Triangle Class 10 Important Question PYQ

Solution:
Let $AE=x$ cm.

Then$$EC=AC-AE=4.8-x$$

Since $DE\parallel BC$, by Thales’ theorem,$$\frac{AD}{DB}=\frac{AE}{EC}$$$$\frac35=\frac{x}{4.8-x}$$$$3(4.8-x)=5x$$$$14.4-3x=5x$$$$8x=14.4$$$$\boxed{x=1.8}$$

Therefore, AE=1.8 cm​

By Thales’ theorem,$$\frac{AD}{BD}=\frac{AE}{CE}$$

Therefore,$$\frac{4x-3}{3x-1} = \frac{8x-7}{5x-3}$$

Cross-multiplying,$$(4x-3)(5x-3)=(3x-1)(8x-7)$$$$20x^2-27x+9=24x^2-29x+7$$$$4x^2-2x-2=0$$$$2x^2-x-1=0$$$$(2x+1)(x-1)=0$$

So,$$x=1\quad\text{or}\quad x=-\frac12$$

Since a length cannot be negative, x=1​

Solution:

By Thales’ theorem,$$\frac{AD}{AB}=\frac{AE}{AC}$$$$\frac{1.7}{6.8}=\frac{AE}{9}$$

Therefore,$$AE=\frac{1.7\times9}{6.8}$$$$AE=2.25$$

Hence, AE=2.25 cm​

Solution:$$\frac{AD}{AB} = \frac{1.4}{5.6} = \frac14$$

and$$\frac{AE}{AC} = \frac{1.8}{7.2} = \frac14$$

Therefore,$$\frac{AD}{AB}=\frac{AE}{AC}$$

Hence, by the converse of Thales’ theorem, DE∥BC​

Solution:

Given,$$\frac{AD}{DB}=\frac{AE}{EC}$$

Therefore, by the converse of Thales’ theorem,$$DE\parallel BC$$

Hence,$$\angle ADE=\angle ABC$$

because they are corresponding angles.

But given,$$\angle ADE=\angle ACB$$

Therefore,$$\angle ABC=\angle ACB$$

Angles opposite equal sides are equal, so$$AB=AC$$

Hence, △ABC is an isosceles triangle​

In ΔPOQ, DE || OQ.

∴$$\frac{PD}{DO}=\frac{PE}{EQ} \qquad …(i)\quad [\text{by Thales’ theorem}]$$

In ΔPOR, DF || OR.

∴$$\frac{PD}{DO}=\frac{PF}{FR} \qquad …(ii)$$

From (i) and (ii), we get:$$\frac{PE}{EQ}=\frac{PF}{FR} \qquad \left[\text{each equal to }\frac{PD}{DO}\right].$$

Thus, in ΔPQR, E and F are points on PQ and PR respectively such that$$\frac{PE}{EQ}=\frac{PF}{FR}.$$

Hence, EF || QR [by the converse of Thales’ theorem].

SOLUTION

In ΔOAB, PQ || AB.

∴$$\frac{OP}{PA}=\frac{OQ}{QB} \qquad …(i)\quad [\text{by the Thales’ theorem}]$$

In ΔAOC, PR || AC.

∴$$\frac{OP}{PA}=\frac{OR}{RC} \qquad …(ii)\quad [\text{by Thales’ theorem}]$$

From (i) and (ii), we get:$$\frac{OQ}{QB}=\frac{OR}{RC} \quad\text{in ΔOBC}.$$

Thus, in ΔOBC, O and R are points on OB and OC respectively such that$$\frac{OQ}{QB}=\frac{OR}{RC}.$$

Hence, by the converse of Thales’ theorem, QR || BC.

SOLUTION

In ΔALM, LM || CB.$$\therefore\quad \frac{AB}{AM}=\frac{AC}{AL} \Rightarrow \frac{AM}{AB}=\frac{AL}{AC} \qquad …(i)$$

In ΔALN, LN || CD.$$\therefore\quad \frac{AC}{AL}=\frac{AD}{AN} \Rightarrow \frac{AL}{AC}=\frac{AN}{AD} \qquad …(ii)$$

From (i) and (ii), we get:$$\boxed{\frac{AM}{AB}=\frac{AN}{AD}}.$$

In ΔBAE, DF || AE.$$\therefore\quad \frac{BD}{DA}=\frac{BF}{FE} \qquad …(i)\quad [\text{by Thales’ theorem}]$$

In ΔBAC, DE || AC.$$\therefore\quad \frac{BD}{DA}=\frac{BE}{EC} \qquad …(ii)\quad [\text{by Thales’ theorem}]$$

From (i) and (ii), we get:$$\frac{BF}{FE}=\frac{BE}{EC} \qquad \left[\text{each equal to }\frac{BD}{DA}\right].$$

SOLUTION

In ΔABC, AB || DE.$$\therefore\quad \frac{CD}{DA}=\frac{CE}{EB} \qquad …(i)\quad [\text{by Thales’ theorem}]$$

In ΔCDB, BD || EF.$$\therefore\quad \frac{CF}{FD}=\frac{CE}{EB} \qquad …(ii)\quad [\text{by Thales’ theorem}]$$

From (i) and (ii), we get:$$\frac{CD}{DA}=\frac{CF}{FD}$$$$\Rightarrow\quad \frac{DA}{DC}=\frac{FD}{CF} \qquad [\text{taking reciprocals}]$$$$\Rightarrow\quad \frac{DA}{DC}+1=\frac{FD}{CF}+1$$$$\Rightarrow\quad \frac{DA+DC}{DC}=\frac{FD+CF}{CF}$$$$\Rightarrow\quad \frac{AC}{DC}=\frac{DC}{CF}$$ ⇒DC2=CF×AC​.

