Class 10 Mathematics: Triangles important questions
Triangles is a chapter of Class 10 Mathematics for the CBSE. These 18 questions are marked Very Important, Important or Practice, and each answer has been checked by our team. Use them to revise, not as a prediction of what will come in the exam.
Practice · 1 marks
1. ΔABC ~ ΔPQR. AM and PN are altitudes of ΔABC and ΔPQR, and AB² : PQ² = 4 : 9. Then AM : PN = (a) 3 : 2 (b) 16 : 81 (c) 4 : 9 (d) 2 : 3
(d) 2 : 3 AB : PQ = 2 : 3, and altitudes of similar triangles are in the same ratio as their sides.
Practice · 1 marks
2. In ΔABC, D is on AB and E is on AC with DE ∥ BC. If AE = a, EC = b, DE = x and BC = y, which of these is true? (a) x = (a + b)/ay (b) y = ax/(a + b) (c) x = ay/(a + b) (d) x/y = a/b
(c) x = ay/(a + b) ΔADE ~ ΔABC, so DE/BC = AE/AC: x/y = a/(a + b).
Practice · 1 marks
3. ABCD is a trapezium with AD ∥ BC and AD = 4 cm. If the diagonals AC and BD meet at O such that AO/OC = DO/OB = 1/2, then BC = (a) 6 cm (b) 7 cm (c) 8 cm (d) 9 cm
(c) 8 cm ΔAOD ~ ΔCOB, so AD/BC = AO/OC = 1/2 and BC = 8 cm.
Important · 2 marks
4. In ΔACD, B is a point on AC and E is a point on AD. If AD/AE = AC/BD and ∠BAD = ∠BDA, show that ΔBAE ~ ΔCAD.
∠BAD = ∠BDA, so AB = BD … (i) Given AD/AE = AC/BD; using (i), AD/AE = AC/AB, i.e. AB/AC = AE/AD. In ΔBAE and ΔCAD, ∠A is common and the sides including it are proportional, so ΔBAE ~ ΔCAD (SAS).
Very Important · 5 marks
5. Prove that if a line is drawn parallel to one side of a triangle intersecting the other two sides at distinct points, the other two sides are divided in the same ratio. Using this, prove that a line through the point of intersection of the diagonals of a trapezium and parallel to its bases divides the non-parallel sides in the same ratio.
Theorem (BPT): in ΔABC with DE ∥ BC, ar(ADE)/ar(BDE) = AD/DB and ar(ADE)/ar(DEC) = AE/EC; ar(BDE) = ar(DEC) (same base DE, same parallels), so AD/DB = AE/EC. Application: in trapezium ABCD with DC ∥ AB, let EF ∥ AB pass through the intersection of the diagonals, with E on AD and F on BC. EF meets AC at G. In ΔABC, GF ∥ AB, so CG/GA = CF/FB … (1) In ΔADC, EG ∥ DC, so DE/EA = CG/GA … (2) From (1) and (2): DE/EA = CF/FB.
Practice · 1 marks
6. In ΔABC, D is on AC and E is on BC with DE ∥ AB. If AB = a, DE = x, BE = b and EC = c, then x in terms of a, b and c is: (a) ac/b (b) ac/(b + c) (c) ab/c (d) ab/(b + c)
(b) ac/(b + c) ΔCDE ~ ΔCAB, so DE/AB = CE/CB: x/a = c/(b + c).
Important · 2 marks
7. ABCD is a parallelogram. P divides AB in the ratio 2 : 3 and Q divides DC in the ratio 4 : 1. PQ meets the diagonal AC at O. Prove that OC is half of OA.
Let AB = DC = a. Then AP = 2a/5 and CQ = a/5 (since DQ : QC = 4 : 1). AB ∥ DC, so in ΔAPO and ΔCQO, ∠PAO = ∠QCO (alternate angles) and ∠AOP = ∠COQ (vertically opposite). So ΔAPO ~ ΔCQO (AA). OC/OA = CQ/AP = (a/5)/(2a/5) = 1/2, so OC = ½ OA.
Very Important · 5 marks
8. (a) State and prove the Basic Proportionality Theorem. (b) In ΔABC, D is on AB and E is on AC, and BE meets DC at F. If ∠CEF = ∠CFE and F is the midpoint of DC, prove that AB/BD = AE/FD.
(a) If a line is drawn parallel to one side of a triangle to meet the other two sides at distinct points, it divides those sides in the same ratio. Proof: in ΔABC with DE ∥ BC, ar(ADE)/ar(BDE) = AD/DB and ar(ADE)/ar(DEC) = AE/EC; ar(BDE) = ar(DEC) (same base DE, between the same parallels), so AD/DB = AE/EC. (b) Draw DG ∥ BE meeting AC at G. In ΔABE, AB/BD = AE/GE (BPT). In ΔCDG, F is the midpoint of DC and FE ∥ DG, so E is the midpoint of CG: GE = CE. ∠CEF = ∠CFE, so CF = CE. And CF = FD. Hence GE = FD, and AB/BD = AE/FD.
Practice · 1 marks
9. O is the point of intersection of two chords AB and CD of a circle. If ∠BOC = 80° and OA = OD, then ΔODA and ΔOBC are: (A) equilateral and similar (B) isosceles and similar (C) isosceles but not similar (D) not similar
(B) isosceles and similar OA × OB = OC × OD and OA = OD give OB = OC, so both are isosceles; the vertical angles at O are equal and the sides are proportional, so they are similar.
