CRE 2 - Similarity of Triangles | Geometry - Triangles
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1. CRE 2 - Similarity of Triangles | Geometry - Triangles
A triangle has sides a, b and c and the measure of the angles opposite to these sides are X, Y and Z respectively. Find the measures of the angles for the triangle with sides 2a, 2b and 2c.
(a) X/2, Y/2 and Z/2
(b) 2X, 2Y and 2Z
(c) X, Y and Z
(d) 4X, 4Y and 4Z
Answer: Option
C
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Explanation :
The sides of the two triangles are a, b, c and 2a, 2b and 2c respectively.
∴ The corresponding sides for the two triangles are in the ratio 1 : 2.
∴ The two triangles are similar.
∴ The corresponding angles of both the triangles are equal.
∴ The measures of the angles in the second triangle are X, Y and Z respectively.
Hence, option (c).
2. CRE 2 - Similarity of Triangles | Geometry - Triangles
A line DE is drawn by joining the midpoints of the sides AB and AC of ∆ABC with area 12 sq. units. What is the area of ∆ADE (in sq. units)?
Answer: Option
D
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Explanation :
∵ DE joins the midpoints of AB and AC
∴ According to the basic proportionality theorem, side DE of ∆ADE is parallel to side BC of ∆ABC and
In ∆ADE and ∆ABC,
∠ADE = ∠ABC (Corresponding angles) and ∠A is common to both the triangles.
∴ ∆ADE ∼ ∆ABC ...(By A-A test of similarity)
The ratio of the areas of similar triangles is equal to ratio of the square of the their respective sides.
A(ΔADE) : A(ΔABC) = (DE)2 : (BC)2 = 1/4
∴ A(ΔADE)/12 = 1/4
∴ A(ΔADE) = 12/4 = 3 sq. units
Hence, option (d).
3. CRE 2 - Similarity of Triangles | Geometry - Triangles
In ∆ABC, AB = 15, AC = 10 and BC = 6. D and E are points on AB and AC respectively such that DE || BC. If DB = 10, DE =?
Answer: 2
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Explanation :
In ∆ABC and ∆ ADE
∠ABC = ∠ADE (Corresponding angles) and ∠A is common to both the triangles.
∴ ∆ABCE ∼ ∆ADE ...(By A-A test of similarity)
⇒ AD/AB=DE/BC
⇒ (15 - 10)/15 = DE/6
⇒ DE = 2
Hence, 2.
4. CRE 2 - Similarity of Triangles | Geometry - Triangles
In the figure below, AB ∥ CD. Find the value of x.
(a) 2
(b) 3
(c) 1/2
(d) None of these
Answer: Option
B
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Explanation :
Given, AB || CD.
Consider ∆AOB and ∆COD
∠AOB = ∠COD (Vertically opposite angles) and
∠ABD = ∠CDB (alternate interior angles)
∴ ∆AOB ~ ∆COD
⇒ AO/CO = BO/DO
⇒ 4/(4x - 4) = (2x - 1)/(2x + 4)
⇒ 1/(x - 1) = (2x - 1)/(2x + 4)
⇒ 2x + 4 = 2x2 - 3x + 1
⇒ 2x2 – 5x – 3 = 0
⇒ (2x + 1)(x - 3) = 0
⇒ x = -1/2 or 3
Since x has to be positive, x = 3
Hence, option (b).
5. CRE 2 - Similarity of Triangles | Geometry - Triangles
ΔABC and ΔDEF are similar and their areas be respectively 81 cm2 and 144 cm2 . If EF = 9 cm, BC is?
(a) 10.5
(b) 10.8
(c) 12
(d) 6.75
(e) None of these
Answer: Option
D
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Explanation :
Given, ΔABC ~ ΔDEF
∴ The ratio of the areas of similar triangles is equal to ratio of the square of the their respective sides.
In the two similar triangles side BC in ∆ABC is corresponding to side EF in ∆DEF.
⇒ A r e a ∆ A B C A r e a ∆ D E F = B C E F 2
⇒ 81 144 = B C 9 2
⇒ 9 12 = B C 9
⇒ BC = 6.75
Hence, option (d).
6. CRE 2 - Similarity of Triangles | Geometry - Triangles
Two isosceles triangles have equal vertical angles and their areas are in the ratio 9 : 16. Find the ratio of their corresponding heights.
(a) 3 : 4
(b) 4 : 3
(c) 9 : 16
(d) Can't be determined
Answer: Option
A
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Explanation :
Let the triangles be ∆ABC and ∆DEF.
Given vertical angles are equal, ∠A = ∠D = x (say)
In Isosceles ∆ABC, ∠B = ∠C = (180 - x)/2
Similarly, in Isosceles ∆DEF, ∠E = ∠F = (180 - x)/2
In ΔABC ~ ΔDEF
∠A = ∠D (vertically opposite angles)
∠B = ∠E = (180 - x)/2
∴ ∆ABCE ∼ ∆ADE ...(By A-A test of similarity)
⇒ The ratio of the areas of similar triangles is equal to ratio of the square of the their respective heights.
⇒ h 1 h 2 2 = A r e a ∆ A B C A r e a ∆ D E F
⇒ h 1 h 2 2 = 9 16
⇒ h1 : h2 = 3 : 4
Hence, option (a).
7. CRE 2 - Similarity of Triangles | Geometry - Triangles
In ∆ABC, DE is a line segment intersecting AB at D and AC at E such that DE ∥ BC. DE divides ∆ ABC in two parts equal in area. Find BD/AB?
