Exercise 6.2 answers: Triangles

Class 10 Maths10 questions

Exercise 6.2

10 questions · page 84 of the book

Question 1

“In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).” · p. 84

Open NCERT p. 84Matches NCERT’s answer

(i)

  1. In triangle ABC, D is on AB with AD = 1.5 cm and DB = 3 cm; E is on AC with AE = 1 cm.
  2. Since DE || BC, the Basic Proportionality Theorem gives AD/DB = AE/EC.
  3. So 1.5/3 = 1/EC, which gives EC = 1 × (3/1.5) = 2 cm.

AnswerEC = 2 cm

(ii)

  1. In triangle ABC, D is on AB and E is on AC, with DB = 7.2 cm, AE = 1.8 cm and EC = 5.4 cm.
  2. Since DE || BC, the Basic Proportionality Theorem gives AD/DB = AE/EC.
  3. So AD/7.2 = 1.8/5.4 = 1/3, which gives AD = 7.2/3 = 2.4 cm.

AnswerAD = 2.4 cm

Watch this explained “The statement, word by word”, 4:08 into A line drawn parallel to a side cuts the other two in matching ratios

Question 2

“E and F are points on the sides PQ and PR … state whether EF || QR” · p. 84

Open NCERT p. 84Matches NCERT’s answer

(i) PE = 3.9 cm, EQ = 3 cm …

  1. Check PE/EQ against PF/FR.
  2. PE/EQ = 3.9/3 = 1.3; PF/FR = 3.6/2.4 = 1.5.
  3. The two ratios are different, so by the converse of the Basic Proportionality Theorem, EF is not parallel to QR.

AnswerNo, EF is not parallel to QR.

(ii) PE = 4 cm, QE = 4.5 cm …

  1. Check PE/QE against PF/RF.
  2. PE/QE = 4/4.5 = 8/9; PF/RF = 8/9.
  3. The ratios are equal, so by the converse of the Basic Proportionality Theorem, EF || QR.

AnswerYes, EF || QR.

(iii) PQ = 1.28 cm, PR = 2.56 cm …

  1. First find QE = PQ − PE = 1.28 − 0.18 = 1.10 cm and RF = PR − PF = 2.56 − 0.36 = 2.20 cm.
  2. Check PE/QE against PF/RF: PE/QE = 0.18/1.10 = 9/55; PF/RF = 0.36/2.20 = 9/55.
  3. The ratios are equal, so by the converse of the Basic Proportionality Theorem, EF || QR.

AnswerYes, EF || QR.

Watch this explained “Manufacturing a parallel line”, 13:58 into Turning it around: matching ratios force the line to be parallel

Question 3

“In Fig. 6.18, if LM || CB and LN || CD, prove that …” · p. 84

Open NCERT p. 84One way to think about it

  1. In triangle ABC, L lies on AC and M lies on AB, with LM || CB.
  2. By the Basic Proportionality Theorem, AM/AB = AL/AC.
  3. In triangle ACD, L lies on AC and N lies on AD, with LN || CD.
  4. By the Basic Proportionality Theorem, AN/AD = AL/AC.
  5. Both AM/AB and AN/AD equal the same ratio AL/AC.

In shortHence AM/AB = AN/AD.

Watch this explained “One diagonal, two triangles”, 12:28 into Turning it around: matching ratios force the line to be parallel

Question 4

“In Fig. 6.19, DE || AC and DF || AE. Prove that …” · p. 84

Open NCERT p. 84One way to think about it

  1. In triangle ABE, D lies on AB and F lies on BE, with DF || AE.
  2. By the Basic Proportionality Theorem, BF/FE = BD/DA.
  3. In triangle ABC, D lies on AB and E lies on BC, with DE || AC.
  4. By the Basic Proportionality Theorem, BD/DA = BE/EC.
  5. So BF/FE = BD/DA = BE/EC.

In shortHence BF/FE = BE/EC.

Watch this explained “The statement, word by word”, 4:08 into A line drawn parallel to a side cuts the other two in matching ratios

Question 5

“In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.” · p. 85

Open NCERT p. 85One way to think about it

  1. In triangle POQ, D lies on PO and E lies on PQ, with DE || OQ.
  2. By the Basic Proportionality Theorem, PD/DO = PE/EQ.
  3. In triangle POR, D lies on PO and F lies on PR, with DF || OR.
  4. By the Basic Proportionality Theorem, PD/DO = PF/FR.
  5. So PE/EQ = PD/DO = PF/FR, i.e. PE/EQ = PF/FR.
  6. By the converse of the Basic Proportionality Theorem in triangle PQR, EF || QR.

