CBSE Class 10 Triangles Sure Shot Questions Set 02

Question. O is the point of intersection of two equal chords AB and CD such that OB = OD, then triangles OAC and ODB are
(a) Equilateral but not similar
(b) Isosceles but not similar
(c) Equilateral and similar
(d) Isosceles and similar
Answer: (D)
Explanation: Since O is the point of intersection of two equal chords AB and CD such that OB = OD,
As chords are equal and OB = OD, so AO will also be equal to OC
Also \( \angle AOC = \angle DOB = 45^\circ \)
Now in triangles OAC and ODB
\( \frac{AO}{OB} = \frac{CO}{OD} \)
And \( \angle AOC = \angle DOB = 45^\circ \)
So triangles are isosceles and similar

Question. D and E are respectively the midpoints on the sides AB and AC of a triangle ABC and BC = 6 cm. If DE || BC, then the length of DE (in cm) is
(a) 2.5
(b) 3
(c) 5
(d) 6
Answer: B
Explanation: By midpoint theorem,
If D and E are respectively the midpoints on the sides AB and AC of a triangle ABC, DE||BC and BC = 6 cm
So, DE will be half of BC i.e. 3cm

Question. In triangle PQR, if PQ = 6 cm, PR = 8 cm, QS = 3 cm, and PS is the bisector of angle QPR, what is the length of SR?
(a) 2
(b) 4
(c) 6
(d) 8
Answer: (B)
Explanation: Since, PS is the angle bisector of angle QPR
So, by angle bisector theorem,
\( \frac{QS}{SR} = \frac{PQ}{PR} \)
\( \Rightarrow \frac{3}{SR} = \frac{6}{8} \)
\( \Rightarrow SR = \frac{(3 \times 8)}{6} \text{ cm} = 4 \text{ cm} \)

Question. The lengths of the diagonals of a rhombus are 16 cm and 12cm. Then, the length of the side of the rhombus is
(a) 9 cm
(b) 10 cm
(c) 8 cm
(d) 20 cm
Answer: (B)
Explanation: The diagonals of rhombus bisect each other at right angle, so side of rhombus is the hypotenuse for the triangles formed.
Therefore,
By Pythagoras theorem
\( (\frac{16}{2})^2 + (\frac{12}{2})^2 = \text{Side}^2 \)
\( \Rightarrow 8^2 + 6^2 = \text{Side}^2 \)
\( \Rightarrow 64 + 36 = \text{Side}^2 \)
\( \Rightarrow \text{Side} = 10 \text{ cm} \)

Question. A flag pole 18 m high casts a shadow 9.6 m long. Find the distance of the top of the pole from the far end of the shadow.
(a) 25.6
(b) 20.4
(c) 23.7
(d) 32.5
Answer: (B)
Explanation: According to given question, the far end of shadow is represented by point A, There fore we need to Find AC
By Pythagoras theorem,
\( (18)^2 + (9.6)^2 = (AC)^2 \)
\( \Rightarrow AC^2 = 416.16 \)
\( \Rightarrow AC = 20.4 \text{ m (approx)} \)

Question. Diagonals of a trapezium PQRS intersect each other at the point O, PQ || RS and PQ = 3 RS, Then the ratio of areas of triangles POQ and ROS is:
(a) 1:9
(b) 9:1
(c) 3:1
(d) 1:3
Answer: (B)
Explanation: According to given Question Since SR || PQ,
So, \( \angle OSR = \angle OQP \) (alternate interior angles)
Also \( \angle SOR = \angle POQ \) (vertically opposite angles)
So triangles SOR and POQ are similar,
Therefore,
\( \frac{ar(POQ)}{ar(SOR)} = (\frac{PQ}{SR})^2 \)
\( \frac{ar(POQ)}{ar(SOR)} = (\frac{3 SR}{SR})^2 \)
\( \frac{ar(POQ)}{ar(SOR)} = \frac{9}{1} \)

Question. ABCD is a trapezium in which AB|| DC and P, Q are points on AD and BC respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD.
(a) 55cm
(b) 57cm
(c) 60cm
(d) 62cm
Answer: (C)
Explanation: According to question ABCD is a trapezium in which AB || DC and P and Q are points on AD and BC, respectively such that PQ || DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm,
In triangle ABD
\( \frac{DP}{AP} = \frac{OD}{OB} \)
In triangle BDC
\( \frac{BQ}{QC} = \frac{OB}{OD} \)
This implies
\( \frac{DP}{AP} = \frac{QC}{BQ} \)
\( \frac{18}{AP} = \frac{15}{35} \)
\( AP = (18 \times 35)/15 \)
\( AP = 42 \)
Therefore, \( AD = AP + DP = 42 + 18 = 60\text{cm} \)

