TRIANGLES-Exercise 6.3
Maths - Exercise
Q1. State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :
Solution:
Q2. In Fig. 6.35, \(\triangle ODC \sim \triangle OBA\), \(\angle BOC = 125^\circ\) and \(\angle CDO = 70^\circ.\) Find \(\angle DOC, \angle DCO\) and \(\angle OAB.\)
Solution:
$$\scriptsize\begin{aligned}\triangle ODC&\sim \triangle OBA\quad\text{(Given)}\\\\ \angle ODC&=\angle OBA\\ \scriptsize\text{ (Vertically }&\text{Opposite Angle)}\\ \Rightarrow \angle OBA&=70^{\circ }\\\\ \angle DOA=\angle COB&=125^{\circ }\\ \tiny\angle DOA+\angle COB+\angle DOC+\angle AOB&=360^{\circ }\\ \angle DOC&=\angle AOB\\ \scriptsize\text{ (Vertically }&\text{Opposite Angle)}\\ 125^\circ+125^{\circ }+2\angle DOC&=360^{\circ}\\ \angle DOC=\dfrac{360^{\circ }-250^{\circ}}{2}&=55^{\circ }\\\\ \angle OAB+\angle AOB+\angle OBA&=180^\circ\\ \angle OAB+55^\circ+70^{\circ}&=180^{\circ }\\ \angle OAB&=180^{\circ }-\left( 125^{\circ }\right) \\ \angle OAB&=55^{\circ }\\\\ \angle DOC=55^{\circ},\ \angle DCO&=55^{\circ },\ \angle OAB=55^{\circ}\end{aligned}$$
Q3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \(\dfrac{OA}{OC}=\dfrac{OB}{OD}\)
Solution:
3. Given: \[AB\parallel DC\]In \(\triangle DCO\) and \(\triangle BAO\)
$$\begin{aligned}\angle ODC&=\angle OBA\\ \scriptsize\text{ (Altern}&\scriptsize\text{ate Angle)}\\ \angle DCO&=\angle BAO\\ \scriptsize\text{ (Altern}&\scriptsize\text{ate Angle)}\\ \angle COD&=\angle AOD\\ \scriptsize\text{ (Vertically }&\scriptsize\text{Opposite Angle)}\\\\ \therefore \triangle DCO&\sim \triangle BAO\\\\ \dfrac{CO}{AO}&=\dfrac{OD}{OB}\text{ (By BPT)}\\ \Rightarrow \dfrac{OA}{OC}&=\dfrac{OB}{OD}\end{aligned}$$
Q In Fig. 6.36, \(\dfrac{QR}{QS}=\dfrac{QT}{PR}\) and \(\angle 1 = \angle 2\). Show that \(\triangle PQS \sim \triangle TQR.\)
Solution:
Given: $$\dfrac{QR}{QS}=\dfrac{QT}{PR}$$ \(\angle 1=\angle 2\)To Prove: \[\triangle PQS\sim \triangle TQR\] Proof:
$$\begin{aligned} \angle 1&=\angle 2\quad\text{ (Given)}\\ QP&=PR\quad\tiny\text{Sides Opposite to equal Angles} \\ \dfrac{QR}{QS}&=\dfrac{QT}{PR}\quad\text{ (Given)}\\ \dfrac{QR}{QS}&=\dfrac{QT}{QP}\quad\text{ (QP=PR)}\\ \angle TQR&=\angle TQR\quad\text{(Common)}\\ \\ \triangle PQS&\sim \triangle TQR\quad\small\text{(SAS Criterion)}\end{aligned}$$ Hence Proved.
Q5. S and T are points on sides PR and QR of \(\triangle PQR\) such that \(\angle P = \angle RTS\). Show that \(\triangle RPQ \sim \triangle RTS.\)
Solution:
\(\text{In } \triangle RPQ \text{ and }\triangle RTS\) $$\begin{aligned}\angle P&=\angle RTS\quad\text{ (Given)}\\ \angle PRQ&=\angle SRT\quad\text{ (Common)}\\ \triangle QRP&\sim \triangle RTS\end{aligned}$$ By AA Criterion
Q6. In Fig. 6.37, if \(\triangle ABE \cong \triangle ACD\), show that \(\triangle ADE \sim \triangle ABC.\)
Solution:
$$\scriptsize\begin{align} \angle A&=\angle A\quad\text{(Common)}\\ AD&=AE\quad\text{(By CPCT)}\tag{1} \\ AB&=AC\quad\text{(By CPCT)}\tag{2} \\ \text{Dividi}&\text{ng Eqn-(1) by eqn-(2)}\\ \Rightarrow \dfrac{AD}{AB}&=\dfrac{AE}{AC}\\\\ \triangle ADE&\sim \triangle ABC\\ \text{(By SAS}&\text{ Criterion)}\end{align}$$ Hence Proved.