In shortHence EF || QR.

Watch this explained “The statement”, 5:11 into Turning it around: matching ratios force the line to be parallel

Question 6

“A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR” · p. 85

Open NCERT p. 85One way to think about it

  1. In triangle OPQ, A lies on OP and B lies on OQ, with AB || PQ.
  2. By the Basic Proportionality Theorem, OA/AP = OB/BQ.
  3. In triangle OPR, A lies on OP and C lies on OR, with AC || PR.
  4. By the Basic Proportionality Theorem, OA/AP = OC/CR.
  5. So OB/BQ = OA/AP = OC/CR, i.e. OB/BQ = OC/CR.
  6. By the converse of the Basic Proportionality Theorem in triangle OQR, BC || QR.

In shortHence BC || QR.

Watch this explained “The statement”, 5:11 into Turning it around: matching ratios force the line to be parallel

Question 7

“prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side” · p. 85

Open NCERT p. 85One way to think about it

  1. Let ABC be a triangle, D the midpoint of AB, and DE drawn parallel to BC meeting AC at E.
  2. Since D is the midpoint of AB, AD = DB, so AD/DB = 1.
  3. Since DE || BC, Theorem 6.1 (the Basic Proportionality Theorem) in triangle ABC gives AD/DB = AE/EC.
  4. So AE/EC = 1, i.e. AE = EC.
  5. So E is the midpoint of AC.

In shortHence the line through the midpoint of AB, parallel to BC, bisects the third side AC.

Watch this explained “The statement, word by word”, 4:08 into A line drawn parallel to a side cuts the other two in matching ratios

Question 8

“prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side” · p. 85

Open NCERT p. 85One way to think about it

  1. Let ABC be a triangle, D the midpoint of AB and E the midpoint of AC.
  2. Since D is the midpoint of AB, AD = DB, so AD/DB = 1.
  3. Since E is the midpoint of AC, AE = EC, so AE/EC = 1.
  4. So AD/DB = AE/EC (both equal 1).
  5. By Theorem 6.2 (the converse of the Basic Proportionality Theorem), DE || BC.

In shortHence the line DE joining the two midpoints is parallel to the third side BC.

Watch this explained “The statement”, 5:11 into Turning it around: matching ratios force the line to be parallel

Question 9

“ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O” · p. 85

Open NCERT p. 85One way to think about it

  1. Through O draw OE parallel to AB, meeting AD at E. Since AB || DC, OE is parallel to DC as well.
  2. In ΔADC, E lies on AD and O lies on AC, with EO || DC. By the Basic Proportionality Theorem, AE/ED = AO/OC.
  3. In ΔDAB, E lies on DA and O lies on DB, with EO || AB. By the Basic Proportionality Theorem, DE/EA = DO/OB, so AE/ED = BO/OD.
  4. So AO/OC = BO/OD.
  5. Cross-multiplying, AO × OD = BO × OC. Dividing both sides by BO × OD gives AO/BO = CO/DO.

In shortHence AO/BO = CO/DO.

Watch this explained “One diagonal, two triangles”, 12:28 into Turning it around: matching ratios force the line to be parallel

Question 10

“The diagonals of a quadrilateral ABCD intersect each other at the point O such that …” · p. 85

Open NCERT p. 85One way to think about it

  1. From AO/BO = CO/DO, cross-multiplying gives AO × DO = BO × CO, so AO/CO = BO/DO.
  2. Through O draw OE parallel to AB, meeting AD at E.
  3. In ΔDAB, E lies on DA and O lies on DB, with EO || AB. By the Basic Proportionality Theorem, DE/EA = DO/OB, so AE/ED = BO/DO.
  4. With step 1, AE/ED = BO/DO = AO/CO. So AE/ED = AO/OC.
  5. In ΔADC, E lies on AD and O lies on AC, and they cut these two sides in the same ratio. By the converse of the Basic Proportionality Theorem, EO || DC.
  6. AB || EO and DC || EO, so AB || DC.
  7. A quadrilateral with one pair of opposite sides parallel is a trapezium.

In shortHence ABCD is a trapezium.

Watch this explained “The statement”, 5:11 into Turning it around: matching ratios force the line to be parallel

Every question here was solved twice, separately, by two different AI models, and each answer was put back into the question by a computer program to check it. Where the two disagreed, a stronger model solved it again and the computer check had to pass on its answer. A question about reasoning rather than a number is shown as “one way to think about it”, and anything not yet proven says so instead of guessing. Each answer links to the moment in the video that teaches it.

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