SOLUTION$$PA\perp AC\text{ and }QB\perp AC \Rightarrow QB\parallel PA.$$

Thus, in ΔPAC, QB || PA. So, ΔQBC ∼ ΔPAC.$$\therefore\quad \frac{QB}{PA}=\frac{BC}{AC} \Rightarrow \frac{z}{x}=\frac{b}{a+b} \qquad …(i)\quad [\text{by the property of similar }\triangle]$$

In ΔRAC, QB || RC. So, ΔQBC ∼ ΔRAC.$$\therefore\quad \frac{QB}{RC}=\frac{AB}{AC} \Rightarrow \frac{z}{y}=\frac{a}{a+b} \qquad …(ii)\quad [\text{by the property of similar }\triangle]$$

From (i) and (ii), we get:$$\frac{z}{x}+\frac{z}{y} = \left(\frac{b}{a+b}+\frac{a}{a+b}\right)=1$$$$\Rightarrow\quad \frac{z}{x}+\frac{z}{y}=1$$you can show the steps clearly by taking z common:$$\Rightarrow z\left(\frac{1}{x}+\frac{1}{y}\right)=1$$$$\Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{1}{z} \qquad [\text{dividing both sides by }z]$$

SOLUTION

GIVEN A trap. $ABCD$ in which $AB\parallel DC$. E and F are points on $AD$ and $BC$ respectively such that $EF\parallel AB$.

TO PROVE$$\frac{AE}{ED}=\frac{BF}{FC}.$$

CONSTRUCTION Join $AC$, intersecting $EF$ at $G$.

PROOF $EF\parallel AB$ and $AB\parallel DC\Rightarrow EF\parallel DC$.

Now, in $\triangle ADC$, $EG\parallel DC$.$$\therefore\quad \frac{AE}{ED}=\frac{AG}{GC} \qquad …(i)\quad [\text{by Thales’ theorem}]$$

Similarly, in $\triangle CAB$, $GF\parallel AB$.$$\therefore\quad \frac{AG}{GC}=\frac{BF}{FC} \qquad …(ii)\quad \left[\therefore\frac{GC}{AG}=\frac{FC}{BF}\text{ by Thales’ theorem}\right]$$

From (i) and (ii), we get: EDAE​=FCBF​​.

SOLUTION

GIVEN A trapezium $ABCD$ in which $AB\parallel DC$ and its diagonals $AC$ and $BD$ intersect at O.

TO PROVE$$\frac{AO}{OC}=\frac{BO}{OD}.$$

CONSTRUCTION Through O, draw $EO\parallel AB$, meeting $AD$ at E.

PROOF In $\triangle ADC$, $EO\parallel DC$
$[\therefore EO\parallel AB\parallel DC]$$$\therefore\quad \frac{AE}{ED}=\frac{AO}{OC} \qquad …(i)\quad [\text{by Thales’ theorem}]$$

In $\triangle DAB$, $EO\parallel AB$.$$\therefore\quad \frac{AE}{ED}=\frac{BO}{OD} \qquad …(ii)$$$$\text{From (i) and (ii), we get:}$$$$\boxed{\frac{AO}{OC}=\frac{BO}{OD}}.$$

SOLUTION

GIVEN A quadrilateral $ABCD$ whose diagonals $AC$ and $BD$ intersect at a point O such that$$\frac{AO}{OC}=\frac{BO}{OD}.$$

TO PROVE $ABCD$ is a trapezium, i.e., $AB\parallel DC$.

CONSTRUCTION Draw $EO\parallel DC$, meeting $AD$ at E.

PROOF In $\triangle ACD$, $EO\parallel DC$.$$\therefore\quad \frac{AO}{OC}=\frac{AE}{ED} \qquad …(i)\quad [\text{by Thales’ theorem}]$$

But,$$\frac{AO}{OC}=\frac{BO}{OD}\quad\text{(given)}$$$$\therefore\quad \frac{BO}{OD}=\frac{AE}{ED} \quad\text{in }\triangle DAB.$$

So, $EO\parallel AB$ [by the converse of Thales’ theorem].

But, $EO\parallel DC$.

Hence,$$\boxed{AB\parallel DC}.$$

SOLUTION

We know that $AB\parallel DC$ in trap. $ABCD$ and its diagonals intersect at O. Then, we have:$$\frac{AO}{OC}=\frac{BO}{OD} \Rightarrow \frac{3x-1}{5x-3}=\frac{2x+1}{6x-5}$$$$\Rightarrow (3x-1)(6x-5)=(2x+1)(5x-3)$$$$\Rightarrow 18x^2-21x+5=10x^2-x-3$$$$\Rightarrow 8x^2-20x+8=0$$$$\Rightarrow 2x^2-5x+2=0$$$$\Rightarrow (x-2)(2x-1)=0$$$$\Rightarrow x=2\text{ or }x=\frac12.$$

But, $x=\frac12$ will make$$OC=(5x-3)\text{ cm} =\left(5\times\frac12-3\right)\text{ cm}$$$$=-\frac12\text{ cm}.$$

And, the distance cannot be negative.$$\therefore\quad x\ne\frac12.$$

Hence, x=2​.