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https://www.urbanpro.com/ncert-solutions/class-10/maths-triangles-exercise-6-3 # Learn Exercise 6.3 with Free Lessons & Tips

##### All Exercises - Chapter 6 - Triangles

If AD and PM are medians of triangles ABC and PQR, respectively where , prove that .

It is given that ΔABC ∼ ΔPQR

We know that the corresponding sides of similar triangles are in proportion.

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

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

From equations (1) and (3), we obtain

In ΔABD and ΔPQM,

∠B = ∠Q [Using equation (2)]

(From equation iv)

∴ ΔABD ∼ ΔPQM (By SAS similarity criterion)

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Comments

State which pairs of triangles in the given figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

(i) ∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

Therefore, ΔABC ∼ ΔPQR [By AAA similarity criterion]

(iii)The given triangles are not similar as the corresponding sides are not proportional.

(iv) In ? MNL and ? QPR,

v)The given triangles are not similar as the corresponding sides are not proportional.

(vi) In ΔDEF,

∠D +∠E +∠F = 180º

(Sum of the measures of the angles of a triangle is 180º.)

70º + 80º +∠F = 180º

∠F = 30º

Similarly, in ΔPQR,

∠P +∠Q +∠R = 180º

(Sum of the measures of the angles of a triangle is 180º.)

∠P + 80º +30º = 180º

∠P = 70º

In ΔDEF and ΔPQR,

∠D = ∠P (Each 70°)

∠E = ∠Q (Each 80°)

∠F = ∠R (Each 30°)

∴ ΔDEF ∼ ΔPQR [By AAA similarity criterion]

∠M = ∠Q = 70°

∴?MNL ~ ?QPR [By SAS similarity criterion]

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Comments In the figure, , and . Find and .

DOB is a straight line.

∴ ∠DOC + ∠COB = 180°

⇒ ∠DOC = 180° − 125°

= 55°

In ΔDOC,

∠DCO + ∠CDO + ∠DOC = 180°

(Sum of the measures of the angles of a triangle is 180º.)

⇒ ∠DCO + 70º + 55º = 180°

⇒ ∠DCO = 55°

It is given that ΔODC ∼ ΔOBA.

∴ ∠OAB = ∠ OCD [Corresponding angles are equal in similar triangles.]

⇒ ∠OAB = 55°

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Comments Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that .

n ΔDOC and ΔBOA,

∠CDO = ∠ABO [Alternate interior angles as AB || CD]

∠DCO = ∠BAO [Alternate interior angles as AB || CD]

∠DOC = ∠BOA [Vertically opposite angles]

∴ ΔDOC ∼ ΔBOA [AAA similarity criterion]

(Corresponding sides are proportional)

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Comments In the figure, and . Show that .

In ΔPQR, ∠PQR = ∠PRQ

∴ PQ = PR (i)

Given,

Using (i) we obtain

using (ii)

(SAS similarity)

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Comments S and T are points on sides PR and QR of such that . Show that .

In ΔRPQ and ΔRST,

∠RTS = ∠QPS (Given)

∠R = ∠R (Common angle)

∴ ΔRPQ ∼ ΔRTS (By AA similarity criterion)

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Comments In the figure, if , show that .

It is given that ΔABE ≅ ΔACD.

∴ AB = AC [By CPCT] (1)

And, AD = AE [By CPCT] (2)

In ΔADE and ΔABC,

Dividing (2) by (1)

∠A = ∠A [Common angle]

∴ ΔADE ∼ ΔABC [By SAS similarity criterion]

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Comments In the figure, altitudes AD and CE of intersect each other at the point P.

Show that:

(i) AEP ~ Ã¢?? CDP

(ii) Ã¢??ABD ~ Ã¢?? CBE

(iii) Ã¢??AEP ~ Ã¢??ADB

(iv) Ã¢?? PDC ~ Ã¢?? BEC

(i)

In ΔAEP and ΔCDP,

∠AEP = ∠CDP (Each 90°)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by using AA similarity criterion,

ΔAEP ∼ ΔCDP

(ii)

In ΔABD and ΔCBE,

∠ADB = ∠CEB (Each 90°)

∠ABD = ∠CBE (Common)

Hence, by using AA similarity criterion,

ΔABD ∼ ΔCBE

(iii)

