# Class 10 NCERT Solutions- Chapter 6 Triangles – Exercise 6.3 | Set 2

Last Updated : 03 Apr, 2024

Content of this article has been merged with the article, Class 10 NCERT Solutions- Chapter 6 Triangles â€“ Exercise 6.3

### Question 11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD âŠ¥ BC and EF âŠ¥ AC, prove that Î”ABD ~ Î”ECF.

Solution:

Given, ABC is an isosceles triangle.

Since, the sides of an isosceles triangle are equal, we have,

âˆ´ AB = AC

â‡’ âˆ ABD = âˆ ECF

In Î”ABD and Î”ECF,

Since, each of the following angles are 90Â°.

By AA similarity criterion, we have,

âˆ´ Î”ABD ~ Î”ECFÂ

### Question 12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of Î”PQR (see Fig 6.41). Show that Î”ABC ~ Î”PQR.

Solution:

Given that in Î”ABC and Î”PQR,Â

AB is proportional to PQ

BC is proportional to QR

That is, AB/PQ = BC/QR = AD/PM

We know,

Since, D is the midpoint of BC and M is the midpoint of QR

By SSS similarity criterion, we have,

â‡’ Î”ABD ~ Î”PQMÂ

Now, since the corresponding angles of two similar triangles are equal, we obtain,

âˆ´ âˆ ABD = âˆ PQMÂ

â‡’ âˆ ABC = âˆ PQR

Now,

In Î”ABC and Î”PQR

AB/PQ = BC/QR â€¦â€¦â€¦â€¦.(i)

âˆ ABC = âˆ PQR â€¦â€¦â€¦â€¦â€¦(ii)

From equation (i) and (ii), we get,

By SAS similarity criterion, we have,

Î”ABC ~ Î”PQRÂ

### Question 13. D is a point on the side BC of a triangle ABC such that âˆ ADC = âˆ BAC. Show that CA2 = CB.CD

Solution:

Given, D is a point on the side BC of a triangle ABC such that âˆ ADC = âˆ BAC.

âˆ ACD = âˆ BCA (Common angles)

By AA similarity criterion, we have,

Since, the corresponding sides of similar triangles are in proportion, we obtain,

âˆ´ CA/CB = CD/CA

That is, CA2 = CB.CD.

Hence, proved.

### Question 14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that Î”ABC ~ Î”PQR.

Solution:

Given, Two triangles Î”ABC and Î”PQR in which AD and PM are medians such that;

Now,

Construction,Â

Produce AD further to E so that AD = DE. Now, join CE.

Similarly, produce PM further until N such that PM = MN. Join RN.

In Î”ABD and Î”CDE, we have

From the construction done,

Now, since AP is the median,

BD = DC

and, âˆ ADB = âˆ CDE [Vertically opposite angles are equal]

By SAS criterion,

âˆ´ Î”ABD â‰… Î”CDEÂ

By CPCT, we have,

â‡’ AB = CEâ€¦â€¦â€¦â€¦â€¦..(i)

In Î”PQM and Î”MNR,

From the construction done,

PM = MN

Now, since PM is the median,

QM = MRÂ

and, âˆ PMQ = âˆ NMR [Vertically opposite angles are equal]

By SAS criterion,

âˆ´ Î”PQM = Î”MNRÂ

By CPCT,Â

â‡’ PQ = RN â€¦â€¦â€¦â€¦â€¦â€¦â€¦(ii)

Now, AB/PQ = AC/PR = AD/PM

From equation (i) and (ii), we conclude,

â‡’ CE/RN = AC/PR = 2AD/2PM

We know,Â

2AD = AE and 2PM = PN.

â‡’ CE/RN = AC/PR = AE/PNÂ

By SSS similarity criterion,

âˆ´ Î”ACE ~ Î”PRNÂ

Thus,

âˆ 2 = âˆ 4

And, âˆ 1 = âˆ 3

âˆ´ âˆ 1 + âˆ 2 = âˆ 3 + âˆ 4

â‡’ âˆ A = âˆ P â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(iii)

Now, in Î”ABC and Î”PQR, we have

AB/PQ = AC/PR [Given]

From equation (iii), we have,Â

âˆ A = âˆ P

By SAS similarity criterion,

âˆ´ Î”ABC ~ Î”PQRÂ

### Question 15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution:

Given, Length of the vertical pole = 6m

Length of shadow of the tower = 28 m

Shadow of the pole = 4 m

Let us assume the height of tower to be h m.

In Î”ABC and Î”DEF,

âˆ C = âˆ E (By angular elevation of sum)

Since, the following angles are equivalent to 90Â°

âˆ B = âˆ FÂ

By AA similarity criterion, we have,

âˆ´ Î”ABC ~ Î”DEFÂ

Since, if two triangles are similar corresponding sides are proportional

âˆ´ AB/DF = BC/EFÂ

Substituting values,Â

âˆ´ 6/h = 4/28

â‡’h = (6Ã—28)/4

â‡’ h = 6 Ã— 7

â‡’ h = 42 m

Hence, the height of the tower specified is 42 m.

### Question 16. If AD and PM are medians of triangles ABC and PQR, respectively where Î”ABC ~ Î”PQR prove that AB/PQ = AD/PM.

Solution:

Given, Î”ABC ~ Î”PQR

Since, the corresponding sides of similar triangles are in proportion.

âˆ´ AB/PQ = AC/PR = BC/QRâ€¦â€¦â€¦â€¦â€¦(i)

And, âˆ A = âˆ P, âˆ B = âˆ Q, âˆ C = âˆ R â€¦â€¦â€¦â€¦.â€¦..(ii)

Since, AD and PM are the medians, they will divide their opposite sides correspondingly.

âˆ´ BD = BC/2 and QM = QR/2 â€¦â€¦â€¦â€¦.(iii)

From equations (i) and (iii), we obtain,

AB/PQ = BD/QM â€¦â€¦â€¦â€¦â€¦.(iv)

In Î”ABD and Î”PQM,

From equation (ii), we have

âˆ B = âˆ Q

From equation (iv), we have,

AB/PQ = BD/QM

By SAS similarity criterion, we have,

âˆ´ Î”ABD ~ Î”PQMÂ

That is, AB/PQ = BD/QM = AD/PM

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