Question. Areas of two similar triangles are 36 cm2 and 100 cm2. If the length of a side of the larger triangle is 20 cm, then the length of the corresponding side of the smaller triangle is:
(a) 12cm
(b) 13cm
(c) 14cm
(d) 15cm
Answer: (A)
Explanation: Let the side of smaller triangle be x cm.
\( \frac{ar(\text{Larger Triangle})}{ar(\text{Smaller Triangle})} = (\frac{\text{side of larger triangle}}{\text{side of smaller triangle}})^2 \)
\( \frac{100}{36} = (20/x)^2 \)
\( x = \sqrt{144} \)
\( X = 12 \text{ cm} \)

Question. In the figure if \( \angle ACB = \angle CDA \), AC = 8 cm and AD = 3 cm, find BD.
(a) 53/3 cm
(b) 55/3 cm
(c) 64/3 cm
(d) 35/7 cm
Answer: (B)
Explanation: In triangle ACB and ADC
\( \angle A = \angle A \)
\( \angle ACB = \angle CDA \)
There fore triangle ACB and ADC are similar,
Hence
\( \frac{AC}{AD} = \frac{AB}{AC} \)
\( AC^2 = AD \times AB \)
\( 8^2 = 3 \times AB \)
\( \Rightarrow AB = 64/3 \)
This implies,
\( BD = 64/3 – AD \)
\( \Rightarrow BD = 55/3 \)

Question. If ABCD is parallelogram, P is a point on side BC and DP when produced meets AB produced at L, then select the correct option
(a) DP/BL = DC/PL
(b) DP/PL = DC/BL
(c) DP/PL = BL/DC
(d) DP/PL = AB/DC
Answer: (B)
Explanation: In \( \Delta ALD \), we have
BP || AD
\( \therefore LB/BA = LP/PD \)
\( \Rightarrow BL/AB = PL/DP \)
\( \Rightarrow BL/DC = PL/DP \) [\( \because AB = DC \)]
\( \Rightarrow DP/PL = DC/BL \)

Question. In the figure given below DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, the value of x is:
(a) 4
(b) 8
(c) 16
(d) 32
Answer: (A)
Explanation: In triangle ABC, we have DE || BC
\( \therefore AD/DB = AE/EC \) (By Thale’s Theorem)
\( \Rightarrow x/x - 2 = (x + 2)/(x - 1) \)
\( \Rightarrow x (x - 1) = (x - 2)(x + 2) \)
\( \Rightarrow x^2 – x = x^2 – 4 \)
\( \Rightarrow x = 4 \)

Question. The length of altitude of an equilateral triangle of side 8cm is
(a) √3 cm
(b) 2√3 cm
(c) 3√3 cm
(d) 4√3 cm
Answer: (D)
Explanation: The altitude divides the opposite side into two equal parts, Therefore, BD = DC = 4 cm
In triangle ABD
\( AB^2 = AD^2 + BD^2 \)
\( 8^2 = AD^2 + 4^2 \)
\( AD^2 = 64 – 16 \)
\( AD^2 = 48 \)
\( AD = 4\sqrt{3} \text{ cm} \)

Question. If \( \Delta ABC \sim \Delta DEF \), AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, find the perimeter of ABC.
(a) 18 cm
(b) 20 cm
(c) 21 cm
(d) 22 cm
Answer: (A)
Explanation: According to question, \( \Delta ABC \sim \Delta DEF \), AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm,
Therefore,
\( AB/DE = BC/EF = AC/DF \)
\( 4/6 = BC/9 = AC/12 \)
\( \Rightarrow 4/6 = BC/9 \)
\( \Rightarrow BC = 6 \text{ cm} \)
And
\( 4/6 = AC/12 \)
\( \Rightarrow AC = 8 \text{ cm} \)
Perimeter = AB + BC + CA = 4 + 6 + 8 = 18 cm

