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Class 9 Maths Chapter 6 Question Answers - Triangles

Very Short Answer Type Questions

Q1.In the given figure, DE y BC such that AC = 9 cm, AB = 7.2 cm and AD = 1.8 cm. Find AE.
Class 9 Maths Chapter 6 Question Answers - Triangles

Sol. In Δ ABC DE y BC
∴ Using the Basic Proportionality Theorem, we have:
Class 9 Maths Chapter 6 Question Answers - Triangles
Thus, the required value of AE = 2.25 cm.


Q2. In the figure, DE y BC such that AD = 4.8 cm, AE = 6.4 cm and EC = 9.6 cm, find DB.
Class 9 Maths Chapter 6 Question Answers - Triangles

Sol. Since DE y BC
∴In Δ ABC, we have
= Class 9 Maths Chapter 6 Question Answers - Triangles
[using the Basic Prop. Theorem]
Class 9 Maths Chapter 6 Question Answers - Triangles
⇒ DB = = 7.2 cm
Thus, the required value of DB = 7.2 cm.

Q3. In the given figure, DE y BC such that
AD = (7x - 4) cm, AE = (5x - 2) cm. If EC = 3x and DB = (3x + 4) cm, then find the value of x.

Class 9 Maths Chapter 6 Question Answers - Triangles

Sol. Since, in Δ ABC, DE y BC.
∴ Using Basic Proportionality Theorem, we have
= Class 9 Maths Chapter 6 Question Answers - Triangles
⇒ 3x (7x - 4) = (5x - 2) (3x + 4)
⇒ 21x2 - 12x = 15x2 + 20x - 6x - 8
⇒ 21x2 - 15x2 - 12x - 20x + 6x = - 8
⇒ 6x2 - 26x + 8 = 0
⇒ 3x2 - 13x + 4 = 0
⇒ 3x2 - 12x - x + 4 = 0
⇒ 3x (x - 4) - 1 (x - 4) = 0
⇒ (3x - 1) (x - 4) = 0
⇒ x = or 4.

Q4. In the given figure, DE y BC such that Class 9 Maths Chapter 6 Question Answers - Triangles . If AB = 4.8 cm then find AD.
Class 9 Maths Chapter 6 Question Answers - Triangles

Sol. Since DE y BC
∴ Using Basic Proportionality Theorem in D ABC, we have
= Class 9 Maths Chapter 6 Question Answers - Triangles
Class 9 Maths Chapter 6 Question Answers - Triangles

Q5.In the given figure, if DE y BC and, then find AC such that AE = 2.1 cm.
Class 9 Maths Chapter 6 Question Answers - Triangles

Sol. Œ DE y BC
∴In Δ ABC, using the Basic Proportionality Theorem, we have:
Class 9 Maths Chapter 6 Question Answers - Triangles
Now, AC = AE + EC = (2.1 + 3.5) cm
= 5.6 cm.

Q6. In the figure, D ABC ~ D DEF and AB/DE=3. If BC = 4 cm then find EF.
Class 9 Maths Chapter 6 Question Answers - Triangles

Sol. Œ Δ ABC ~ Δ DEF
Class 9 Maths Chapter 6 Question Answers - Triangles .

Q7. The areas of two similar triangles ABC and DEF are 81 cm2 and 100 cm2 respectively.
 If EF = 5 cm, then find BC.
 Sol.
Œ Δ ABC ~ Δ DEF
Class 9 Maths Chapter 6 Question Answers - Triangles

Q8. The area of Δ PQR = 64 cm2. Find the area of Δ LMN, if Class 9 Maths Chapter 6 Question Answers - Triangles and Δ PQR ~ Δ LMN.
 Sol. 
Œ Δ PQR ~ Δ LMN
Class 9 Maths Chapter 6 Question Answers - Triangles

Q9. In the figure, <B = 90. Find AC.
 Sol.
Œ< B = 90°
Class 9 Maths Chapter 6 Question Answers - Triangles

∴Δ ABC is a right angled triangle.
∴Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
= (4.5)2 + 42
= 20.25 + 16
= 36.25
⇒ AC = = 6.02 cm (approx.)

Q10. In the figure, if <A = 30° then find the measure of <C.
Class 9 Maths Chapter 6 Question Answers - Triangles

Sol. In Δ ABC, we have:
52 = 32 + 42
∴Using the converse of Pythagoras Theorem, we have
Δ ABC is right-angled at B; i.e., <B = 90°
Now using the angle-sum-property of a triangle, we get
90° + 30° + <C = 180°
⇒ <C = 180° - 90° - 30° = 60°.

