CBSE Class 10 Mathematics Triangles Assignment Set 15

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Exercise-1 Subjective: Introduction to Similar Triangles and Thales Theorem

 

Question 1. Any point O inside \( \triangle ABC \) is joined to its vertices. From a point D on AO, DE is drawn so that \( DE \parallel AB \) (E on OB) and \( EF \parallel BC \) (F on OC), as shown. Prove that \( DF \parallel AC \).
OCADFEBAnswer:
In \( \triangle OAB \), \( DE \parallel AB \Rightarrow \dfrac{OD}{DA} = \dfrac{OE}{EB} \). In \( \triangle OBC \), \( EF \parallel BC \Rightarrow \dfrac{OE}{EB} = \dfrac{OF}{FC} \).
So \( \dfrac{OD}{DA} = \dfrac{OF}{FC} \), and by the converse of the basic proportionality theorem in \( \triangle OAC \), \( DF \parallel AC \).
In simple words: Chain two proportions through the common ratio OE : EB.

Teacher's Note:
1. Apply the basic proportionality theorem twice.
2. Then use the converse in the third triangle.

 

Question 2. A kitchen garden of Ms. Sanjana is in the form of a triangle PQR. She wants to divide it into two parts, one triangle and one trapezium, by a line EF. She takes PE = 4 m, QE = 4.5 m, PF = 8 m and RF = 9 m. Is EF parallel to QR? Justify your answer.
PFEQRAnswer:
\( \dfrac{PE}{EQ} = \dfrac{4}{4.5} = \dfrac{8}{9} \) and \( \dfrac{PF}{FR} = \dfrac{8}{9} \).
The two ratios are equal, so by the converse of the basic proportionality theorem Yes, EF \( \parallel \) QR.
In simple words: Equal ratios on two sides mean the line is parallel to the third.

Teacher's Note:
1. Compare PE : EQ with PF : FR.

 

Question 3. In the figure, \( AB \parallel DC \). Find the value of x.
DABCx-2x+3x+5x-1OAnswer:
The diagonals meet at O. \( \triangle ODC \sim \triangle OBA \) (alternate angles), so \( \dfrac{OD}{OB} = \dfrac{OC}{OA} \).
\( \dfrac{x - 2}{x - 1} = \dfrac{x + 3}{x + 5} \Rightarrow (x - 2)(x + 5) = (x + 3)(x - 1) \Rightarrow x^2 + 3x - 10 = x^2 + 2x - 3 \Rightarrow x = \) 7.
In simple words: Parallel sides make the diagonal pieces proportional.

Teacher's Note:
1. OD, OC are on one side; OB, OA on the other.
2. The x2 terms cancel.

 

Question 4. In the figure, two triangles ABC and DBC lie on the same side of the base BC. P is a point on BC such that \( PQ \parallel BA \) and \( PR \parallel BD \). Prove that \( QR \parallel AD \).
BADCPRQAnswer:
In \( \triangle CAB \), \( PQ \parallel BA \Rightarrow \dfrac{CQ}{QA} = \dfrac{CP}{PB} \). In \( \triangle CDB \), \( PR \parallel BD \Rightarrow \dfrac{CR}{RD} = \dfrac{CP}{PB} \).
So \( \dfrac{CQ}{QA} = \dfrac{CR}{RD} \), and in \( \triangle CAD \) the converse of the basic proportionality theorem gives \( QR \parallel AD \).
In simple words: Both ratios equal CP : PB.

Teacher's Note:
1. Q lies on CA and R on CD.
2. Compare the two ratios through CP : PB.

 

Exercise-1 Subjective: Criteria for Similarity of Triangles

 

Question 5. In \( \triangle LMN \), \( \angle L = 50^\circ \) and \( \angle N = 60^\circ \). If \( \triangle LMN \sim \triangle PQR \), then find \( \angle Q \).
Answer:
\( \angle M = 180^\circ - 50^\circ - 60^\circ = 70^\circ \). Q corresponds to M, so \( \angle Q = \) 70°.
In simple words: Match the letters in order.

Teacher's Note:
1. L, M, N match P, Q, R.

 

Question 6. In the figure, \( DE \parallel BC \) in \( \triangle ABC \), where BC = 8 cm, AB = 6 cm and DA = 1.5 cm. Find DE.
BCADEAnswer:
\( \triangle ADE \sim \triangle ABC \), so \( \dfrac{DE}{BC} = \dfrac{AD}{AB} = \dfrac{1.5}{6} = \dfrac{1}{4} \).
\( DE = \dfrac{8}{4} = \) 2 cm.
In simple words: DE is one quarter of BC.

