NCERT Solutions Class 10 Maths Chapter 6: Triangles

#### **Q1. In Fig. 6.17, (i) and (ii), $DE \parallel BC$. Find $EC$ in (i) and $AD$ in (ii).** *(Fig. (i): $AD=1.5, DB=3, AE=1$ | Fig. (ii): $DB=7.2, AE=1.8, EC=5.4$)* #### **A1. Solution:** **(i)** Since $DE \parallel BC$, by BPT: $\frac{AD}{DB} = \frac{AE}{EC}$ $\frac{1.5}{3} = \frac{1}{EC}$ $1.5 \times EC = 3$ $EC = \frac{3}{1.5} = 2$ cm. **Ans: $EC = 2$ cm** **(ii)** Since $DE \parallel BC$, by BPT: $\frac{AD}{DB} = \frac{AE}{EC}$ $\frac{AD}{7.2} = \frac{1.8}{5.4}$ $\frac{AD}{7.2} = \frac{1}{3}$ $3 \times AD = 7.2$ $AD = \frac{7.2}{3} = 2.4$ cm. **Ans: $AD = 2.4$ cm** #### **Q2. E and F are points on the sides PQ and PR respectively of a $\triangle PQR$. For each of the following cases, state whether $EF \parallel QR$:** **(i) $PE = 3.9$ cm, $EQ = 3$ cm, $PF = 3.6$ cm and $FR = 2.4$ cm** **(ii) $PE = 4$ cm, $QE = 4.5$ cm, $PF = 8$ cm and $RF = 9$ cm** **(iii) $PQ = 1.28$ cm, $PR = 2.56$ cm, $PE = 0.18$ cm and $PF = 0.36$ cm** #### **A2. Solution:** **(i)** $\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$ $\frac{PF}{FR} = \frac{3.6}{2.4} = 1.5$ Since $\frac{PE}{EQ} \neq \frac{PF}{FR}$, **$EF$ is not parallel to $QR$.** **(ii)** $\frac{PE}{QE} = \frac{4}{4.5} = \frac{8}{9}$ $\frac{PF}{RF} = \frac{8}{9}$ Since $\frac{PE}{QE} = \frac{PF}{RF}$, by Converse of BPT, **$EF \parallel QR$.** **(iii)** $EQ = PQ - PE = 1.28 - 0.18 = 1.10$ cm $FR = PR - PF = 2.56 - 0.36 = 2.20$ cm $\frac{PE}{EQ} = \frac{0.18}{1.10} = \frac{18}{110} = \frac{9}{55}$ $\frac{PF}{FR} = \frac{0.36}{2.20} = \frac{36}{220} = \frac{9}{55}$ Since $\frac{PE}{EQ} = \frac{PF}{FR}$, by Converse of BPT, **$EF \parallel QR$.** #### **Q3. In Fig. 6.18, if $LM \parallel CB$ and $LN \parallel CD$, prove that $\frac{AM}{AB} = \frac{AN}{AD}$.** #### **A3. Solution:** In $\triangle ABC$, $LM \parallel CB$. By BPT, $\frac{AM}{AB} = \frac{AL}{AC}$ ...(1) *(Note: Using the corollary of BPT)* In $\triangle ADC$, $LN \parallel CD$. By BPT, $\frac{AN}{AD} = \frac{AL}{AC}$ ...(2) From (1) and (2), we get: $\frac{AM}{AB} = \frac{AN}{AD}$ **Hence Proved.** #### **Q4. In Fig. 6.19, $DE \parallel AC$ and $DF \parallel AE$. Prove that $\frac{BF}{FE} = \frac{BE}{EC}$.** #### **A4. Solution:** In $\triangle ABE$, $DF \parallel AE$. By BPT, $\frac{BF}{FE} = \frac{BD}{DA}$ ...(1) In $\triangle ABC$, $DE \parallel AC$. By BPT, $\frac{BE}{EC} = \frac{BD}{DA}$ ...(2) From (1) and (2), we get: $\frac{BF}{FE} = \frac{BE}{EC}$ **Hence Proved.** #### **Q5. In Fig. 6.20, $DE \parallel OQ$ and $DF \parallel OR$. Show that $EF \parallel QR$.** #### **A5. Solution:** In $\triangle POQ$, $DE \parallel OQ$. By BPT, $\frac{PE}{EQ} = \frac{PD}{DO}$ ...(1) In $\triangle POR$, $DF \parallel OR$. By BPT, $\frac{PF}{FR} = \frac{PD}{DO}$ ...(2) From (1) and (2), we get: $\frac{PE}{EQ} = \frac{PF}{FR}$ In $\triangle PQR$, since the sides are divided in the same ratio, by Converse of BPT: $EF \parallel QR$ **Hence Proved.