The document Additional Questions Solution- Triangles Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.

All you need of Class 10 at this link: Class 10

**Additional Questions Solved Very Short Answer Type Questions**

**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‡’ 21x^{2} - 12x = 15x^{2} + 20x - 6x - 8

â‡’ 21x^{2} - 15x^{2} - 12x - 20x + 6x = - 8

â‡’ 6x^{2} - 26x + 8 = 0

â‡’ 3x^{2} - 13x + 4 = 0

â‡’ 3x^{2} - 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 cm ^{2} 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

Sol.

âˆ´Î” ABC is a right angled triangle.

âˆ´Using Pythagoras theorem, we have:

AC^{2} = AB^{2} + BC^{2}

= (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°.

**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 cm ^{2}, 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 cm^{2}.**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,

PR^{2} = PA^{2} + AR^{2}

= 62 + 82 = 36 + 64 = 100

But length cannot be negative

âˆ´ PR = 10 cm

Now,

PR^{2} + PQ^{2} = 102 + 24^{2}

= 100 + 576 = 676

= 262 = QR^{2}

â‡’ PR^{2} + PQ^{2} = QR^{2}

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]

<B = <ADE

[Corresponding angles]

âˆ´Using AA similarity, we have:

Î” ADE ~ Î” ABC

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‡’

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!

178 videos|268 docs|103 tests

### Similarity of Triangles and Basic Proportionality (Hindi)

- Video | 07:53 min
### Example Thales Theorem- 1

- Video | 11:55 min
### Long Answer Type Questions- Triangles

- Doc | 7 pages
### Example Thales theorem- 2

- Video | 10:46 min
### Example Thales Theorem- 3

- Video | 10:55 min
### Example Thales Theorem- 4

- Video | 09:20 min

- Test: Converse Of BPT
- Test | 10 ques | 10 min
- Proof Converse of Pythagoras Theorem
- Video | 03:57 min