CBSE Class 10 Triangles Sure Shot Questions Set D

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.

C. Long Answer Type Questions

Question. ABCD is a quadrilateral; P, Q, R and S are the points of trisection of sides AB, BC, CD and DA respectively and are adjacent to A and C; prove that PQRS is a parallelogram.
Answer: Proved.

Question. In \( \Delta ABC \), the bisector of \( \angle B \) meets AC at D. A line PQ || AC meets AB, BC and BD at P, Q and R respectively. Show that
(i) PR \(\cdot\) BQ = QR\(\cdot\) BP
(ii) AB \(\times\) CQ = BC \(\times\) AP.
Answer: Proved using similarity and basic proportionality theorem.

Question. In fig. CD and GH are respectively the medians of \( \Delta ABC \) and \( \Delta EFG \). If \( \Delta ABC \sim \Delta FEG \). Prove that
(i) \( \Delta ADC \sim \Delta FHG \) (ii) \( \frac{CD}{GH} = \frac{AB}{FE} \) (iii) \( \Delta CDB \sim \Delta GHE \)
Answer: (i) Since \( \Delta ABC \sim \Delta FEG \), \( \frac{AB}{FE} = \frac{AC}{FG} \). Since D and H are midpoints, \( \frac{2AD}{2FH} = \frac{AC}{FG} \implies \frac{AD}{FH} = \frac{AC}{FG} \). Also \( \angle A = \angle F \). By SAS criteria, \( \Delta ADC \sim \Delta FHG \). Ratio of sides gives (ii). Similarly (iii).

Question. ABC is an isosceles triangle with AB = AC and D is a point on AC such that \( BC^2 = AC \times CD \). Prove that BD = BC.
Answer: Given \( \frac{BC}{AC} = \frac{CD}{BC} \). In \( \Delta BDC \) and \( \Delta ABC \), \( \frac{BC}{AC} = \frac{CD}{BC} \) and \( \angle C = \angle C \). By SAS criteria, \( \Delta BDC \sim \Delta ABC \). Thus \( \frac{BD}{AB} = \frac{BC}{AC} \). Since AB = AC, BD = BC. Proved.

Question. In \( \Delta PQR \), QM \( \perp \) PR and \( PR^2 – PQ^2 = QR^2 \). Prove that \( QM^2 = PM \times MR \).
Answer: \( PR^2 = PQ^2 + QR^2 \) implies \( \angle PQR = 90^\circ \). In right \( \Delta PQR \), QM \( \perp \) hypotenuse PR. Thus \( \Delta PQM \sim \Delta RQM \implies \frac{QM}{RM} = \frac{PM}{QM} \implies QM^2 = PM \times MR \). Proved.

Question. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.
Answer: \( AC^2 + BD^2 = 2(AB^2 + BC^2) \). Use Apollonius theorem or Pythagoras on altitudes. Proved.

Short Answer Type Questions

Question. For a triangle ABC, the true statement is –
(a) \( AC^2 = AB^2 + BC^2 \)
(b) \( AC = AB + BC \)
(c) \( AC > AB + BC \)
(d) \( AC < AB + BC \)
Answer: (d) \( AC < AB + BC \)

Question. The distance between the tops of two trees 20 m and 28 m high is 17 m. Find the horizontal distance between the trees.
Answer: 15 m

Question. Triangle ABC is such that AB = 3 cm, BC = 2 cm and CA = 2.5 cm. Triangle DEF is similar to \( \Delta ABC \). If EF = 4 cm, then find the perimeter of \( \Delta DEF \).
Answer: 15 cm

Question. In \( \Delta ABC \), AB = 3 cm, AC = 4 cm and AD is the bisector of \( \angle A \). Then find BD : DC.
Answer: 3 : 4

Question. In an equilateral triangle ABC, if AD \( \perp \) BC, then prove that \( 3AB^2 = 4AD^2 \).
Answer: Proved.

Question. ABC is a triangle and DE is drawn parallel to BC cutting the other sides at D and E. If AB = 3.6 cm, AC = 2.4 cm and AD = 2.1 cm, then find AE.
Answer: 1.4 cm

Question. In a right angled triangle, one of the angles is 60º. Find the side opposite to this angle.
Answer: \( \frac{\sqrt{3}}{2} \times \text{Hypotenuse} \)

Question. If in the \( \Delta \)'s ABC and DEF, angle A is equal to angle E, both are equal to 40º, AB : ED = AC : EF and angle F is 65º, then find angle B.
Answer: 75°

Question. The ratio of the corresponding sides of two similar triangles is 1 : 3. Find the ratio of their corresponding heights.
Answer: 1 : 3

Question. The areas of two similar triangles are 49 \( \text{cm}^2 \) and 64 \( \text{cm}^2 \) respectively. Find the ratio of their corresponding sides.
Answer: 7 : 8

Question. The areas of two similar triangles are 12 \( \text{cm}^2 \) and 48 \( \text{cm}^2 \). If the height of the smaller one is 2.1 cm, then find the corresponding height of the bigger one.
Answer: 4.2 cm

Question. In \( \Delta ABC \), D and E are points on AB and AC respectively such that DE || BC. If AE = 2 cm, EC = 3 cm and BC = 10 cm, then find DE.
Answer: 4 cm

Question. If D, E, F are respectively the mid points of the sides BC, CA and AB of \( \Delta ABC \) and the area of \( \Delta ABC \) is 24 sq. cm, then find the area of \( \Delta DEF \).
Answer: 6 \( \text{cm}^2 \)

Question. A 25 m long ladder is placed against a vertical wall inside a room such that the foot of the ladder is 7 m from the foot of the wall. If the top of the ladder slides 4 m downwards, then find the foot of the ladder will slide by how much.
Answer: 8 m