{3}$ cm.

$OP$ is a tangent to the circle with centre $O'$ and $O'P$ is a tangent to the circle with centre $O$.

So $\angle OPO' = 90^\circ$.

$OO' = \sqrt{OP^2 + O'P^2} = \sqrt{3^2 + 4^2} = 5$ cm.

$PQ$ is the common chord and $OO'$ bisects $PQ$ at right angles at $M$.

Area of $\triangle OPO' = \dfrac12 \times OP \times O'P = \dfrac12 \times OO' \times PM$

$\dfrac12 \times 3 \times 4 = \dfrac12 \times 5 \times PM \Rightarrow PM = \dfrac{12}{5} = 2.4$ cm

$PQ = 2 \times PM = 4.8$ cm

Let $\triangle ABC$ be given and let $AD$, $BE$, $CF$ be the internal bisectors of $\angle A$, $\angle B$, $\angle C$, meeting the opposite sides at $D\in BC$, $E\in CA$, $F\in AB$.

Since $AD$ bisects $\angle A$, by the Angle Bisector Theorem $\dfrac{BD}{DC}=\dfrac{AB}{AC}$ … (1)

Since $BE$ bisects $\angle B$, $\dfrac{CE}{EA}=\dfrac{BC}{BA}$ … (2)

Since $CF$ bisects $\angle C$, $\dfrac{AF}{FB}=\dfrac{CA}{CB}$ … (3)

Multiplying (1), (2) and (3): $\dfrac{BD}{DC}\times\dfrac{CE}{EA}\times\dfrac{AF}{FB}=\dfrac{AB}{AC}\times\dfrac{BC}{BA}\times\dfrac{CA}{CB}$

All the factors cancel, so $\dfrac{BD}{DC}\times\dfrac{CE}{EA}\times\dfrac{AF}{FB}=1$.

Since $D$, $E$, $F$ are interior points of the sides and the product of the ratios is $1$, by the converse of Ceva's Theorem the cevians $AD$, $BE$, $CF$ are concurrent.

Hence the three angle bisectors of a triangle meet at a single point (the incentre $I$ of the triangle).

$D$, $E$, $F$ lie on $BC$, $CA$, $AB$ of $\triangle ABC$ and they are collinear (the strip is a straight line).

By Menelaus' theorem, $\dfrac{BD}{DC}\times\dfrac{CE}{EA}\times\dfrac{AF}{FB} = 1$ (in magnitude).

Given $BD = 3$, $DC = 10$, $CE = 4$, $EA = 3$, $AF = 5$.

$\dfrac{3}{10}\times\dfrac{4}{3}\times\dfrac{5}{FB} = 1$

$\dfrac{2}{5}\times\dfrac{5}{FB} = 1$

$\dfrac{2}{FB} = 1 \Rightarrow FB = 2$ cm

Step 1: With centre $O$, draw a circle of radius $5$ cm.

Step 2: Draw a line $OP$ of length $10$ cm.

Step 3: Draw the perpendicular bisector of $OP$, cutting $OP$ at $M$.

Step 4: With $M$ as centre and $MO$ as radius, draw a circle cutting the first circle at $A$ and $B$.

Step 5: Join $PA$ and $PB$. These are the required tangents.

Verification: In right $\triangle OAP$, $PA^2 = OP^2 - OA^2 = 100 - 25 = 75$

$PA = \sqrt{75} = 8.7$ cm (approximately)

Step 1: With centre $O$ and radius $4$ cm, draw a circle.

Step 2: Mark a point $P$ such that $OP = 11$ cm and join $OP$.

Step 3: Draw the perpendicular bisector of $OP$; let it meet $OP$ at $M$ (so $MO = MP = 5.5$ cm).

Step 4: With $M$ as centre and $MO$ as radius, draw a circle which cuts the given circle at $A$ and $B$.

Step 5: Join $PA$ and $PB$. Since $\angle OAP = \angle OBP = 90^\circ$ (angle in a semicircle), $PA$ and $PB$ are the required tangents.

Verification: $\angle OAP = 90^\circ$, so $PA^2 = OP^2 - OA^2 = 11^2 - 4^2 = 121 - 16 = 105$.

$PA = PB = \sqrt{105} \approx 10.25$ cm, which agrees with the measured length ($\approx 10.3$ cm).

Diameter $= 6$ cm, so radius $= 3$ cm.

Step 1: With centre $O$, draw a circle of radius $3$ cm.

Step 2: Draw a line $OP$ of length $5$ cm.

Step 3: Draw the perpendicular bisector of $OP$, cutting $OP$ at $M$.

Step 4: With $M$ as centre and $MO$ as radius, draw a circle cutting the first circle at $A$ and $B$.

Step 5: Join $PA$ and $PB$. These are the required tangents.

Verification: $PA^2 = OP^2 - OA^2 = 25 - 9 = 16$, so $PA = 4$ cm.

Step 1: Draw a circle with centre $O$ and radius $3.6$ cm.

Step 2: Mark a point $P$ on the plane with $OP = 7.2$ cm and join $OP$.

Step 3: Draw the perpendicular bisector of $OP$; let it meet $OP$ at its mid-point $M$ (so $OM = MP = 3.6$ cm).

Step 4: With $M$ as centre and $MO = 3.6$ cm as radius, draw a circle; it cuts the given circle at $A$ and $B$.

Step 5: Join $PA$ and $PB$. Since $\angle OAP = \angle OBP = 90^\circ$ (angle in a semicircle), $PA$ and $PB$ are the required tangents.

Verification: $PA^2 = OP^2 - OA^2 = (7.2)^2 - (3.6)^2 = 51.84 - 12.96 = 38.88$.

So $PA = \sqrt{38.88} \approx 6.24$ cm, i.e. about $6.2$ cm, and $PB = PA$ (equal tangents from an external point).