Teacher's Note:
1. The small triangle at the vertex A is similar to the whole one.

 

Question 7. In the figure, \( DB \perp BC \), \( DE \perp AB \) and \( AC \perp BC \). Prove that \( \dfrac{BE}{DE} = \dfrac{AC}{BC} \).
ABCDEAnswer:
DB and AC are both perpendicular to BC, so \( DB \parallel AC \) and \( \angle DBE = \angle BAC \) (alternate angles).
In \( \triangle DEB \) and \( \triangle BCA \): \( \angle DEB = \angle BCA = 90^\circ \) and \( \angle DBE = \angle BAC \), so \( \triangle DEB \sim \triangle BCA \) (AA).
Hence \( \dfrac{BE}{CA} = \dfrac{DE}{BC} \), that is \( \dfrac{BE}{DE} = \dfrac{AC}{BC} \).
In simple words: Two lines perpendicular to the same line are parallel.

Teacher's Note:
1. Alternate angles on parallel lines are equal.
2. Match D to B, E to C, B to A.

 

Question 8. In \( \triangle ABC \), D and E are points on AB and AC respectively such that \( DE \parallel BC \). If AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm, find BD and CE.
Answer:
\( \triangle ADE \sim \triangle ABC \): \( \dfrac{AD}{AB} = \dfrac{AE}{AC} = \dfrac{DE}{BC} = \dfrac{2}{5} \).
\( AB = 2.4 \times \dfrac{5}{2} = 6 \) cm, so \( BD = 6 - 2.4 = \) 3.6 cm. \( AC = 3.2 \times \dfrac{5}{2} = 8 \) cm, so \( CE = 8 - 3.2 = \) 4.8 cm.
In simple words: The scale factor from small to large triangle is 5/2.

Teacher's Note:
1. DE : BC = 2 : 5 gives the scale.
2. BD = AB − AD and CE = AC − AE.

 

Question 9. Given \( \angle GHE = \angle DFE = 90^\circ \), DH = 8, DF = 12, DG = 3x - 1 and DE = 4x + 2. Find the lengths of segments DG and DE.
EHDFGAnswer:
In \( \triangle DHG \) and \( \triangle DFE \): \( \angle H = \angle F = 90^\circ \) and \( \angle D \) is common, so \( \triangle DHG \sim \triangle DFE \) (AA).
\( \dfrac{DH}{DF} = \dfrac{DG}{DE} \Rightarrow \dfrac{8}{12} = \dfrac{3x - 1}{4x + 2} \Rightarrow 8x + 4 = 9x - 3 \Rightarrow x = 7 \).
DG = 20 units, DE = 30 units.
In simple words: Two right triangles sharing the angle D are similar.

Teacher's Note:
1. Match D to D, H to F, G to E.
2. Substitute x = 7 back.

 

Question 10. In the figure, \( \angle QPS = \angle RPT \) and \( \angle PST = \angle PQR \). Prove that \( \triangle PST \sim \triangle PQR \) and hence find the ratio ST : PT, if PR : QR = 4 : 5.
QSPTRxxAnswer:
The rays from P in order are PS, PQ, PT, PR. \( \angle SPT = \angle SPQ + \angle QPT = \angle RPT + \angle QPT = \angle QPR \).
With \( \angle PST = \angle PQR \), we get \( \triangle PST \sim \triangle PQR \) (AA), with S to Q and T to R.
\( \dfrac{ST}{QR} = \dfrac{PT}{PR} \Rightarrow \dfrac{ST}{PT} = \dfrac{QR}{PR} = \dfrac{5}{4} \). So ST : PT = 5 : 4.
In simple words: Add the equal angles to the common angle QPT.

Teacher's Note:
1. Add the same angle to both sides of the given equality.
2. Match S to Q and T to R.

 

Question 11. In the figure, ABC and DBC are two right triangles with the common hypotenuse BC and with their sides AC and DB intersecting at P. Prove that \( AP \times PC = DP \times PB \).
PADBCAnswer:
In \( \triangle APB \) and \( \triangle DPC \): \( \angle A = \angle D = 90^\circ \) and \( \angle APB = \angle DPC \) (vertically opposite).
So \( \triangle APB \sim \triangle DPC \), \( \dfrac{AP}{DP} = \dfrac{PB}{PC} \Rightarrow AP \times PC = DP \times PB \).
In simple words: Right angles plus vertical angles give similarity.

Teacher's Note:
1. Vertically opposite angles are equal.

