Chapter 6 · Triangles

NCERT Class 10 Maths Exercise 6.2

All 10 questions of NCERT Class 10 Maths Exercise 6.2 (Triangles), in the book's order, each with its answer and a step-by-step solution.

1. Q1
   In Fig. 6.17, ​(i) and (ii), DE || BC. Find EC in ​(i) and AD ​in (ii).

   1. (i)​Find $EC$, ​given $AD=1.5$ cm, $DB=3$ ​cm, $AE=1$ cm (Fig. 6.17 (i)).
   2. (ii)​Find $AD$, given $DB=7.2$ cm, $AE=1.8$ cm, $EC=5.4$ cm (Fig. 6.17 (ii)).

   Show answer

   (i) $EC=2$ cm. (ii) $AD=2.4$ cm.

   Step-by-step solution →
2. 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 || QR:

   1. (i)PE ​= 3.9 cm, EQ = 3 cm, PF = 3.6 cm and ​FR = 2.4 cm
   2. (ii)PE = 4 cm, QE = 4.5 cm, ​PF ​= ​8 cm and RF = 9 cm
   3. (iii)PQ = ​1.28 cm, PR = 2.56 cm, PE = 0.18 cm ​and PF = 0.36 cm
   Show answer

   (i) $EF\nparallel QR$ (since $1.3\ne 1.5$); (ii) And (iii) $EF\parallel QR$.

   Step-by-step solution →
3. Q3
   In ​Fig. 6.18, ​if $LM\parallel CB$ and $LN\parallel CD$, prove ​that $\frac{AM}{AB}=\frac{AN}{AD}$.

   Show answer

   Proved: $\frac{AM}{AB}=\frac{AN}{AD}$, using the Basic Proportionality Theorem in $△ABC$ and $△ACD$.

   Step-by-step solution →
4. Q4
   In ​Fig. 6.19, $DE\parallel AC$ and $DF\parallel AE$. ​Prove ​that $\frac{BF}{FE}=\frac{BE}{EC}$.

   Show answer

   Proved. By the Basic Proportionality Theorem in $△ABE$ (where $DF\parallel AE$), $\frac{BF}{FE}=\frac{BD}{DA}$; by the same theorem in $△ABC$ (where $DE\parallel AC$), $\frac{BE}{EC}=\frac{BD}{DA}$. Hence $\frac{BF}{FE}=\frac{BE}{EC}$.

   Step-by-step solution →
5. Q5
   In ​Fig. ​6.20, $DE\parallel OQ$ ​and $DF\parallel OR$. Show ​that $EF\parallel QR$.

   Show answer

   Proved: $EF\parallel QR$, using BPT in $△POQ$ and $△POR$ and the converse of BPT in $△PQR$.

   Step-by-step solution →
6. 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 $BC\parallel QR$.

   Show answer

   Since $\frac{OA}{AP}=\frac{OB}{BQ}$ and $\frac{OA}{AP}=\frac{OC}{CR}$, we get $\frac{OB}{BQ}=\frac{OC}{CR}$, so by the converse of the Basic Proportionality Theorem, $BC\parallel QR$.

   Step-by-step solution →
7. Q7
   Using Theorem 6.1, ​prove ​that a line drawn through the mid-point of one side ​of a triangle parallel to another ​side bisects the third side. (Recall that you have ​proved it in Class IX).
   Show answer

   Since $MN\parallel BC$, the Basic Proportionality Theorem (Theorem 6.1) gives $\frac{AM}{MB}=\frac{AN}{NC}$. As $M$ is the mid-point of $AB$, $AM=MB$, so the ratio is $1$ and therefore $AN=NC$. Hence $N$ is the mid-point of $AC$, i.e. the line bisects the third side.

   Step-by-step solution →
8. Q8
   Using ​Theorem 6.2, prove that the line joining ​the mid-points of any two sides of a ​triangle is parallel ​to the third side. (Recall that you have done it in Class ​IX).
   Show answer

   Since $D$ and $E$ are mid-points, $\frac{AD}{DB}=\frac{AE}{EC}=1$; by Theorem 6.2 (the converse of the Basic Proportionality Theorem), the line $DE$ joining the mid-points is parallel to the third side, so $DE\parallel BC$.

   Step-by-step solution →
9. Q9
   ABCD is a trapezium in ​which $AB\parallel DC$ and ​its diagonals intersect each ​other at ​the point O. Show ​that $\frac{AO}{BO}=\frac{CO}{DO}$.
   Show answer

   Proved: $\frac{AO}{BO}=\frac{CO}{DO}$, using $△AOB\sim △COD$ (AA similarity).

   Step-by-step solution →
10. 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.
    Show answer

    Proved: $AB\parallel DC$, so $ABCD$ is a trapezium (one pair of parallel sides).

    Step-by-step solution →