TRIANGLES-Exercise 6.2
Maths - Exercise
Q1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Solution:
$$ DE\parallel BC$$ To Find EC $$\begin{aligned}\dfrac{AD}{DB}&=\dfrac{AE}{EC}\quad\left( \text{By BPT}\right) \\ \dfrac{1.5}{3}&=\dfrac{1}{EC}\\ EC&=\dfrac{3}{1.5}\\ &=2cm\end{aligned}$$TO Find AD $$\begin{aligned}\dfrac{AE}{EC}&=\dfrac{AD}{BD}\quad\left( \text{By BPT}\right) \\ \dfrac{1.8}{5.4}&=\dfrac{AD}{7.2}\\ AD&=\dfrac{1.8\times 7.2}{5.4}\\ &=2.4\end{aligned}$$
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 :
- PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
- PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
- PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm
Solution:
- PE=3.9cm, EA=3cm, PF=3.6cm, FR=2.4cm Let us assume that EF || QR, therefore, by BPT $$\begin{aligned}\dfrac{PE}{EQ}&=\dfrac{PF}{FR}\\ \dfrac{PE}{EQ}&=\dfrac{39}{3}=1.3\\\\ \dfrac{PF}{FR}&=\dfrac{3.6}{2\cdot 4}=\dfrac{3}{2}\\\\ \dfrac{PE}{EQ}&\neq \dfrac{PF}{FR}\end{aligned}$$ Hence, EF is not parallel to QR
- PE=4cm, QE=4.5cm, PF=8cm, RF=9cm Let us assume that EF || QR, therefore, by BPT $$\begin{aligned}\dfrac{PE}{QE}&=\dfrac{PF}{RF}\\ \dfrac{PE}{QE}&=\dfrac{4}{4.5}\\ &=\dfrac{40}{45}&\\ &=\dfrac{8}{9}\\\\ \dfrac{PF}{RF}&=\dfrac{8}{9}\\\\ \dfrac{PE}{QE}&=\dfrac{PF}{RF}=\dfrac{8}{9}\end{aligned}$$ Hence $$ EF\parallel QR$$
- PQ=1.28cm, PR=2.56cm, PE=0.18cm, PF=0.36cm Let us assume that EF || QR, therefore, by BPT $$\begin{aligned}\dfrac{PE}{PQ}&=\dfrac{PF}{PR}\\ \dfrac{PE}{PQ}&=\dfrac{0.18}{1.28}\\&=\dfrac{9}{64}\\\\ \dfrac{PF}{PR}&=\dfrac{0.36}{2.56}\\&=\dfrac{9}{64}\\\\ \dfrac{PE}{PQ}=&\dfrac{PF}{PR}=\dfrac{9}{64}\end{aligned}$$ Hence, $$ EF\parallel QR$$
Q3. In Fig. 6.18, if LM || CB and LN || CD, prove that \(\dfrac{AM}{AB}=\dfrac{AN}{AD}\)
Solution:
Given: $$\begin{aligned}LM\parallel CB\\ LN\parallel CD\end{aligned}$$ To Prove $$\begin{aligned}\dfrac{AM}{AB}=\dfrac{AN}{AD}\end{aligned}$$ Proof: In \(\triangle ABC\) \[LM\parallel BC \quad\text{( Given)}\] therefore by BPT \[\dfrac{AM}{MB}=\dfrac{AL}{LC}\tag{1}\] In triangle \(\triangle ADC\) \[LN\parallel CD\quad\text{( Given)}\] therefore by BPT \[\dfrac{AN}{ND}=\dfrac{AL}{LC}\tag{2}\] From equation-1 and equation-2 \[\dfrac{MB}{AM}=\dfrac{ND}{AN}\] or \[\dfrac{AM}{MB}=\dfrac{AN}{ND}\] Adding 1 to both sides of the equation $$\begin{aligned} \dfrac{MB}{AM}+1&=\dfrac{ND}{AN}+1\\\\ \dfrac{MB+AM}{AM}&=\dfrac{ND+AN}{AN}\\\\ \dfrac{AB}{AM}&=\dfrac{AD}{AN}\\\\ &\text{or}\\\\ \dfrac{AM}{AB}&=\dfrac{AN}{AD}\\ \end{aligned}$$ Thus Proved
Q4. In Fig. 6.19, DE || AC and DF || AE. Prove that
\(\dfrac{BF}{FE}=\dfrac{BE}{EC}\)
Solution:
Given: $$\begin{aligned}DE\parallel AC\\ DF\parallel AE\end{aligned}$$ To Prove: $$\dfrac{BF}{FE}=\dfrac{BE}{EC}$$Proof:
In triangle \(\triangle ABC\)
In triangle \(\triangle ABE\)
$$ \Delta F\parallel AE\quad\text{(Given)}$$ therefore by BPT $$\dfrac{BD}{DA}=\dfrac{BF}{FE}\tag{2}$$ From equation-1 and equation-2 $$\dfrac{BF}{FE}=\dfrac{BE}{EC}$$ Hence Proved.