Question. A 5 m long ladder is placed leaning towards a vertical wall such that it reaches the wall at a point 4 m high. If the foot of the ladder is moved 1.6 m towards the wall, then the distance by which the top of the ladder would slide upwards on the wall is:
(a) 2 m
(b) 1.2 m
(c) 0.8 m
(d) 0.5 m
Answer: (C)
Explanation: Let AC be the ladder of length 5m and BC = 4m be the height of the wall where ladder is placed. If the foot of the ladder is moved 1.6m towards the wall i.e. AD = 1.6 m, then the ladder is slided upward to position E i.e. CE = x m.
In right triangle ABC
\( AC^2 = AB^2 + BC^2 \)
\( \Rightarrow 5^2 = AB^2 + 4^2 \)
\( \Rightarrow AB = 3\text{m} \)
\( \Rightarrow DB = AB – AD = 3 – 1.6 = 1.4\text{m} \)
In right angled \( \Delta EBD \)
\( ED^2 = EB^2 + BD^2 \)
\( \Rightarrow 5^2 = EB^2 + (1.4)^2 \)
\( \Rightarrow EB = 4.8\text{m} \)
EC = EB – BC = 4.8 – 4 = 0.8m
Hence the top of the ladder would slide upwards on the wall at distance 0.8 m.

Question. Corresponding sides of two similar triangles are in the ratio of 2 : 3. If the area of the smaller triangle is 48 cm2, then the area of the larger triangle is:
(a) 108 m2
(b) 107 m2
(c) 106 m2
(d) 230 m2
Answer: (A)
Explanation: According to given Question
\( ar(\text{Larger Triangle})/ar(\text{Smaller Triangle}) = (\text{side of larger triangle}/\text{side of larger triangle})^2 \)
\( ar(\text{Larger Triangle})/48 = (3/2)^2 \)
\( ar(\text{Larger Triangle}) = (9 \times 48 )/4 \)
\( ar(\text{Larger Triangle}) = 108 \text{ cm}^2 \)

A. Very Short Answer Type Questions

Question. In a \( \Delta ABC \), D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 6cm, DB = 9cm and AE = 8 cm, find AC.
Answer: 12 cm

Question. In a \( \Delta ABC \), D and E are points on the sides AB and AC respectively such that DE || BC. If \( \frac{AD}{DB} = \frac{3}{4} \) and AC = 15 cm find AE.
Answer: 6.43 cm

Question. In a \( \Delta ABC \), D and E are points on the sides AB and AC respectively such that DE || BC. If \( \frac{AD}{DB} = \frac{2}{3} \) and AC = 18 cm, find AE.
Answer: 7.2 cm

Question. In a \( \Delta ABC \), D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 4 cm, AE = 8 cm, DB = x – 4 and EC = 3x – 19, find x.
Answer: 11 cm

Question. In a \( \Delta ABC \), D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 8 cm, AB = 12 cm and AE = 12 cm, find CE.
Answer: 6 cm

Question. In a \( \Delta ABC \), D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 4 cm, DB = 4.5 cm and AE = 8 cm, find AC.
Answer: 17 cm

Question. In a \( \Delta ABC \), D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 2 cm, AB = 6 cm and AC = 9 cm, find AE.
Answer: 3 cm

Question. In a \( \Delta ABC \), D and E are points on the sides AB and AC respectively such that DE || BC. If \( \frac{AD}{BD} = \frac{4}{5} \) and EC = 2.5 cm, find AE.
Answer: 2 cm

Question. In a \( \Delta ABC \), D and E are points on the sides AB and AC respectively such that DE || BC. If AD = x, DB = x – 2, AE = x + 2 and EC = x – 1, find the value of x.
Answer: x = 4

Question. In a \( \Delta ABC \), D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 8x – 7, DB = 5x – 3, AE = 4x – 3 and EC = (3x – 1), find the value of x.
Answer: x = 1

Question. In a \( \Delta ABC \), D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1 and CE = 5x – 3, find the value of x.
Answer: x = 1

Question. In a \( \Delta ABC \), D and E are points on the sides AB and AC respectively. For each of the following cases show that DE || BC :
(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm.
(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm
(iii) AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm.
(iv) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm
Answer: DE || BC in all cases as \( \frac{AD}{DB} = \frac{AE}{EC} \).

