Samacheer Kalvi 10th Maths Exercise 4.2 Solutions

$DE \parallel BC$, so by Thales theorem $\dfrac{AD}{DB}=\dfrac{AE}{EC}$.

$\dfrac{AE}{EC}=\dfrac{3}{4}$, so $EC=\dfrac{4}{3}AE$.

$AE+EC=AC=15$

$AE+\dfrac{4}{3}AE=15$

$\dfrac{7}{3}AE=15$

$AE=\dfrac{45}{7}=6.43$ cm

By Thales theorem $\dfrac{AD}{DB}=\dfrac{AE}{EC}$.

$\dfrac{8x-7}{5x-3}=\dfrac{4x-3}{3x-1}$

$(8x-7)(3x-1)=(4x-3)(5x-3)$

$24x^2-29x+7=20x^2-27x+9$

$4x^2-2x-2=0$

$2x^2-x-1=0$

$(2x+1)(x-1)=0$

$x=-\dfrac{1}{2}$ or $x=1$. Lengths must be positive, so $x=1$.

In trapezium $ABCD$, $AB \parallel PQ \parallel DC$.

So $P$ and $Q$ divide $AD$ and $BC$ in the same ratio:

$\dfrac{AP}{PD}=\dfrac{BQ}{QC}$

$\dfrac{AP}{18}=\dfrac{35}{15}$

$AP=\dfrac{18\times 35}{15}=42$ cm

$AD=AP+PD=42+18=60$ cm

$DB=AB-AD=12-8=4$ cm

$EC=AC-AE=18-12=6$ cm

$\dfrac{AD}{DB}=\dfrac{8}{4}=2$

$\dfrac{AE}{EC}=\dfrac{12}{6}=2$

So $\dfrac{AD}{DB}=\dfrac{AE}{EC}$.

By the converse of Basic Proportionality Theorem, $DE \parallel BC$.

Let each side of the rhombus $PQRB$ be $x$ cm.

In a rhombus opposite sides are parallel, so $PQ \parallel BR$, that is $PQ \parallel BC$.

In $\Delta ABC$, $PQ \parallel BC$, so $\dfrac{AP}{AB}=\dfrac{PQ}{BC}$.

$AP=AB-PB=12-x$

$\dfrac{12-x}{12}=\dfrac{x}{6}$

$6(12-x)=12x$

$72=18x$ so $x=4$

Join $AC$; let it cut $EF$ at $G$.

In $\Delta ADC$, $EG \parallel DC$ (since $EF \parallel AB \parallel DC$).

By BPT, $\dfrac{AE}{ED}=\dfrac{AG}{GC}$ … (1)

In $\Delta ABC$, $GF \parallel AB$.

By BPT, $\dfrac{AG}{GC}=\dfrac{BF}{FC}$ … (2)

From (1) and (2), $\dfrac{AE}{ED}=\dfrac{BF}{FC}$.

In $\Delta ABC$, $DE \parallel BC$.

By BPT, $\dfrac{AD}{AB}=\dfrac{AE}{AC}$ … (1)

In $\Delta ADC$, $FE \parallel DC$.

By BPT, $\dfrac{AF}{AD}=\dfrac{AE}{AC}$ … (2)

From (1) and (2), $\dfrac{AD}{AB}=\dfrac{AF}{AD}$.

$AD^2=AB\times AF$

In $\Delta PQR$, $\angle QPR = 90^\circ$ and $PS$ is the bisector of $\angle QPR$, with $S$ on $QR$.

By the Angle Bisector Theorem, $\dfrac{QS}{SR}=\dfrac{PQ}{PR}$.

Adding $1$ to both sides, $\dfrac{QS+SR}{SR}=\dfrac{PQ+PR}{PR}$, that is $\dfrac{QR}{SR}=\dfrac{PQ+PR}{PR}$, so $\dfrac{SR}{QR}=\dfrac{PR}{PQ+PR}$ … (1)

Since $\angle QPR=90^\circ$, $QP \perp PR$; also $ST \perp PR$, hence $ST \parallel QP$.

In $\Delta RQP$, $S$ lies on $RQ$, $T$ lies on $RP$ and $ST \parallel QP$, so $\Delta RST \sim \Delta RQP$ (AA: $\angle R$ common, $\angle RTS = \angle RPQ = 90^\circ$).

Hence $\dfrac{ST}{QP}=\dfrac{RS}{RQ}$ … (2)

Substituting (1) in (2), $\dfrac{ST}{PQ}=\dfrac{PR}{PQ+PR}$.

Cross multiplying, $ST \times (PQ+PR) = PQ \times PR$. Hence proved.

In $\Delta ABC$, $AE$ is the bisector of $\angle BAC$.

By Angle Bisector Theorem, $\dfrac{CE}{EB}=\dfrac{AC}{AB}$ … (1)

In $\Delta ACD$, $AF$ is the bisector of $\angle CAD$.

By Angle Bisector Theorem, $\dfrac{CF}{FD}=\dfrac{AC}{AD}$ … (2)

Given $AB=AD$, so from (1) and (2), $\dfrac{CE}{EB}=\dfrac{CF}{FD}$.