Q11. In the figure, ST y QR, PS = 2 cm and QS = 3 cm. What is the ratio of the area of Δ PQR to the area of Δ PST.
Class 9 Maths Chapter 6 Question Answers - Triangles

Sol. In Δ PQR and Δ PST
Œ <P = <P [Common]
And <PQR = <PST [Œ ST y QR]
∴ ΔPQR ~ Δ PST
[By AA similarity rule]
Class 9 Maths Chapter 6 Question Answers - Triangles

Q12. In the figure, DE y AB, find the length of AC.
Class 9 Maths Chapter 6 Question Answers - Triangles

 

Sol. In Δ ABC, we have DE y AB
∴Using the Basic Proportionality Theorem, we have:
= Class 9 Maths Chapter 6 Question Answers - Triangles
Now AC = AD + DC
= 4.5 + 3 = 7.5 cm
Thus, AC = 7.5 cm.

Q13. In < LMN, <L = 50° and <N = 60°. If Δ LMN ~ Δ PQR, then find <Q.
 Sol.
By the angle-sum property,
<L + <M + <N = 180°
⇒ 50° + <M + 60° = 180°
⇒ <M = 180° - 60° - 50°
= 70°.
Since, Δ LMN ~ Δ PQR
∴<L = <P
<M = <Q
<N = <R
Now, <Q = <M = 70°
⇒ <Q = 70°.

Q14. In the figure, <M = <N = 46°. Express x in terms of a, b and c where a, b, and c are lengths of LM, MN and NK respectively. 
Class 9 Maths Chapter 6 Question Answers - Triangles 

 

Sol. In Δ KML and Δ KNP
<M = <N = 46° [Given]
<MKL = <NKP [Common]
⇒ Δ KML ~ Δ KNP
[AA similarity rule]
∴Their corresponding sides are proportional.
Class 9 Maths Chapter 6 Question Answers - Triangles

Q15. If the areas of two similar triangles are in the ratio 25 : 64, write the ratio of their corresponding sides. 
Class 9 Maths Chapter 6 Question Answers - Triangles
Class 9 Maths Chapter 6 Question Answers - Triangles

 

Sol. Here, Δ ABC ~ Δ DEF
Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides,
Class 9 Maths Chapter 6 Question Answers - Triangles

Q16. In a Δ ABC, DE y BC. If DE = BC and area of Δ ABC = 81 cm2, find the area of Δ ADE.
Class 9 Maths Chapter 6 Question Answers - Triangles

 

Sol. Since DE y BC
∴<1 = <2 [Corresponding angle]
Also <3 = <4 [Corresponding angles]
⇒ Δ DAE ~ Δ BAC [By AA criterion]
Class 9 Maths Chapter 6 Question Answers - Triangles
Thus, ar D ADE = 36 cm2.

Q17. In the figure, P and Q are points on the sides AB and AC respectively of D ABC such that AP = 3.5 cm, PB = 7 cm, AQ = 3 cm and
 QC = 6 cm. If PQ = 4.5 cm, find BC.

Class 9 Maths Chapter 6 Question Answers - Triangles

 

Sol. We have:
Class 9 Maths Chapter 6 Question Answers - Triangles
In Δ AQP and Δ ACD
Class 9 Maths Chapter 6 Question Answers - Triangles

⇒ Δ AQP ~ Δ ACB
[SAS similarity rule]
Class 9 Maths Chapter 6 Question Answers - Triangles [Œ In similar Ds, the ratio of corresponding sides are equal]
⇒ BC = 4.5 × 3 = 13.5 cm.

Q18. In the figure, PQ = 24 cm, QR = 26 cm, <PQR = 90°, PA = 6 cm and AR = 8 cm. Find <QPR. 
Class 9 Maths Chapter 6 Question Answers - Triangles

Sol.  In right Δ PAR,
By the Pythagoras theorem,
PR2 = PA2 + AR2
= 62 + 82 = 36 + 64 = 100
Class 9 Maths Chapter 6 Question Answers - Triangles
But length cannot be negative
∴ PR = 10 cm
Now,
PR2 + PQ2 = 102 + 242
= 100 + 576 = 676
= 262 = QR2
⇒ PR2 + PQ2 = QR2
⇒ <QPR = 90°

