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# Additional Questions Solution- Triangles Class 10 Notes | EduRev

## Class 10 : Additional Questions Solution- Triangles Class 10 Notes | EduRev

The document Additional Questions Solution- Triangles Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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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. Å’ In Î” ABC DE y BC
âˆ´ 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
âˆ´In Î” ABC, we have
=
[using the Basic Prop. Theorem]

â‡’ 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 Î” ABC, DE y BC.
âˆ´ Using Basic Proportionality Theorem, we have
=
â‡’ 3x (7x - 4) = (5x - 2) (3x + 4)
â‡’ 21x2 - 12x = 15x2 + 20x - 6x - 8
â‡’ 21x2 - 15x2 - 12x - 20x + 6x = - 8
â‡’ 6x2 - 26x + 8 = 0
â‡’ 3x2 - 13x + 4 = 0
â‡’ 3x2 - 12x - x + 4 = 0
â‡’ 3x (x - 4) - 1 (x - 4) = 0
â‡’ (3x - 1) (x - 4) = 0
â‡’ 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
âˆ´ 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. Å’ DE y BC
âˆ´In Î” 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. Å’ Î” ABC ~ Î” 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.
Å’ Î” ABC ~ Î” DEF
âˆ´

Q8. The area of D PQR = 64 cm2. Find the area of D LMN, if  and D PQR ~ D LMN.
Sol.
Å’ Î” PQR ~ Î” LMN

Q9. In the figure, <B = 90. Find AC.
Sol.
Å’< B = 90Â°

âˆ´Î” ABC is a right angled triangle.
âˆ´Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
= (4.5)2 + 42
= 20.25 + 16
= 36.25
â‡’ AC = = 6.02 cm (approx.)

Q10. In the figure, if <A = 30Â° then find the measure of <C.

Sol. In Î” ABC, we have:
52 = 32 + 42
âˆ´Using the converse of Pythagoras Theorem, we have
Î” ABC is right angled at B; i.e., <B = 90Â°
Now using the angle-sum-property of a triangle, we get
90Â° + 30Â° + <C = 180Â°
â‡’ <C = 180Â° - 90Â° - 30Â° = 60Â°.

Q11. In the figure, ST y QR, PS = 2 cm and QS = 3 cm. What is the ratio of the area of D PQR to the area of D PST.

Sol. In Î” PQR and Î” PST
Å’ <P = <P [Common]
And <PQR = <PST [Å’ ST y QR]
âˆ´ Î”PQR ~ Î” PST
[By AA similarity rule]

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

Sol. In Î” ABC, we have DE y AB
âˆ´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 = 50Â° and <N = 60Â°. If Î” LMN ~ Î” PQR, then find <Q.
Sol.
By the angle-sum property,
<L + <M + <N = 180Â°
â‡’ 50Â° + <M + 60Â° = 180Â°
â‡’ <M = 180Â° - 60Â° - 50Â°
= 70Â°.
Since, Î” LMN ~ Î” PQR
âˆ´<L = <P
<M = <Q
<N = <R
Now, <Q = <M = 70Â°
â‡’ <Q = 70Â°.

Q14. In the figure, <M = <N = 46Â°. Express x in terms of a, b and c where a, b, and c are lengths of LM, MN and NK respectively.

Sol. In Î” KML and Î” KNP
<M = <N = 46Â° [Given]
<MKL = <NKP [Common]
â‡’ Î” KML ~ Î” KNP
[AA similarity rule]
âˆ´Their corresponding sides are proportional.
â‡’

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

Sol. Here, Î” ABC ~ Î” DEF
Since the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides,
âˆ´

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

Sol. Since DE y BC
âˆ´<1 = <2 [Corresponding angle]
Also <3 = <4 [Corresponding angles]
â‡’ Î” DAE ~ Î” BAC [By AA criterion]
âˆ´
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 Î” AQP and Î” ACD

â‡’ Î” AQP ~ Î” ACB
[SAS similarity rule]
[Å’ In similar Ds, ratio of
corresponding sides are equal]
â‡’ BC = 4.5 Ã— 3 = 13.5 cm.

Q18. In the figure, PQ = 24 cm, QR = 26 cm, <PQR = 90Â°, PA = 6 cm and AR = 8 cm. Find <QPR.

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

But length cannot be negative
âˆ´ PR = 10 cm
Now,
PR2 + PQ2 = 102 + 242
= 100 + 576 = 676
= 262 = QR2
â‡’ PR2 + PQ2 = QR2
â‡’ <QPR = 90Â°

Q19. D, E and F are the mid-points of the sides BC, AC and AB respectively of D ABC. Find  .

Sol. Å’ E and F are the mid points of CA
and AB [Given]
âˆ´ By mid point 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 â‡’ PB = 2k
âˆ´ AB = AP + PB = k + 2k = 3k
Å’ PQ y BC
âˆ´ <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
âˆ´EC = AC - AE
= (4.8 - x) cm
Å’ DE y BC [Given]
âˆ´
[By the Basic Proportionality Theorem]
â‡’

Q22. In Î” ABC (shown in the figure), DE y BC. If BC = 8 cm, DE = 6 cm and area of Î” ADE = 45 cm2, then what is the area of Î” ABC?

Sol. Å’ DE y BC
âˆ´In Î” ADE and Î” ABC,
<D = <B and <E = <C
[Corresponding angles]
âˆ´ Î” ADE ~ Î” ABC

Q23. In the figure, DE y BC and AD = 1 cm, BD = 2 cm. What is the ratio of the areas of Î” ABC to the area of Î” ADE?

Sol. In Î” ABC and Î” ADE
<A = <A [Common]
[Corresponding angles]
âˆ´Using AA similarity, we have:
â‡’
â‡’ The required ratio = 9 : 1.

Q24. In the figure, AC y BD. Is  ?

Sol. In Î” ACE and Î” DBE,
<A = <D
[Alternate Interior Angles]
<C = <B
âˆ´Using AA similarity
Î” ACE ~ Î” DBE
âˆ´Their corresponding sides are proportional.
â‡’

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