** #### **Q6. In Fig. 6.21, A, B and C are points on OP, OQ and OR respectively such that $AB \parallel PQ$ and $AC \parallel PR$. Show that $BC \parallel QR$.** #### **A6. Solution:** In $\triangle OPQ$, $AB \parallel PQ$. By BPT, $\frac{OA}{AP} = \frac{OB}{BQ}$ ...(1) In $\triangle OPR$, $AC \parallel PR$. By BPT, $\frac{OA}{AP} = \frac{OC}{CR}$ ...(2) From (1) and (2), we get: $\frac{OB}{BQ} = \frac{OC}{CR}$ In $\triangle OQR$, by Converse of BPT: $BC \parallel QR$ **Hence Proved.** #### **Q7. Using Theorem 6.1 (BPT), prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.** #### **A7. Solution:** Let $\triangle ABC$ be the triangle where $D$ is the mid-point of $AB$. A line $DE$ is drawn parallel to $BC$. To prove: $E$ is the mid-point of $AC$ (i.e., $AE = EC$). Proof: Since $D$ is the mid-point of $AB$, $AD = DB \Rightarrow \frac{AD}{DB} = 1$ ...(1) In $\triangle ABC$, $DE \parallel BC$. By BPT: $\frac{AD}{DB} = \frac{AE}{EC}$ From (1), $1 = \frac{AE}{EC}$ $AE = EC$. Therefore, $E$ bisects $AC$. **Hence Proved.** #### **Q8. Using Theorem 6.2 (Converse of BPT), prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.** #### **A8. Solution:** Let $\triangle ABC$ be the triangle where $D$ and $E$ are mid-points of $AB$ and $AC$ respectively. To prove: $DE \parallel BC$. Proof: Since $D$ is the mid-point of $AB$, $AD = DB \Rightarrow \frac{AD}{DB} = 1$ ...(1) Since $E$ is the mid-point of $AC$, $AE = EC \Rightarrow \frac{AE}{EC} = 1$ ...(2) From (1) and (2), $\frac{AD}{DB} = \frac{AE}{EC}$. In $\triangle ABC$, since the line $DE$ divides two sides in the same ratio, by Converse of BPT: $DE \parallel BC$ **Hence Proved.** #### **Q9. ABCD is a trapezium in which $AB \parallel DC$ and its diagonals intersect each other at the point O. Show that $\frac{AO}{BO} = \frac{CO}{DO}$.** #### **A9. Solution:** Draw a line $EO$ through $O$ such that $EO \parallel DC$. In $\triangle ADC$, $EO \parallel DC$. By BPT, $\frac{AE}{ED} = \frac{AO}{OC}$ ...(1) Since $AB \parallel DC$ and $EO \parallel DC$, then $EO \parallel AB$. In $\triangle ABD$, $EO \parallel AB$. By BPT, $\frac{ED}{AE} = \frac{DO}{OB} \Rightarrow \frac{AE}{ED} = \frac{OB}{DO}$ ...(2) From (1) and (2): $\frac{AO}{OC} = \frac{OB}{DO}$ Rearranging the terms: $\frac{AO}{BO} = \frac{CO}{DO}$ **Hence Proved.** #### **Q10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that $\frac{AO}{BO} = \frac{CO}{DO}$. Show that ABCD is a trapezium.** #### **A10. Solution:** Given: $\frac{AO}{BO} = \frac{CO}{DO} \Rightarrow \frac{AO}{CO} = \frac{BO}{DO}$ Draw $EO \parallel AB$ intersecting $AD$ at $E$. In $\triangle ABD$, $EO \parallel AB$. By BPT: $\frac{AE}{ED} = \frac{BO}{DO}$ ...(1) But we are given $\frac{AO}{CO} = \frac{BO}{DO}$ ...(2) From (1) and (2), $\frac{AE}{ED} = \frac{AO}{CO}$. In $\triangle ADC$, since $\frac{AE}{ED} = \frac{AO}{CO}$, by Converse of BPT: $EO \parallel DC$. We started with $EO \parallel AB$ and found $EO \parallel DC$. Therefore, $AB \parallel DC$. Since one pair of opposite sides is parallel, ABCD is a trapezium. **Hence Proved.**