Q19. D, E and F are the mid-points of the sides BC, AC and AB respectively of D ABC. Find Class 9 Maths Chapter 6 Question Answers - Triangles .
Class 9 Maths Chapter 6 Question Answers - Triangles 

Sol. Œ E and F are the midpoints of CA
and AB [Given]
∴ By midpoint theorem,
Class 9 Maths Chapter 6 Question Answers - Triangles ...(1)
Similarly,
Class 9 Maths Chapter 6 Question Answers - Triangles ...(2)
=                             ...(3)
From (1), (2) and (3), we get
= Class 9 Maths Chapter 6 Question Answers - Triangles

Q20. In the figure, PQ y BC and AP : BP = 1 : 2. Find Class 9 Maths Chapter 6 Question Answers - Triangles
Class 9 Maths Chapter 6 Question Answers - Triangles

Sol. It is given that
AP : PB = 1 : 2
Let AP = k ⇒ PB = 2k
∴ AB = AP + PB = k + 2k = 3k
Œ PQ y BC
∴ <APQ = <ABC and
<AQP = <ACB [Corresponding Angles]
Class 9 Maths Chapter 6 Question Answers - Triangles

Q21. In the figure, DE y BC and if AC = 4.8 cm, find the length of AE. 
Class 9 Maths Chapter 6 Question Answers - Triangles

 

Sol. Let AE = x
∴EC = AC - AE
= (4.8 - x) cm
Œ DE y BC [Given]
Class 9 Maths Chapter 6 Question Answers - Triangles
[By the Basic Proportionality Theorem]
Class 9 Maths Chapter 6 Question Answers - Triangles

Q22. In Δ ABC (shown in the figure), DE y BC. If BC = 8 cm, DE = 6 cm and area of Δ ADE = 45 cm2, then what is the area of Δ ABC? 
Class 9 Maths Chapter 6 Question Answers - Triangles

 

Sol. Œ DE y BC
∴In Δ ADE and Δ ABC,
<D = <B and <E = <C
[Corresponding angles]
∴ Δ ADE ~ Δ ABC
Class 9 Maths Chapter 6 Question Answers - Triangles

Q23. In the figure, DE y BC and AD = 1 cm, BD = 2 cm. What is the ratio of the areas of Δ ABC to the area of Δ ADE?
Class 9 Maths Chapter 6 Question Answers - Triangles

Sol. In Δ ABC and Δ ADE
<A = <A [Common]
<B = <ADE
[Corresponding angles]
∴Using AA similarity, we have:
Δ ADE ~ Δ ABC
Class 9 Maths Chapter 6 Question Answers - Triangles
⇒ The required ratio = 9: 1.

Q24. In the figure, AC y BD. Is Class 9 Maths Chapter 6 Question Answers - Triangles ?
Class 9 Maths Chapter 6 Question Answers - Triangles 

 

Sol. In Δ ACE and Δ DBE,
<A = <D
[Alternate Interior Angles]
<C = <B
∴Using AA similarity
Δ ACE ~ Δ DBE
∴Their corresponding sides are proportional.
Class 9 Maths Chapter 6 Question Answers - Triangles

The document Class 9 Maths Chapter 6 Question Answers - Triangles is a part of the Class 10 Course Mathematics (Maths) Class 10.
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FAQs on Class 9 Maths Chapter 6 Question Answers - Triangles

1. What are the properties of an equilateral triangle?
Ans. An equilateral triangle has three equal sides and three equal angles. Each angle measures 60 degrees. The sum of the angles in an equilateral triangle is always 180 degrees.
2. How do you find the area of a triangle?
Ans. The formula to find the area of a triangle is given as A = (1/2) × base × height. The base and height are perpendicular to each other, and the base is the length of one side of the triangle.
3. What is the Pythagorean Theorem?
Ans. The Pythagorean Theorem states that in a right-angled triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. It can be written as a^2 + b^2 = c^2, where c represents the length of the hypotenuse.
4. How do you determine if three given sides can form a triangle?
Ans. To determine if three given sides can form a triangle, you can use the Triangle Inequality Theorem. According to this theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. If this condition is satisfied for all three combinations of sides, then the given lengths can form a triangle.
5. What is the difference between an acute triangle and an obtuse triangle?
Ans. An acute triangle is a triangle in which all three angles are less than 90 degrees. On the other hand, an obtuse triangle is a triangle in which one angle is greater than 90 degrees.
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