Question. In a \( \Delta ABC \), AD is the bisector of \( \angle A \), meeting side BC at D. If BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm, find DC.
Answer: 2.1 cm

Question. In a \( \Delta ABC \), AD is the bisector of \( \angle A \), meeting side BC at D. If BD = 2cm, AB = 5 cm and DC = 3 cm, find AC.
Answer: 7.5 cm

Question. In a \( \Delta ABC \), AD is the bisector of \( \angle A \), meeting side BC at D. If AB = 3.5 cm, AC = 4.2 cm and DC = 2.8 cm, find BD.
Answer: 2.3 cm

Question. In a \( \Delta ABC \), AD is the bisector of \( \angle A \), meeting side BC at D. If AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.
Answer: 2.5 cm, 3.5 cm

Question. In a \( \Delta ABC \), AD is the bisector of \( \angle A \), meeting side BC at D. If AC = 4.2 cm, DC = 6 cm, BC = 10 cm, find AB.
Answer: 2.8 cm

Question. In a \( \Delta ABC \), AD is the bisector of \( \angle A \), meeting side BC at D. If AB = 5.6 cm, AC = 6 cm and DC = 3 cm, find BC.
Answer: 5.8 cm

Question. In a \( \Delta ABC \), AD is the bisector of \( \angle A \), meeting side BC at D. If AB = 5.6 cm, BC = 6 cm and BD = 3.2 cm find AC.
Answer: 4.9 cm

Question. In a \( \Delta ABC \), AD is the bisector of \( \angle A \), meeting side BC at D. If AB = 10 cm, AC = 6 cm and BC = 12 cm, find BD and DC.
Answer: 7.5 cm, 4.5 cm

Question. In \( \Delta ABC \), \( \angle B = 2 \angle C \) and the bisector of \( \angle B \) intersects AC at D. Prove that \( \frac{BD}{DA} = \frac{BC}{BA} \).
Answer: In \( \Delta ABC \), let \( \angle C = \theta \), then \( \angle B = 2\theta \). Since BD bisects \( \angle B \), \( \angle ABD = \angle CBD = \theta \). In \( \Delta BCD \), \( \angle CBD = \angle BCD = \theta \), so BD = CD. By angle bisector theorem in \( \Delta ABC \), \( \frac{AB}{BC} = \frac{AD}{DC} \implies \frac{BA}{BC} = \frac{DA}{BD} \implies \frac{BD}{DA} = \frac{BC}{BA} \). Proved.

Question. E is a point on side AD produced of a parallelogram ABCD and BE intersects CD at F. Prove that \( \Delta ABE \sim \Delta CFB \).
Answer: In \( \Delta ABE \) and \( \Delta CFB \), \( \angle A = \angle C \) (opposite angles of parallelogram) and \( \angle AEB = \angle CBF \) (alternate angles as AE || BC). By AA similarity, \( \Delta ABE \sim \Delta CFB \). Proved.

B. Short Answer Type Questions

Question. D, E and F are the points on sides BC, CA and AB respectively of \( \Delta ABC \) such that AD bisects \( \angle A \), BE bisects \( \angle B \) and CF bisects \( \angle C \). If AB = 5 cm, BC = 8 cm and CA = 4 cm, determine AF, CE and BD.
Answer: AF = \( \frac{5}{3} \) cm, CE = \( \frac{32}{13} \) cm, BD = \( \frac{40}{9} \) cm

Question. In a \( \Delta ABC \), D and E are points on sides AB and AC respectively such that BD = CE. If \( \angle B = \angle C \), show that DE || BC.
Answer: Since \( \angle B = \angle C \), AB = AC. Given BD = CE, so AB – BD = AC – CE \(\implies\) AD = AE. Thus \( \frac{AD}{AB} = \frac{AE}{AC} \). By converse of Thales theorem, DE || BC. Proved.

Question. In \( \Delta ABC \), D is the mid-point of BC and ED is the bisector of the \( \angle ADB \) and EF is drawn parallel to BC cutting AC in F. Prove that \( \angle EDF \) is a right angle.
Answer: In \( \Delta ABD \), DE bisects \( \angle ADB \), so \( \frac{AE}{EB} = \frac{AD}{DB} \). In \( \Delta ADC \), DF is the bisector of \( \angle ADC \) if we show \( \frac{AF}{FC} = \frac{AD}{DC} \). Since EF || BC, \( \frac{AE}{EB} = \frac{AF}{FC} \) by Thales theorem. Given D is mid-point, DB = DC. So \( \frac{AD}{DB} = \frac{AD}{DC} \), which implies \( \frac{AE}{EB} = \frac{AF}{FC} \), thus DF is the bisector of \( \angle ADC \). Since \( \angle ADB + \angle ADC = 180^\circ \), the angle between their bisectors DE and DF is \( \frac{1}{2} \times 180^\circ = 90^\circ \). Proved.