Q5. In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
Solution:
Given: $$\begin{aligned}DE\parallel OQ\\ DF\parallel OR\end{aligned}$$ To prove: \[EF\parallel QR\] ProofIn Triangle \(\triangle POQ\)
\[DE\parallel OQ\] Therefore, by BPT \[\dfrac{PE}{EQ}=\dfrac{PD}{DO}\tag{1}\]In triangle \(\triangle POR\)
\[DF\parallel OR\] Therefore by BPT \[\dfrac{PD}{DO}=\dfrac{PF}{FR}\tag{2}\] From equation-1 and equation-2 \[\dfrac{PE}{EQ}=\dfrac{PF}{FR}\] Adding 1 to both sides of the equation $$\begin{aligned}\dfrac{PE}{EQ}+1&=\dfrac{PF}{FR}+1\\\\ \dfrac{PE+EQ}{EQ}&=\dfrac{PF+FR}{FR}\\\\ \dfrac{PQ}{EQ}&=\dfrac{PR}{FR}\end{aligned}$$ By the 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 || PQ and AC || PR. Show that BC || QR.
Solution:
Given: $$\begin{aligned} AB\parallel PQ\\ AC\parallel PR\end{aligned}$$ To Prove: \[BC\parallel QR\]Proof:
In triangle \(\triangle POQ\)
\[ AB\parallel PQ\\ \quad\text{(Given)} \] By BPT\ \[\dfrac{PA}{AO}=\dfrac{QB}{BO}\tag{1}\] In triangle \(\triangle POR\)By BPT \[\dfrac{PA}{AO}=\dfrac{RC}{CO}\tag{2}\] From equation-1 and equation-2 \[\dfrac{QB}{BO}=\dfrac{RC}{CO}\] By Converse of BPT \[BC\parallel QR\] Hence, Proved.
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).
Solution:
Given:E is mid-point of AB
To Prove \[\begin{aligned}&EF\parallel BC\\ &\text{F is mid-point of AC}\end{aligned}\] Proof: \[\begin{align}\dfrac{AE}{EB}=1\tag{1}\\ \text{(E is mid-point)} \end{align}\] \[EF\parallel BC\quad\text{( Given)} \] \[\dfrac{AE}{EB}=\dfrac{AF}{FC}\quad\text{(Theorem 6.1)}\] From equation-(1) $$\begin{aligned}\dfrac{AF}{FC}&=1\\ \Rightarrow AF&=FC \end{aligned}$$ Hence, F is a mid-point of side AC
Q8. Using Theorem 6.2, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
Solution:
\(E\) and \(F\) are midpoints on \(AB\) and \(AC\) respectively
therefore,
Q9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \(\dfrac{AO}{BO}=Dfrac{CO}{DO}\)
Solution:
Given: $$ AB\parallel DC$$ To Prove $$\dfrac{AO}{BO}=\dfrac{CO}{DO}$$
Construction:
Draw a line \(EF\) passing through \(O\) and parallel to \(AB\)
In triangle \(\triangle ACD\)
By BPT $$\dfrac{AO}{OC}=\dfrac{AE}{ED}\tag{1}$$In tringle \(\triangle ABD\)
$$\dfrac{BO}{OD}=\dfrac{AE}{ED}\tag{2}$$ From equation-1 and equation-2 $$\begin{aligned}\dfrac{AO}{OC}&=\dfrac{BO}{OD}\\\\ \Rightarrow \dfrac{AO}{BO}&=\dfrac{CO}{DO}\end{aligned}$$ Hence Proved.
10. The diagonals of a quadrilateral ABCD intersect each other at the point O, such that \(\dfrac{AO}{BO}\dfrac{CO}{DO}) Show that ABCD is a trapezium
Solution:
Given $$\dfrac{AO}{BO}=\dfrac{CO}{DO}$$To Prove:
ABCD is a trapezium
Construction:
Draw a line E F passing through 0 and parallel to AB
