The document Long Answer Type Questions- Triangles Class 10 Notes | EduRev is a part of the Class 10 Course Class 10 Mathematics by VP Classes.

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

** Long Answer Type Questions****Q1.Prove that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides. Using the above result, prove the following: In Î” ABC, XY is parallel to BC and it divides Î” ABC into two parts of equal area. Prove that: = **

Part-II

We have a Î” ABC in which XY y BC such that

ar (Î” ABC) = 2 ar (Î” AXY)

â‡’ ...(1)

Now, in Î” ABC and Î” AXY, we have

<1 = <2 and <3 = <4 [Corresponding angles]

â‡’ Î” ABC ~ Î” AXY

[using the Basic Proportionality Theorem]

[using the above theorem] ...(2)

From (1) and (2), we get**Q2.Prove that the ratio of the areas of two similar triangles is equal to the ratio of the squares on their corresponding sides.Using the above, do the following: The diagonals of a trapezium ABCD, with AB y DC, intersect each other at the point O. If AB = 2 CD, find the ratio of the area of Î” AOB to the area of Î”COD. Sol.**Part-I [See proof of the theorem on the ratio of areas of similar Ds

Part-II

We have a trapezium ABCD such that AB y DC and AB = 2 CD

In Î” AOB and Î” COD

Å’ AB y DC

âˆ´ <1 = < 2

[Int. alternate angles are equal]

<3 = <4

âˆ´ Using AA similarity, we have

But AB = 2 CD

âˆ´

Thus, ar (Î” AOB) : ar (Î” COD) = 4 : 1.

Sol.

âˆ´ SH = 27 m = PC

[Å’ their speeds are equal]

Now SC = (27 - x) m

PC = (27 - x) m

In right D PHC, we have:

PH^{2} + CH^{2} = PC^{2}

[By Pythagoras Theorem]

â‡’ (9)^{2} + (x)2 = (27 - x)^{2}

â‡’ 81 + x^{2} = 729 - 54x + x^{2}

â‡’ 54x = 729 - 81 = 648

â‡’ x = 648/54=12m

Hence, the required distance = 12 m.**Q4. Prove that the ratio of the areas of two similar triangles is equal to the ratio of squares of their corresponding sides. Using the above result, prove the following: If D is a point on the side AB of D ABC such that AD : DB = 3 : 2 and E is a point on BC such that DE y AC; find the ratio of the areas of Î” ABC and Î” BDE. Sol.** Part-I [See proof of the theorem on ratio of areas of similar triangles]

Part-II

In Î” ABC,

Å’ DE y AC [Given]

âˆ´ <1 = <2 [Corresponding Angles]

Let BD = 2 k and AD = 3 k

âˆ´ AB = AD + DB

= 3 k + 2 k = 5 k

Now in D ABC and D BDE

<B = <B [Common]

<1 = <2 [Proved above]

âˆ´ Using AA similarity we have:

Î” ABC ~ Î” BDE

â‡’

â‡’ ar (Î” ABC) : (ar Î” BDE) = 25 : 4.**Q5. Prove that in a right triangle the square on the hypotenuse is equal to the sum of the squares of the other two sides. Making use of the above, prove the following: In a rhombus ABCD, prove that: 4 AB2 = AC**

**Sol.** Part-I . [see proof of Pythagoras Theorem]

Part-II.

We have a rhombus ABCD whose diagonals AC and BD intersect at O.

Since, the diagonals of a rhombus bisect each other at right angles.

âˆ´ AC â‰Œ BD And

OA = OC

OB = OD

In right Î” AOB, we have

AB^{2} = OA^{2} + OB^{2}

[using the above result]**Q6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Using the above, do the following: If D is a point on the side AC of D ABC such that AD : DC = 2 : 3 and E is a point on BC such that DE y AB; then find the ratio of areas of Î” ABC and Î” CDE. Sol.** Part-I [See the proof of the theorem]

Part-II. We have D ABC in which

DE y AB

âˆ´ In

<C = <C [Common]

<1 = <2 [Corresponding

Using AA similarity,

Î” ABC ~ Î” DEC

From (1) and (2), we get

Using the above result, do the following:

In the figure, DE y BC and BD = CE. Prove that D ABC is an isosceles triangle.

Sol.