Q7. In Fig. 6.38, altitudes AD and CE of \(\triangle ABC\) intersect each other at the point P. Show that:
- \(\triangle AEP \sim \triangle CDP\)
- \(\triangle ABD \sim \triangle CBE\)
- \(\triangle AEP \sim \triangle ADB\)
- \(\triangle PDC \sim \triangle BEC\)
Solution:
(i) In \(\triangle AEP\) and \(\triangle CDP\)
$$\begin{aligned}\angle E&=\angle D\left( 90^{\circ }\right) \\ \angle DP&C=\angle EPA \\ \scriptsize\text{ (Vertically }&\scriptsize\text{Opposite Angle)}\\\\ \therefore \triangle AEP&\sim \triangle CDP\end{aligned}$$ By AA criterion of similarity(ii) In \(\triangle ABD\) and \(\triangle CBE\)
$$\begin{aligned}\angle E&=\angle D\left( 90^{v}\right) \\ \angle B&=\angle B\quad\text{(Common)} \\\\ \therefore \triangle ABD&\sim \triangle CBE\end{aligned}$$ AA criterion of similarity(iii) In \(\triangle AEP\) and \(\triangle ADB\)
$$\begin{aligned}\angle E&=\angle D\left( 90^{0}\right) \\ \angle DAB&=\angle EAP\quad\text{(Common)} \\\\ \therefore \triangle AEP&\sim \triangle DB\end{aligned}$$ By AA Criterion of similarity(iv) In \(\triangle PDC\) and \(\triangle BEC\) $$\begin{aligned}\angle D&=\angle E\left( 90^{\circ }\right) \\ \angle DCP&=\angle ECB\quad\text{(Common)} \\\\ \therefore \triangle PDC&\sim \triangle BEC\end{aligned}$$ By AA criterion of Similarity
Q8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that \(\triangle ABE \sim \triangle CFB.\)
Solution:
To Prove $$\triangle ABE\sim \triangle CFB$$ Proof:In \(\triangle ABE\) and \(\triangle CFB\)
$$\begin{aligned}\angle BEA&=\angle FBC\\\scriptsize\text{(Altern}&\text{ate Angle)} \\ \angle EAB&=\angle BCF\\ \scriptsize\text{(Opposite interior }&\scriptsize\text{angle of parallelogram)}\\\\ \therefore \triangle ABE&\sim \triangle CFB\end{aligned}$$ By AA Criterion of similarity
Q9. In Fig. 6.39, ABC and AMP are two right
triangles, right-angled at B and M
respectively. Prove that:
(i) \(\triangle ABC \sim \triangle AMP\)
(ii) \(\dfrac{CA}{PA}=\dfrac{BC}{MB}\)
Solution:
Given:\(\triangle ABC\) and \(\triangle AMP\) are right-angled triangle right angled at B and M
To Prove: $$\triangle ABC\sim \triangle AMP$$ Proof: (i)$$\begin{aligned}\angle A&=\angle A\quad\text{( Common)} \\ \angle ABC&=\angle AMP\left( 90^{\circ }\text{: Given}\right) \\\\ \therefore \triangle ABC&\sim \triangle AMP\end{aligned}$$ By AA Criterion of similarity(ii)$$\begin{aligned}\triangle ABC&\sim \triangle AMP\\ \therefore \dfrac{CA}{PA}&=\dfrac{BC}{MP}\quad\text{(By BPT)}\end{aligned}$$
Q10. CD and GH are respectively the bisectors
