Triangles | Exercise 6.2

Question 1

In Fig. 6.17, (i) and (ii), DE∥BCDE \parallel BC. Find ECEC in (i) and ADAD in (ii).

Question diagram 1Question diagram 2
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Solution
Understand the Question
  • Basic Proportionality Theorem (Thales's Theorem): If a line is drawn parallel to one side of a triangle intersecting the other two sides in distinct points, then it divides the two sides in the same ratio.
  • In ΔABC\Delta ABC, since DE∥BCDE \parallel BC, by Basic Proportionality Theorem: ADDB=AEEC\dfrac{AD}{DB} = \dfrac{AE}{EC}
  • We substitute the given lengths into this proportion to find the unknown segment in each part.

(i) Find ECEC in (i)

Step 1 · Calculate the length of ECEC

Given AD=1.5 cmAD = 1.5\text{ cm}, DB=3 cmDB = 3\text{ cm}, and AE=1 cmAE = 1\text{ cm}.Diagram 1

In ΔABC\Delta ABC, since DE∥BCDE \parallel BC, by Basic Proportionality Theorem (BPT) ADDB=AEEC\dfrac{AD}{DB} = \dfrac{AE}{EC}

Let EC=xEC = x. Substituting the given values 1.53=1x\dfrac{1.5}{3} = \dfrac{1}{x}

1.5×x=3×11.5x=3x=31.5x=2\begin{aligned} 1.5 \times x &= 3 \times 1 \\[0.6em] 1.5x &= 3 \\[0.6em] x &= \dfrac{3}{1.5} \\[0.6em] x &= 2 \end{aligned}

Therefore, EC=2 cmEC = 2\text{ cm}.

Answer

(i) EC=2 cmEC = 2 \text{ cm}

(ii) Find ADAD in (ii)

Step 1 · Calculate the length of ADAD

Given DB=7.2 cmDB = 7.2\text{ cm}, AE=1.8 cmAE = 1.8\text{ cm}, and EC=5.4 cmEC = 5.4\text{ cm}.Diagram 2

In ΔABC\Delta ABC, since DE∥BCDE \parallel BC, by Basic Proportionality Theorem (BPT) ADDB=AEEC\dfrac{AD}{DB} = \dfrac{AE}{EC}

Let AD=xAD = x. Substituting the given values x7.2=1.85.4\dfrac{x}{7.2} = \dfrac{1.8}{5.4}

x=1.8×7.25.4x = \dfrac{1.8 \times 7.2}{5.4}

Simplifying 1.85.4\dfrac{1.8}{5.4}

1.85.4=1854=13\begin{aligned} \dfrac{1.8}{5.4} &= \dfrac{18}{54} \\[0.6em] &= \dfrac{1}{3} \end{aligned}

Substituting back

x=13×7.2=7.23=2.4\begin{aligned} x &= \dfrac{1}{3} \times 7.2 \\[0.6em] &= \dfrac{7.2}{3} \\[0.6em] &= 2.4 \end{aligned}

Therefore, AD=2.4 cmAD = 2.4\text{ cm}.

Answer

(ii) AD=2.4 cmAD = 2.4 \text{ cm}

Common Mistakes
  • Ratio Mismatch: Setting up incorrect ratios such as ADAB=AEEC\dfrac{AD}{AB} = \dfrac{AE}{EC} instead of using the corresponding parts ADDB=AEEC\dfrac{AD}{DB} = \dfrac{AE}{EC}.
  • Decimal Calculation Slip: Making errors when simplifying fractions involving decimals like 1.85.4\dfrac{1.8}{5.4} or 31.5\dfrac{3}{1.5}. Convert them into whole-number fractions (e.g., 1854=13\dfrac{18}{54} = \dfrac{1}{3}) to avoid arithmetic mistakes.

More questions in Exercise 6.2

Q1

In Fig. 6.17, (i) and (ii), DE∥BCDE \parallel BC. Find ECEC in (i) and ADAD in (ii).

Q2

EE and FF are points on the sides PQPQ and PRPR respectively of a ΔPQR\Delta PQR. For each of the following cases, state whether EF∥QREF \parallel QR :

(i) PE=3.9 cmPE = 3.9\text{ cm}, EQ=3 cmEQ = 3\text{ cm}, PF=3.6 cmPF = 3.6\text{ cm}, and FR=2.4 cmFR = 2.4\text{ cm}

(ii) PE=4 cmPE = 4\text{ cm}, QE=4.5 cmQE = 4.5\text{ cm}, PF=8 cmPF = 8\text{ cm}, and RF=9 cmRF = 9\text{ cm}

(iii) PQ=1.28 cmPQ = 1.28\text{ cm}, PR=2.56 cmPR = 2.56\text{ cm}, PE=0.18 cmPE = 0.18\text{ cm}, and PF=0.36 cmPF = 0.36\text{ cm}

Q3

In Fig. 6.18, if LM∥CB\text{LM} \parallel \text{CB} and LN∥CD\text{LN} \parallel \text{CD}, prove that

AMAB=ANAD\dfrac{\text{AM}}{\text{AB}} = \dfrac{\text{AN}}{\text{AD}}

Q4

In Fig. 6.19, DE∥ACDE \parallel AC and DF∥AEDF \parallel AE. Prove that

BFFE=BEEC\dfrac{\text{BF}}{\text{FE}} = \dfrac{\text{BE}}{\text{EC}}

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