(a) 1 : 1
(b) 1 : 2
(c) (√2 - 1)/√2
(d) None of these
Answer: Option
C
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Explanation :
Given, Area(∆ADE) = Area(∎BDEC)
∴ Area(∆ABC) = 2 × Area(∆ADE)
Now, let’s consider ∆ABC and ∆ ADE
∠ABC = ∠ADE (Corresponding angles) and ∠A is common to both the triangles.
∴ ∆ABCE ∼ ∆ADE ...(By A-A test of similarity)
⇒ The ratio of the areas of similar triangles is equal to ratio of the square of the their respective heights.
⇒ A B A D 2 = A r e a ∆ A B C A r e a ∆ D E F = 2 1
⇒ AB = √2 × AD.
⇒ BD = AB – AD = (√2 – 1) × AD.
∴ B D A B = 2 – 1 × AD 2 × AD = 2 - 1 2
8. CRE 2 - Similarity of Triangles | Geometry - Triangles
In the figure below, PS = 32 cm, PQ = 18 cm, SR = 25 cm and ∠QRP = ∠QSR. Find the ratio of the perimeters of the triangle PQR and PRS.
(a) 5 : 4
(b) 4 : 3
(c) 2 : 3
(d) 13 : 11
Answer: Option
B
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Explanation :
In ∆RPS and ∆QPR
∠RPS = ∠QPR (∵ P is common angle)
∠PSR = ∠PRQ (Given)
∴ ∆RPS ~∆QPR
⇒ P S P R = P R P Q = S R R Q
⇒ 32 P R = P R 18 = 25 R Q
Solving first two ratios we get, PR = 24.
Solving last two ratios we get RQ = 75/4 = 18.75
∴ Perimeter of ∆RPS = 24 + 32 + 25 = 81
∴ Perimeter of ∆QPR = 18 + 24 + 18.75 = 60.75
⇒ Ratio of perimeter of ∆RPS and ∆QPR = 81 : 60.75 = 4 : 3.
Hence, option (b).
9. CRE 2 - Similarity of Triangles | Geometry - Triangles
P is the mid-point of side BC of ∆ABC. If Q is mid-point of AP and BQ when produced meets AC at L. Is LA = 1/3(CA)?
Type 1: if your answer is yes.
Answer: 1
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Explanation :
Draw a line from P parallel to BL which meets AC at M.
Now, let’s consider ∆APM and ∆AQL
∠QAL is common in both triangles, and
∠AQL = ∠APM (Corresponding angles ∵ QL || PM)
∴ ∆APM ∼ ∆AQL ...(By A-A test of similarity)
⇒ AQ/AP = AL/AM
⇒ 1/2 = AL/AM
⇒ AL = LM …(1)
Similarly, ∆CPM ~ ∆CBL, and
⇒ CM = LM …(2)
From (1) and (2), we get
CM = ML = LA = 1/3 × AC
Hence, 1.
10. CRE 2 - Similarity of Triangles | Geometry - Triangles
In the figure below, PB and QA are perpendiculars to segment AB. If PO = 5 cm, QO = 7 cm and area ∆POB = 150 sq.cm., find area of ∆QOA.
Answer: 294
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Explanation :
In ∆PBO and ∆QAO
∠B = ∠A = 90°
∠POB = ∠QOA (vertically opposite angles)
∴ ∆PBO ∼ ∆QAO ...(By A-A test of similarity)
⇒ The ratio of the areas of similar triangles is equal to ratio of the square of the their respective sides.
⇒ P O Q O 2 = A r e a ∆ P B O A r e a ∆ Q A O
⇒ 5 7 2 = 150 A r e a ∆ Q A O
⇒ Area(∆QAO) = 294 sq.cm.
Hence, 294.
11. CRE 2 - Similarity of Triangles | Geometry - Triangles
Area of triangle ABC is 144 cm2 . Another triangle is formed by joining the mid-points of sides of ABC. Yet another triangle is formed by joining the mid-points of the new triangle to obtain a third triangle. Find the area of the smallest triangle.
(a) 72
(b) 36
(c) 12
(d) 9
Answer: Option
D
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Explanation :
When mid-points of a triangle is formed, we get 4 smaller triangles and area of each of these smaller triangles is one-fourth the area of the larger triangle.
Thus, if area of triangle ABC is 144, the area of triangle formed by joining the midpoints will be 144/4 = 36.
The area of the third triangle will be 36/4 = 9cm2 .
Hence, option (d).
12. CRE 2 - Similarity of Triangles | Geometry - Triangles
In a right triangle ABC right angled at B, perpendicular BD is drawn from B to AC. AB = 10 and AD = 4, find CD?
(a) 14
(b) 21
(c) 25
(d) Cannot be determined
Answer: Option
B
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Explanation :
In a right triangle, perpendicular drawn from right vertex to hypotenuse divides the triangle in three similar triangles.
⇒ AB2 = AD × AC2 = 4 × AC
∴ CD = AC - AD = 25 - 4 = 21
Hence, option (b).
13. CRE 2 - Similarity of Triangles | Geometry - Triangles
In the figure given below, DE || BC = 1/3 of BC. If area of triangle ADE = 20 cm2 , then what is the area (in cm2 ) of triangle DEC?
(a) 40
(b) 60
(c) 80
(d) 120
Answer: Option
A
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Explanation :
Let us draw AF which is perpendicular to both DE and BC
∆ADE ≈ ∆ABC (since DE || BC)
⇒ AF : AG = DE : BC = 1 : 3
⇒ AF : FG = 1 : 2
For ∆ADE and ∆DEC
∴ Area of ∆DEC will be twice that of ∆ADE
⇒ Area of ∆DEC = 2 × 20 = 40 cm2 .
Hence, option (a).
Solution
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