Class 9 Maths Chapter 6 Question Answers - Triangles

Q1.In the given figure, DE y BC such that AC = 9 cm, AB = 7.2 cm and AD = 1.8 cm. Find AE.

Sol. In Δ ABC DE y BC
∴ Using the Basic Proportionality Theorem, we have:

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.

Sol. Since DE y BC
∴In Δ ABC, we have
=
[using the Basic Prop. Theorem]

⇒ 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.

Sol. Since, in Δ ABC, DE y BC.
∴ Using Basic Proportionality Theorem, we have
=
⇒ 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  . If AB = 4.8 cm then find AD.

Sol. Since DE y BC
∴ Using Basic Proportionality Theorem in D ABC, we have
=

Q5.In the given figure, if DE y BC and, then find AC such that AE = 2.1 cm.

Sol. Œ DE y BC
∴In Δ ABC, using the Basic Proportionality Theorem, we have:

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.

Sol. Œ Δ ABC ~ Δ DEF
.

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

Q8. The area of Δ PQR = 64 cm2. Find the area of Δ LMN, if  and Δ PQR ~ Δ LMN.
Sol.
Œ Δ PQR ~ Δ LMN

Q9. In the figure, <B = 90. Find AC.
Sol.
Œ< B = 90°

∴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.

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.

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]

Q12. In the figure, DE y AB, find the length of AC.

Sol. In Δ ABC, we have DE y AB
∴Using the Basic Proportionality Theorem, we have:
=
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.

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.

Q15. If the areas of two similar triangles are in the ratio 25 : 64, write the ratio of their corresponding sides.

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,

Q16. In a Δ ABC, DE y BC. If DE = BC and area of Δ ABC = 81 cm2, find the area of Δ ADE.

Sol. Since DE y BC
∴<1 = <2 [Corresponding angle]
Also <3 = <4 [Corresponding angles]
⇒ Δ DAE ~ Δ BAC [By AA criterion]

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.

Sol. We have:

In Δ AQP and Δ ACD

⇒ Δ AQP ~ Δ ACB
[SAS similarity rule]
[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.

Sol.  In right Δ PAR,
By the Pythagoras theorem,
PR2 = PA2 + AR2
= 62 + 82 = 36 + 64 = 100

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  .

Sol. Œ E and F are the midpoints of CA
and AB [Given]
∴ By midpoint theorem,
...(1)
Similarly,
...(2)
=                             ...(3)
From (1), (2) and (3), we get
=

Q20. In the figure, PQ y BC and AP : BP = 1 : 2. Find

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]

Q21. In the figure, DE y BC and if AC = 4.8 cm, find the length of AE.

Sol. Let AE = x
∴EC = AC - AE
= (4.8 - x) cm
Œ DE y BC [Given]

[By the Basic Proportionality Theorem]

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?

Sol. Œ DE y BC
∴In Δ ADE and Δ ABC,
<D = <B and <E = <C
[Corresponding angles]
∴ Δ ADE ~ Δ ABC

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?

Sol. In Δ ABC and Δ ADE
<A = <A [Common]
[Corresponding angles]
∴Using AA similarity, we have:

⇒ The required ratio = 9: 1.

Q24. In the figure, AC y BD. Is  ?

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.

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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