In ΔAEP and ΔADB,

∠AEP = ∠ADB (Each 90°)

∠PAE = ∠DAB (Common)

Hence, by using AA similarity criterion,

ΔAEP ∼ ΔADB

(iv)

In ΔPDC and ΔBEC,

∠PDC = ∠BEC (Each 90°)

∠PCD = ∠BCE (Common angle)

Hence, by using AA similarity criterion,

ΔPDC ∼ ΔBEC

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Comments E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that .

n ΔABE and ΔCFB,

∠A = ∠C (Opposite angles of a parallelogram)

∠AEB = ∠CBF (Alternate interior angles as AE || BC)

∴ ΔABE ∼ ΔCFB (By AA similarity criterion)

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Comments In the given figure, ABC and AMP are two right triangles, right angled at B and M respectively.

Prove that:

(i)

(ii)

(i) In ΔABC and ΔAMP,

∠ABC = ∠AMP (Each 90°)

∠A = ∠A (Common)

∴ ΔABC ∼ ΔAMP (By AA similarity criterion)

Corresponding sides of similar triangles are proportional.

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Comments CD and GH are respectively the bisectors of and such that D and H lie on sides AB and FE of and respectively. If , show that:

(i)

(ii)

(iii)

It is given that ΔABC ∼ ΔFEG.

∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE

∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle bisector)

And, ∠DCB = ∠HGE (Angle bisector)

In ΔACD and ΔFGH,

∠A = ∠F (Proved above)

∠ACD = ∠FGH (Proved above)

∴ ΔACD ∼ ΔFGH (By AA similarity criterion)

In ΔDCB and ΔHGE,

∠DCB = ∠HGE (Proved above)

∠B = ∠E (Proved above)

∴ ΔDCB ∼ ΔHGE (By AA similarity criterion)

In ΔDCA and ΔHGF,

∠ACD = ∠FGH (Proved above)

∠A = ∠F (Proved above)

∴ ΔDCA ∼ ΔHGF (By AA similarity criterion)

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Comments In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If and , prove that .

It is given that ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ABD = ∠ECF

In ΔABD and ΔECF,

∠ADB = ∠EFC (Each 90°)

∠ABD = ∠ECF (Proved above)

∴ ΔABD ∼ ΔECF (By using AA similarity criterion)

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Comments Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of (see the figure). Show that

Median divides the opposite side.

and

Given that,

In ΔABD and ΔPQM,

∴ ΔABD ∼ ΔPQM (By SSS similarity criterion)

⇒ ∠ABD = ∠PQM (Corresponding angles of similar triangles)

In ΔABC and ΔPQR,

∠ABD = ∠PQM (Proved above)

∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)

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Comments D is a point on the side BC of a triangle ABC such that . Show that .

n ΔADC and ΔBAC,

∠ADC = ∠BAC (Given)

∠ACD = ∠BCA (Common angle)

∴ ΔADC ∼ ΔBAC (By AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.

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Comments 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 .

Given that,

Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join B to E, C to E, Q to L, and R to L.

We know that medians divide opposite sides.

Therefore, BD = DC and QM = MR

Also, AD = DE (By construction)

And, PM = ML (By construction)

In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.

Therefore, quadrilateral ABEC is a parallelogram.

∴ AC = BE and AB = EC (Opposite sides of a parallelogram are equal)

Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR

It was given that

∴ ΔABE ∼ ΔPQL (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal.

∴ ∠BAE = ∠QPL … (1)

Similarly, it can be proved that ΔAEC ∼ ΔPLR and

∠CAE = ∠RPL … (2)

Adding equation (1) and (2), we obtain

∠BAE + ∠CAE = ∠QPL + ∠RPL

⇒ ∠CAB = ∠RPQ … (3)

In ΔABC and ΔPQR,

since

∠CAB = ∠RPQ [Using equation (3)]

∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)

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

Let AB and CD be a tower and a pole respectively.

Let the shadow of BE and DF be the shadow of AB and CD respectively.

At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.

Therefore, ∠DCF = ∠BAE

And, ∠DFC = ∠BEA

∠CDF = ∠ABE (Tower and pole are vertical to the ground)

∴ ΔABE ∼ ΔCDF (AAA similarity criterion)

Therefore, the height of the tower will be 42 metres.

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Comments Exercise 6.3

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