Similar triangles questions and solutions are provided here to help students learn how to solve questions involving similar triangles. These questions and solutions are useful resources for the students of Class 10. Practising these questions will help students score good marks in their examinations. Also, they can boost their preparation for exams using additional questions provided here.
What are similar triangles?
Similar triangles are those triangles whose angles are equal, and the corresponding sides are in proportion. That means they need not be of the same size. Suppose triangle ABC and triangle PQR are similar triangles, such that AB/PQ = BC/QR = AC/PR and ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R. Also, we have several criteria for the similarity of triangles, and they are:
- AAA similarity criterion
- AA similarity criterion
- SSS similarity criterion
- SAS similarity criterion
Get more information about similar triangles here.
Similar Triangles Questions and Answers
1. Give two examples of similar figures.
Solution:
As we know, similar figures are figures containing the same shape but not necessarily the same size. Some of the examples of similar figures are:
(i) All circles are similar figures
(ii) All equilateral triangles are similar figures
(iii) All right angle isosceles triangles are similar figures
2. What are the criteria for the similarity of triangles?
Solution:
Two triangles are said to be similar, if:
(i) Their corresponding angles are equal and
(ii) Their corresponding sides are in the same ratio (or proportion).
These are the criteria for the similarity of triangles.
3. If a line intersects sides AB and AC of a triangle ABC at D and E, respectively and is parallel to BC, prove that AD/AB = AE/AC.
Solution:
Given,
DE || BC
From the Basic Proportionality Theorem, we have:
AD/DB = AE/EC
⇒ DB/AD = EC/AE
Adding 1 on both sides, we get:
⇒ (DB/AD) + 1 = (EC/AE) + 1
⇒ (DB + AD)/AD = (EC + AE)/AE
⇒ AB/AD = AC/AE
⇒ AD/AB = AE/AC
Hence proved.
4. In the figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD.
Solution:
Given,
LM || CB
In triangle ABC, from the Basic Proportionality Theorem, we have:
AM/AB = AL/AC….(i)
Also, given that LN || CD.
In triangle ADC, from the Basic Proportionality Theorem, we have:
AN/AD = AL/AC….(ii)
From (i) and (ii),
AM/AB = AN/AD
Hence proved.
5. ABCD is a trapezium in which AB || DC, and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.
Solution:
Given,
ABCD is a trapezium, and side AB is parallel to DC .i.e AB || DC.
Diagonals AC and BD intersect each other at point O.
From the point O, draw a line EO touching AD at E, such that,
EO || DC || AB
In ΔADC,
OE || DC
From the Basic Proportionality Theorem,
AE/ED = AO/CO….(i)
Now, In ΔABD,
OE || AB
From the Basic Proportionality Theorem,
DE/EA = DO/BO…(ii)
From equations (i) and (ii),
AO/CO = BO/DO
⇒ AO/BO = CO/DO
6. In the figure, if PQ || RS, prove that ΔPOQ ~ ΔSOR.
Solution:
Given,
PQ || RS
∠P = ∠S (Alternate angles)
∠Q = ∠R
Also, ∠POQ = ∠SOR (Vertically opposite angles)
Therefore, by AAA similarity criterion,
ΔPOQ ~ ΔSOR
Hence proved.
7. In the figure, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB. 
Solution:
Given,
∠BOC = 125° and ∠CDO = 70°
In the figure, DOB is a straight line.
So, ∠DOC + ∠COB = 180°
∠DOC = 180° – 125° = 55°
In ΔDOC,
∠DCO + ∠ CDO + ∠ DOC = 180° {angle sum property of a triangle}
∠DCO + 70º + 55º = 180°
∠DCO = 180º – 70º – 55º = 55°
Also, it is given that ΔODC ~ ΔOBA.
That means the corresponding angles are equal.
∠OAB = ∠OCD
⇒ ∠OAB = 55°
8. Find the value of x for which DE||AB is in the given figure.
Solution:
Given,
DE || AB
Using the Basic Proportionality Theorem,
CD/AD = CE/BE
(x + 3)/(3x + 19) = x/(3x + 4)
(x + 3) (3x + 4) = x (3x + 19)
3×2 + 4x + 9x + 12 = 3×2 + 19x
19x – 13x = 12
6x = 12
x = 12/6 = 2
Therefore, the value of x is 2.
9. D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the areas of ΔDEF and ΔABC.
Solution:
Given,
D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC.
∴ DE || AC
By the midpoint theorem,
DE = (1/2) AC
DE/AC = ½ …. (1)
In ΔBED and ΔBCA,
∠BED = ∠BCA (Corresponding angles)
∠BDE = ∠BAC (Corresponding angles)
∠EBD = ∠CBA (Common angles)
∴ΔBED∼ΔBCA (By AAA similarity criterion)
We know that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.
ar (ΔBED) / ar (ΔBCA) = (DE/AC)2
⇒ar (ΔBED) / ar (ΔBCA) = (1/)2 = (1/4) [From (1)]
⇒ar (ΔBED) = (1/4) ar (ΔBCA)
Similarly,
ar (ΔCFE) = (1/4) ar (CBA)
ar (ΔADF) = (1/4) ar (ΔADF) = (1/4) ar (ΔABC)
Also,
ar (ΔDEF) = ar (ΔABC) − [ar (ΔBED) + ar (ΔCFE) + ar (ΔADF)]
⇒ ar (ΔDEF) = ar (ΔABC) − (3/4) ar (ΔABC) = (1/4) ar (ΔABC)
⇒ ar (ΔDEF) / ar (ΔABC) = 1/4
10. It is given that ∆ABC ~ ∆EDF, such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm. Find the lengths of the remaining sides of the triangles.
Solution:
Given,
AB = 5 cm, AC = 7 cm
DF = 15 cm and DE = 12 cm
∆ABC ~ ∆EDF
From the property of similar triangles, we have:
AB/ED = AC/EF = BC/DF ….(i)
Now, AB/ED = AC/EF
5/12 = 7/EF
⇒ EF = (7 × 12)/5 = 16.8 cm
Also, AB/ED = BC/DF
5/12 = BC/15
⇒ BC = (5 × 15)/12 = 6.25 cm
Therefore, EF = 16.8 cm and BC = 6.25 cm
Practice Questions on Similar Triangles
- E is a point on the side AD produced of a parallelogram ABCD, and BE intersects CD at F. Show that Δ ABE ~ Δ CFB.
- 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.
- Diagonals of a trapezium PQRS intersect each other at the point 0, PQ || RS and PQ = 3 RS. Find the ratio of the areas of Δ POQ and Δ ROS.
- D is a point on the side BC of a triangle ABC, such that ∠ADC = ∠BAC. Show that CA2 = CB.CD.
- Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