Part-II In Î” ABC,

Since, DE y BC [Given]

âˆ´ Using the Basic Proportionality Theorem, we get

=

From (1) and (2) we get

=

From (2) and (3), we get

AD + BD = AE + CE

â‡’ AB = AC

â‡’ Î” ABC is an isosceles triangle.**Q8. Prove that, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Using the above, do the following: Prove that, in a D ABC, if AD is perpendicular to BC, then AB** Part-I See the proof of the Pythagoras Theorem]

Part-II We have Î” ABC in which AD âŠ¥ BC

âˆ´ In right D ADB, <D = 90Â°

âˆ´ Using Pythagoras theorem,

AB^{2} = AD^{2} + BD^{2}

â‡’AD^{2} = AB^{2} - BD^{2} ...(1)

Similarly, in right D ADC

AC^{2} = AD^{2} + CD^{2}

â‡’ AD^{2} = AC^{2} - CD^{2} ...(2)

From (1) and (2), we get

AB^{2} - BD^{2} = AC^{2} - CD^{2}

AB^{2} + CD^{2} = AC^{2} + BD^{2}

**Q9. ABCD is a trapezium with AB y DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB. Show that**

^{= }**Sol.** Join AC such that it meets EF at G.

ABy DC and EFy AB

âˆ´ EF y DC [Q Lines parallel to the same line are parallel to each other]

In Î” ADC

EG y DC [Q EG is a part of EF]

â‡’ = ^{ ...(1) Next in DCAB = }^{ ...(2) From (1) and (2), we get = }

^{Q10. ABCD is a trapezium in which AByDC and P and Q are points on AD and BC, respectively such that PQy DC. If PD = 18 cm, BQ = 35 cm and QC = 15 cm, find AD. [NCERT Exemplar] Sol. Since AByDC and PQyDC â‡’ = } ...(1)

Substituting

PD = 18 cm, BQ = 35 cm

and CQ = 15 cm in (1), we get

^{= }^{â‡’}

Since, AD = AP + PD

âˆ´ AD = 42 cm + 18 cm = 60 cm**Q11. In the following figure if AByDC and AC and PQ intersect each other at the point O, prove that: OAÃ—CQ = OCÃ—AP **

Sol.

âˆ´ <OCQ = <OAB

[Alternate angles]

Again, PQ is also a transversal

âˆ´ <OQC = <APO

[Alternate angles]

Now, in Î” OAP and Î” OQC, we have

<OAP = <OCQ

<APO = <OQC [Proved above]

<AOP = <QOC [Vertically opposite angles]

â‡’ Î” OAP and Î” OQC are equiangular

âˆ´ The triangles are similar

Sol.

Let one of the sides be x cm

âˆ´ Other side = (x + 5)cm

Now, using Pythagoras Theorem,

x2 + (x + 5)2 = 252

â‡’ x

â‡’ 2x

or x

or x

or x(x + 20) â€“ 15 (x + 20)

or (x â€“ 15) (x + 20) = 0

Either x â€“ 15 = 0

â‡’ x = 15

or x + 20 = 0

â‡’ x = â€“ 20

x = â€“ 20

is not required.

Therefore, x = 15

Now, x + 5 = 15 + 5 = 20

The other two sides are 15 cm and 20 cm.

Using the above, do the following:

In a trapezium ABCD, AC and BD are intersecting at O, AB y DC and AB = 2 CD. If the area of Î” AOB = 84 cm2, find the area of Î” COD.

Sol.

Part-II In Î” AOB and Î” COD

<1 = <2

[Alternate interior angles Å’ AB y DC]

<3 = <4

â‡’ Î” AOB ~ Î” COD

[By AA similarity]

â‡’

[using the above theorem]

â‡’

[Å’ It is given that AB = 2 CD]

â‡’

**Sol.**

Part-I [See the converse of the Pythagoras theorem]

Part-II In right Î” QPR

Using Pythagoras theorem,

PR^{2} + PQ^{2} = QR^{2}

â‡’ PR2 + (24)^{2} = 262

â‡’ PR2 = 262 - 242

= (26 - 24) (26 + 24)

= 2 Ã— 50 = 100 ...(1)

Now in D PKR,

PK^{2} + KR^{2} = 82 + 62

= 64 + 36 = 100 ...(2)

From (1) and (2),

PK^{2} + KR^{2} = PR^{2}

which is according to the above theorem.

âˆ´Angle opposite PR must be 90Â°

â‡’ <PKR = 90Â°

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

132 docs

### Facts That Matter- Triangles

- Doc | 1 pages
### Short Answer Type Questions (Part 2) - Triangles

- Doc | 5 pages
### Ex 6.6 NCERT Solutions - Triangles

- Doc | 7 pages
### Additional Questions Solution- Triangles

- Doc | 6 pages

- Ex 6.5 NCERT Solutions - Triangles
- Doc | 6 pages
- Short Answer Type Questions (Part 1) - Triangles
- Doc | 4 pages