Important · 3 marks
10. In ΔABC, D, E and F are the midpoints of BC, CA and AB respectively. Prove that ΔFBD ~ ΔDEF and ΔDEF ~ ΔABC.
By the midpoint theorem, EF ∥ BC, DF ∥ AC and DE ∥ AB. BDEF is a parallelogram, so ∠B = ∠FED (opposite angles) and the diagonal FD makes equal alternate angles: ∠BFD = ∠FDE and ∠BDF = ∠DFE. Hence ΔFBD ~ ΔDEF (AA). Similarly AFDE and DCEF are parallelograms, so ∠A = ∠EDF and ∠C = ∠DFE (and ∠B = ∠FED). Hence ΔDEF ~ ΔABC (AA).
Important · 3 marks
11. In ΔABC, P and Q are points on AB and AC such that PQ ∥ BC. Prove that the median AD drawn from A to BC bisects PQ.
Let AD meet PQ at R. PQ ∥ BC, so ΔAPR ~ ΔABD, giving AP/AB = PR/BD … (i) and ΔAQR ~ ΔACD, giving AQ/AC = RQ/DC … (ii) By BPT, AP/AB = AQ/AC … (iii) From (i), (ii) and (iii): PR/BD = RQ/DC. Since BD = DC, PR = RQ, so AD bisects PQ.
Very Important · 4 marks
12. In a cabinet design, ΔABC is the largest triangle and ΔDEF the smallest, with a shelf PQ parallel to EF (P on DE, Q on DF). (i) Show that ΔDPQ ~ ΔDEF. (ii) If DP = 50 cm and PE = 70 cm, find PQ/EF. (iii) (A) If 2AB = 5DE and ΔABC ~ ΔDEF, show that (perimeter of ΔABC)/(perimeter of ΔDEF) is constant. OR (B) If AM and DN are medians of ΔABC and ΔDEF (with ΔABC ~ ΔDEF), prove that ΔABM ~ ΔDEN.
(i) PQ ∥ EF gives ∠DPQ = ∠DEF (corresponding angles), and ∠D is common, so ΔDPQ ~ ΔDEF (AA). (ii) DE = 50 + 70 = 120 cm. PQ/EF = DP/DE = 50/120 = 5/12. (iii) (A) AB/DE = 5/2 = BC/EF = AC/DF, so perimeter ABC / perimeter DEF = 5/2, a constant. OR (B) AB/DE = BC/EF = (BC/2)/(EF/2) = BM/EN and ∠B = ∠E, so ΔABM ~ ΔDEN (SAS).
Practice · 1 marks
13. If ΔABC ~ ΔDEF, AB = 6 cm, DE = 9 cm, EF = 6 cm and FD = 12 cm, the perimeter of ΔABC is: (A) 28 cm (B) 28.5 cm (C) 18 cm (D) 23 cm
(C) 18 cm Perimeter of DEF = 27 cm. Perimeters are in the ratio AB : DE = 2 : 3, so the perimeter of ABC = 27 × 2/3 = 18 cm.
Important · 2 marks
14. AP and DQ are medians of triangles ABC and DEF, where ΔABC ~ ΔDEF. Prove that AB/DE = AP/DQ.
ΔABC ~ ΔDEF gives AB/DE = BC/EF and ∠B = ∠E. Since P and Q are midpoints, BP = BC/2 and EQ = EF/2, so AB/DE = BP/EQ. In ΔABP and ΔDEQ: AB/DE = BP/EQ and ∠B = ∠E, so ΔABP ~ ΔDEQ (SAS). Hence AB/DE = AP/DQ.
Very Important · 5 marks
15. Prove that a line drawn parallel to one side of a triangle, meeting the other two sides at distinct points, divides them in the same ratio. Hence, in ΔPQR, a line LM ∥ QR meets PQ at L and PR at M. If PL = 5.7 cm, PQ = 15.2 cm and MR = 5.5 cm, find PM.
Proof (Basic Proportionality Theorem): in ΔABC with DE ∥ BC, ar(ADE)/ar(BDE) = AD/DB and ar(ADE)/ar(DEC) = AE/EC, and ar(BDE) = ar(DEC) (same base DE, between the same parallels), so AD/DB = AE/EC. Application: LQ = 15.2 − 5.7 = 9.5 cm. PL/LQ = PM/MR, so 5.7/9.5 = PM/5.5 and PM = 3.3 cm.
Practice · 1 marks
16. The perimeters of two similar triangles are 56 cm and 70 cm. If one side of the first triangle is 14 cm, the corresponding side of the second (in cm) is: (A) 5 (B) 7.5 (C) 10 (D) 17.5
(D) 17.5 The ratio of perimeters equals the ratio of corresponding sides: 56/70 = 14/x, so x = 17.5 cm.
Important · 2 marks
17. In right triangle ABC, right-angled at B, BD is drawn perpendicular to the hypotenuse AC. Prove that ΔBCD ~ ΔACB.
In ΔBCD and ΔACB: ∠BDC = ∠ABC = 90° ∠BCD = ∠ACB (common angle) So ΔBCD ~ ΔACB by the AA criterion.
Very Important · 5 marks
18. Prove that a line drawn parallel to one side of a triangle, meeting the other two sides at distinct points, divides those sides in the same ratio. Using this, show that if PQRS is a trapezium with PQ ∥ SR, and X and Y lie on PS and QR with XY ∥ PQ, then PX/XS = QY/YR.
Proof of the theorem (Basic Proportionality Theorem): in ΔABC with DE ∥ BC, join BE and CD and draw the heights. ar(ADE)/ar(BDE) = AD/DB and ar(ADE)/ar(DEC) = AE/EC. Since ΔBDE and ΔDEC are on the same base DE between the same parallels, their areas are equal, so AD/DB = AE/EC. Application: join PR, meeting XY at M. In ΔPSR, XM ∥ SR, so PX/XS = PM/MR … (i) In ΔPQR, MY ∥ PQ, so QY/YR = PM/MR … (ii) From (i) and (ii), PX/XS = QY/YR.
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