of \(\angle ACB\) and \(\angle EGF\) such that D and H lie
on sides AB and FE of \(\triangle ABC\) and \(\triangle EFG\)
respectively. If \(\triangle ABC \sim \triangle FEG\), show that:
(i) \(\dfrac{CD}{GH}=\dfrac{AC}{FG}\)
(ii) \(\triangle DCB \sim \triangle HGE\)
(iii) \(\triangle DCA \sim \triangle HGF\)
Solution:
Given:
$$\triangle ABC\sim \triangle FEG$$(i) In \(\triangle CDA\) and \(\triangle GHF\)
$$\scriptsize\begin{aligned}\angle ACD&=\angle FGH\quad\text{(Angle Bisector)} \\ \angle DAC&=\angle HFG\quad\text{(Common)} \\\\ \therefore \triangle CDA&\sim \triangle GHF\quad\text{(AA Criterion)}\\ \dfrac{CD}{GH}&=\dfrac{AC}{FG}\quad\text{(By BPT)}\end{aligned}$$(ii) To Prove $$\triangle DCB\sim \triangle HGE$$ Proof:
In \(\triangle DCB\) and \(\triangle HGE\)
$$\scriptsize\begin{aligned}\angle DCB&=\angle HGE\quad\text{(Angle Bisector)} \\ \angle CBD&=\angle GEH\quad\text{(Corresponding Angle)} \\ \triangle DCB&\sim \triangle HGE\end{aligned}$$ By AA Criterion of similarity(iii)To Prove $$\triangle DCA\sim \triangle HGF$$ Proof:
In \(\triangle DCA\) and \(\triangle HGF\)
$$\scriptsize\begin{aligned}\angle A&=\angle F\quad\text{(Common)} \\ \angle DCA&=\angle HGF\quad\text{(Angle Bisector)}\\\\ \therefore \triangle DCA&\sim \triangle HGF\end{aligned}$$ By AA criterion of Similarity
Q11. In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If \(AD \perp BC\) and \(EF \perp AC\), prove that \(\triangle ABD \sim \triangle ECF.\)
Solution:
Given:
ABC is an isosceles triangle
E is a point on CB
AB = AC
AD ⟂ BC
EF ⟂ AC
Q12. Sides AB and BC and median AD of a triangle ABC are respectively propor tional to sides PQ and QR and median PM of \(\triangle PQR\) (see Fig. 6.41). Show that \(\triangle ABC \sim \triangle PQR.\)
Solution:
Given: $$\dfrac{AB}{PQ}=\dfrac{BC}{QR}=\dfrac{AD}{PM}$$ To Prove: $$\triangle ABC\sim \triangle PQR$$ Proof:\(AD\) and \(PM\) are medians of \(\triangle ABC\) and \(\triangle PQR\)
$$\dfrac{BD}{QM}=\dfrac{BC/2}{QR/2}=\dfrac{BC}{QR}$$In \(\triangle ABD\) and \(\triangle PQM\)
$$\scriptsize\begin{aligned}\dfrac{AB}{PQ}&=\dfrac{BD}{QM}=\dfrac{AD}{PM}\\\\ \therefore \triangle ABD\sim &\triangle PQM\quad\text{(SSS Criterion)}\end{aligned}$$In \(\triangle ABC\) and \(\triangle PQR\)
$$\scriptsize\begin{aligned}\dfrac{AB}{PQ}&=\dfrac{BC}{QR}\\ \angle ABC&=\angle PQR\\\\ \therefore \triangle ABC\sim \triangle PQR&\quad\text{(SAS Criterrion)}\end{aligned}$$ By SAS criterion of similarity
Q13. D is a point on the side BC of a triangle ABC such that \(\angle ADC = \angle BAC\). Show that \(CA^2 = CB.CD\).