In $\Delta CBD$, $E$ lies on $CB$ and $F$ lies on $CD$ dividing them in the same ratio.

By the converse of BPT, $EF \parallel BD$.

Draw a line segment $PQ = 4.5$ cm.

At $P$, draw $PE$ such that $\angle QPE = 35^\circ$.

At $P$, draw $PF \perp PE$ (i.e. $\angle EPF = 90^\circ$).

Draw the perpendicular bisector of $PQ$; it meets $PF$ at $O$ and $PQ$ at $G$.

With $O$ as centre and $OP$ as radius, draw a circle. The major arc holds the angle $35^\circ$.

With $G$ as centre and radius $6$ cm, draw an arc cutting the circle at $R$ (and at $S$).

Join $PR$ and $QR$. Then $\Delta PQR$ is the required triangle.

Draw a line segment $QR = 5$ cm.

At $Q$, draw $QE$ such that $\angle RQE = 40^\circ$.

At $Q$, draw $QF \perp QE$.

Draw the perpendicular bisector of $QR$; it meets $QF$ at $O$ and $QR$ at $G$.

With $O$ as centre and $OQ$ as radius, draw a circle.

With $G$ as centre and radius $4.4$ cm, draw an arc cutting the circle at $P$.

Join $PQ$ and $PR$. Then $\Delta PQR$ is the required triangle.

From $P$ draw $PM \perp QR$ and measure $PM$.

Draw a line segment $QR = 6.5$ cm.

At $Q$, draw ray $QE$ such that $\angle RQE = 60^\circ$ (this $QE$ is the tangent to the required circumcircle at $Q$).

At $Q$, draw $QF \perp QE$, i.e. $\angle EQF = 90^\circ$; the centre of the circle lies on $QF$.

Draw the perpendicular bisector $XY$ of $QR$; let it cut $QF$ at $O$ and $QR$ at $G$.

With $O$ as centre and $OQ$ as radius, draw a circle. Every point on the major arc $QR$ sees $QR$ at $60^\circ$ (angle in the alternate segment).

On $XY$, on the same side of $QR$ as the major arc (the side of $O$), mark the point $M$ with $GM = 4.5$ cm.

Through $M$ draw the line $AB \parallel QR$ (i.e. $AB \perp XY$ at $M$); it cuts the circle at $P$ and $S$ (possible, since the farthest point of the arc is about $5.6$ cm from $QR$).

Join $PQ$ and $PR$. Then $\Delta PQR$ is the required triangle (taking $S$ instead of $P$ gives the mirror-image triangle).

Draw a line segment $AB = 5.5$ cm.

At $A$, draw a ray $AE$ making $\angle BAE = 25^\circ$ with $AB$ (so $AE$ will be the tangent at $A$; by the tangent–chord property the angle in the alternate segment is $25^\circ$).

At $A$, draw $AF \perp AE$, i.e. $\angle EAF = 90^\circ$, on the side of $AB$ opposite to $AE$.

Draw the perpendicular bisector $XY$ of $AB$; let it meet $AF$ at $O$ and $AB$ at $G$.

With $O$ as centre and $OA$ ($=OB$) as radius, draw a circle; every point of the major arc $AB$ on the side of $AB$ containing $O$ sees $AB$ at $25^\circ$.

On $XY$, on the same side of $AB$ as $O$, mark the point $M$ with $GM = 4$ cm (this is the required height).

Through $M$ draw the line $MN \parallel AB$; it cuts the circle at $C$ (it meets the circle in two points, either one gives the same triangle, the two being mirror images in $XY$).

Join $CA$ and $CB$. Then $\triangle ABC$ is the required triangle, since $AB = 5.5$ cm, the distance of $C$ from $AB$ is $4$ cm, and $\angle ACB = \angle BAE = 25^\circ$ (angle in the alternate segment).

Draw a line segment $BC = 5.6$ cm.

At $B$, draw $BE$ such that $\angle CBE = 40^\circ$.

At $B$, draw $BF \perp BE$.

Draw the perpendicular bisector of $BC$; it meets $BF$ at $O$ and $BC$ at $G$.

With $O$ as centre and $OB$ as radius, draw a circle.

From $C$, mark an arc of $4$ cm on $BC$ to get $D$ (so $BD = 1.6$ cm).

The perpendicular bisector meets the circle at $I$ (on the side opposite to the major arc). Join $ID$.

Produce $ID$ to meet the circle at $A$. Join $AB$ and $AC$. Then $\Delta ABC$ is the required triangle.

Draw a line segment $PQ = 6.8$ cm.

At $P$, draw $PE$ such that $\angle QPE = 50^\circ$.

At $P$, draw $PF \perp PE$.

Draw the perpendicular bisector of $PQ$; it meets $PF$ at $O$ and $PQ$ at $G$.

With $O$ as centre and $OP$ as radius, draw a circle.

From $P$, mark an arc of $5.2$ cm on $PQ$ to get $D$.

The perpendicular bisector meets the circle at $I$. Join $ID$ and produce it to meet the circle at $R$.

Join $RP$ and $RQ$. Then $\Delta PQR$ is the required triangle, and $RD$ bisects $\angle R$.