 

Question 12. In the figure, ABC is an isosceles triangle in which AB = AC. E is a point on the side CB produced such that \( FE \perp AC \). If \( AD \perp CB \), prove that \( AB \times FE = AD \times EC \).
CABDEFAnswer:
In \( \triangle ADC \) and \( \triangle EFC \): \( \angle ADC = \angle EFC = 90^\circ \) and \( \angle C \) is common, so \( \triangle ADC \sim \triangle EFC \).
\( \dfrac{AD}{EF} = \dfrac{AC}{EC} \Rightarrow AD \times EC = EF \times AC = FE \times AB \) (since AB = AC).
In simple words: Replace AC by AB using the isosceles property.

Teacher's Note:
1. The angle at C is common to both right triangles.

 

Question 13. A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.
Answer:
After 4 s she is \( 1.2 \times 4 = 4.8 \) m from the post. Let the shadow be s m long.
Similar triangles (post and girl): \( \dfrac{3.6}{4.8 + s} = \dfrac{0.9}{s} \Rightarrow 3.6s = 4.32 + 0.9s \Rightarrow 2.7s = 4.32 \Rightarrow s = \) 1.6 m.
In simple words: The lamp-post and the girl form similar right triangles.

Teacher's Note:
1. Change 90 cm to 0.9 m before using the ratio.

 

Question 14. In the figure, PQRS is a parallelogram with PQ = 16 cm and QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N. Find the lengths of PN and RM.
NSPMLRQAnswer:
\( RS \parallel PQ \): \( \triangle RLM \sim \triangle PLQ \), so \( \dfrac{RM}{PQ} = \dfrac{RL}{LP} = \dfrac{2}{3} \Rightarrow RM = \dfrac{2}{3} \times 16 = \) 10.67 cm (\( \dfrac{32}{3} \) cm).
\( PN \parallel QR \): \( \triangle PLN \sim \triangle RLQ \), so \( \dfrac{PN}{QR} = \dfrac{PL}{LR} = \dfrac{3}{2} \Rightarrow PN = \dfrac{3}{2} \times 10 = \) 15 cm.
In simple words: Use the parallel sides to get two pairs of similar triangles.

Teacher's Note:
1. Opposite sides of a parallelogram are parallel.
2. The vertically opposite angles at L are equal.

 

Question 15. In a triangle PQR, L and M are two points on the base QR such that \( \angle LPQ = \angle QRP \) and \( \angle RPM = \angle RQP \). Prove that (i) \( \triangle PQL \sim \triangle RPM \) (ii) \( QL \times RM = PL \times PM \) (iii) \( PQ^2 = QR \times QL \).
QLMRPAnswer:
(i) In \( \triangle PQL \) and \( \triangle RPM \): \( \angle Q = \angle RPM \) (given) and \( \angle LPQ = \angle R \) (given). By AA, \( \triangle PQL \sim \triangle RPM \), with P to R, Q to P, L to M.
(ii) \( \dfrac{QL}{PM} = \dfrac{PL}{RM} \Rightarrow QL \times RM = PL \times PM \).
(iii) In \( \triangle QPL \) and \( \triangle QRP \): \( \angle Q \) is common and \( \angle QPL = \angle QRP \), so \( \triangle QPL \sim \triangle QRP \). Then \( \dfrac{QP}{QR} = \dfrac{QL}{QP} \Rightarrow PQ^2 = QR \times QL \).
In simple words: Use the given angle equalities in different pairs of triangles.

Teacher's Note:
1. Parts (ii) and (iii) come from the proportions of the similarity in (i) and a second similarity.
2. Write the vertices in matching order.

 

Question 16. In the figure, M is the mid-point of side CD of a parallelogram ABCD. The line BM is drawn intersecting AC at L and AD produced at E. Prove that \( EL = 2BL \).
EDACLBMAnswer:
In \( \triangle EDM \) and \( \triangle BCM \): \( \angle EDM = \angle BCM \) (alternate angles, \( AD \parallel BC \)), DM = CM, \( \angle DME = \angle CMB \). So \( \triangle EDM \cong \triangle BCM \) and ED = BC = AD.
So AE = AD + DE = 2BC. Now \( AE \parallel BC \) gives \( \triangle ALE \sim \triangle CLB \): \( \dfrac{EL}{BL} = \dfrac{AE}{BC} = 2 \).
Hence \( EL = 2BL \).
In simple words: E is as far beyond D as A is, so AE is twice BC.

Teacher's Note:
1. Congruent triangles give ED = BC.
2. Then compare AE with BC through the similar triangles.

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Download Practice Assignments: Class 10 Mathematics Chapter 06 Triangles

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