Question. The bisectors of the angles B and C of a triangle ABC, meet the opposite side in D and E respectively. If DE || BC, prove that the triangle is isosceles.
Answer: Proved.

Question. If CD and GH (D and H lie on AB and FE) are respectively bisectors of \( \angle ACB \) and \( \angle EGF \) and \( \Delta ABC \sim \Delta FEG \), prove that
(i) \( \Delta DCA \sim \Delta HGF \)
(ii) \( \frac{CD}{GH} = \frac{AC}{FG} \)
(iii) \( \Delta DCB \sim \Delta HGE \)
Answer: Since \( \Delta ABC \sim \Delta FEG \), \( \angle A = \angle F \), \( \angle B = \angle E \), \( \angle ACB = \angle FGE \). Then \( \frac{1}{2}\angle ACB = \frac{1}{2}\angle FGE \implies \angle DCA = \angle HGF \). By AA criteria, \( \Delta DCA \sim \Delta HGF \). Ratio of sides gives (ii). Similar logic for (iii).

Question. If \( \Delta ABC \), if AD \( \perp \) BC and \( AD^2 = BD \times DC \), prove that \( \angle BAC = 90^\circ \).
Answer: \( AD^2 = BD \times DC \implies \frac{AD}{DC} = \frac{BD}{AD} \). In right triangles \( \Delta ABD \) and \( \Delta ADC \), \( \angle ADB = \angle ADC = 90^\circ \) and \( \frac{AD}{DC} = \frac{BD}{AD} \). By SAS similarity, \( \Delta ABD \sim \Delta CAD \). Then \( \angle BAD = \angle ACD \) and \( \angle DAC = \angle ABD \). Since \( \angle ABD + \angle BAD = 90^\circ \), then \( \angle DAC + \angle BAD = 90^\circ \implies \angle BAC = 90^\circ \). Proved.

Question. ABC is an isosceles right triangle, right angled at C. Prove that \( AB^2 = 2 AC^2 \).
Answer: In \( \Delta ABC \), by Pythagoras: \( AB^2 = AC^2 + BC^2 \). Since it is isosceles, AC = BC. \( \therefore AB^2 = AC^2 + AC^2 = 2 AC^2 \). Proved.

Question. In an isosceles triangle ABC, with AB = AC, BD is perpendicular from B to the side AC. Prove that \( BD^2 – CD^2 = 2 CD \cdot AD \).
Answer: \( BD^2 = BC^2 - CD^2 \). Also \( BC^2 = AC^2 + AB^2 - \dots \) (Using Pythagoras in \( \Delta BDC \) and \( \Delta BDA \)): \( BD^2 = AB^2 - AD^2 \). Since AB = AC, \( BD^2 = AC^2 - AD^2 = (AD + CD)^2 - AD^2 = AD^2 + CD^2 + 2AD \cdot CD - AD^2 = CD^2 + 2 CD \cdot AD \). Hence \( BD^2 - CD^2 = 2 CD \cdot AD \). Proved.

Question. In a \( \Delta ABC \), the angles at B and C are acute. If BE and CF be drawn perpendiculars on AC and AB respectively, prove that
(i) \( BC^2 = AB \times BF + AC \times CE \).
(ii) \( AC^2 = AB^2 + BC^2 – 2AB \cdot BF \)
(iii) \( AB^2 = BC^2 + AC^2 – 2AC \cdot CF \)
Answer: Proved by projections of sides or Pythagoras theorem.

Question. ABC is a right triangle, right angled at C and AC = \( \sqrt{3} \) BC. prove that \( \angle ABC = 60^\circ \).
Answer: \( \tan B = \frac{AC}{BC} = \frac{\sqrt{3} BC}{BC} = \sqrt{3} \implies B = 60^\circ \). Proved.

Question. In a right-angled triangle if a perpendicular is drawn from the right angle to the hypotenuse, prove that the square of the perpendicular is equal to the area of rectangle contained by the two segments of the hypotenuse.
Answer: Let \( \Delta ABC \) be right angled at A, AD \( \perp \) BC. \( \Delta ABD \sim \Delta CAD \implies \frac{AD}{CD} = \frac{BD}{AD} \implies AD^2 = BD \cdot CD \). Proved.