Solution:
Given: $$\angle ADC=\angle BAC$$ To Prove:\[CA^2 = CB\cdot CD\] Proof:
In \(\triangle ABC\) and \(\triangle ADC\)
$$\scriptsize\begin{aligned}\angle ADC&=\angle BAC\quad\text{(Given)} \\ \angle C&=\angle C\quad\text{(Common)} \\\\ \therefore \triangle BAC&\sim \triangle ADC\quad\text{(AA Criterion)}\\\\ \dfrac{CA}{CB}&=\dfrac{CD}{CA}\\\\ \Rightarrow CA^{2}&=CB\cdot AD\end{aligned}$$
Q14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that \(\triangle ABC \sim \triangle PQR.\)
Solution:
Given that $$\dfrac{AB}{PQ}=\dfrac{AC}{PR}=\dfrac{AD}{PM}$$ To Prove $$\triangle ABC\sim APQR$$Construction: \(AD\) is produced to \(E\) such that \(AD = DE\), and produce \(PM\) to \(S\) such that \(PM = PS\)
Proof:In triangles \(\triangle ADC\) and \(\triangle BDE\)
$$\scriptsize\begin{aligned} BD &= DC \quad\text{(AD is Median)}\\ AD &= AE \quad\text{(By Construction)}\\ \angle BDE&=\angle ADC\\ \scriptsize\text{ (Vertically }&\scriptsize\text{Opposite Angle)}\end{aligned}$$ $$\scriptsize\begin{align}\triangle ADC&\cong \triangle BDE\\ BE&=AC\quad\text{(By CPCT)} \tag{1}\end{align}$$In triangle \(\triangle PRM\) and \(\triangle SQM\)
$$\scriptsize\begin{aligned} QR &=MR \quad\text{(PM median)}\\ PM &= PS \quad\text{(By Construction)}\\ \angle QMS&=\angle PMR\\ \scriptsize\text{ (Vertically }&\scriptsize\text{Opposite Angle)}\end{aligned}$$ $$\scriptsize\begin{align}\triangle PRM&\cong \triangle SQM\\ PR&=QS\quad\text{(By CPCT)} \tag{2}\end{align}$$ $$\scriptsize\begin{aligned}\dfrac{AB}{PQ}&=\dfrac{AC}{PR}=\dfrac{AD}{PM}\quad\text{(Given)}\\ \dfrac{AB}{PQ}&=\dfrac{AC}{PR}=\dfrac{2AD}{2PM}\\ \dfrac{AB}{PQ}&=\dfrac{BE}{QS}=\dfrac{AE}{PS}\\\\ \therefore \triangle ABE&\sim \triangle PQS\quad\text{(SSS Criterion)} \\ \angle BAD&=\angle QPM\end{aligned}$$ Similarly $$\scriptsize\begin{align}\angle CAD&=\angle RPM\\ \angle BAD+\angle CAD&=\angle QPM+\angle MPR\\ \angle BAC&=\angle QPR\tag{3} \end{align}$$ $$\scriptsize\begin{aligned}\dfrac{AB}{PQ}&=\dfrac{AC}{PR}\quad\text{(Given)}\\ \angle BAC&=\angle QPR\\\\ \triangle ABC &\sim \triangle PQR\quad\text{(SAS Criterion)} \end{aligned}$$ Hence Proved
Q15. A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same a tower casts a shadow 28 m long. Find the height of the tower.
Solution:
length of the Pole = 6m
shadow of the first pole on the ground = 4m
Shadow of tower = 28m
tower and pole will form similar Triangle on the ground with their shadows,
hence,
Q16. If AD and PM are medians of triangles ABC and PQR, respectively, where \(\triangle ABC \sim \triangle PQR\), prove that \(\dfrac{AB}{PQ}=\dfrac{AD}{PM}\)
Solution:
Given $$\triangle ABC\sim \triangle PQR$$ To Prove $$\dfrac{AB}{PQ}=\dfrac{AD}{PM}$$ Proof:In \(\triangle ABD\) and \(\triangle PQM\)
$$\scriptsize\begin{aligned}\triangle ABC&\sim \triangle PQR\\\\ \dfrac{AB}{PQ}&=\dfrac{BD}{QM}\\\\ \text{Multiply and }&\text{Divide RHS by 2}\\\\ \dfrac{AB}{PQ}&=\dfrac{2BD}{2QM}\\\\ 2BD&=BC\quad\text{(AD is Bisector)}\\ 2QM&=QR\quad\text{(PQ is Bisector)}\\\\ \text{Hence,}\\ \dfrac{AB}{PQ}&=\dfrac{BC}{QR}\\ \angle B&=\angle Q\quad\text{(By BPT)}\\\\ \therefore \triangle ABD&\sim \triangle PQM\quad\text{(SAS Criterion)}\\\\ \Rightarrow \dfrac{AB}{PQ}&=\dfrac{AD}{PM}\end{aligned}$$ Hence Proved.
