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Question 1:

If AD and PM are medians of triangles ABC and PQR, respectively where $\triangle ABC\sim \triangle PQR$ , prove that $\frac{AB}{PQ}=\frac{AD}{PM}$ .

Solution 1:

It is given that ΔABC ∼ ΔPQR

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

$\therefore \frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}........(i)$

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

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

$\therefore BD=\frac{BC}{2}, QM=\frac{QR}{2}..........(iii)$

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

$\frac{AB}{PQ}=\frac{BD}{QM}.......(iv)$

In ΔABD and ΔPQM,

∠B = ∠Q [Using equation (2)]

$\frac{AB}{PQ}=\frac{BD}{QM}.......(iv)$ (From equation iv)

∴ ΔABD ∼ ΔPQM (By SAS similarity criterion)

Comments

Tasneem

Question 2:

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 :

Solution 2:

(i) ∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

Therefore, ΔABC ∼ ΔPQR [By AAA similarity criterion]

$(ii)\frac{AB}{QR}=\frac{BC}{RP}=\frac{CA}{PQ}$

$\therefore \Delta ABC\sim \Delta QRP$

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

(iv) In ? MNL and ? QPR,

$\frac{MN}{QP}=\frac{ML}{QR}=\frac{1}{2}$

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]

$\frac{MN}{QP} = \frac{ML}{QR} = \frac{1}{2} ∠ M = ∠ Q = 70 ° ∴ ∆ MNL ~ ∆ QPR [By SAS similarity criterion]$

Comments

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Question 3:

In the figure, $\bigtriangleup ODC\sim \bigtriangleup OBA$ , $\angle BOC=125^{\circ}$ and $\angle CDO=70^{\circ}$ . Find $\angle DOC,\angle DCO$ and $\angle OAB$ .

Solution 3:

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°

Comments

Tasneem

Question 4:

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 $\frac{OA}{OC}=\frac{OB}{OD}$ .

Solution 4:

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]

$\therefore \frac{DO}{BO}=\frac{OC}{OA}$ (Corresponding sides are proportional)

$\therefore \frac{OA}{OC}=\frac{OB}{OD}$

Comments

Tasneem

Question 5:

In the figure, $\frac{QR}{QS}=\frac{QT}{PR}$ and $\angle 1=\angle 2$ . Show that $\triangle PQS\sim \triangle TQR$ .

Solution 5:

In ΔPQR, ∠PQR = ∠PRQ

∴ PQ = PR (i)

Given,

$\frac{QR}{QS}=\frac{QT}{PR}$

Using (i) we obtain

$\frac{QR}{QS}=\frac{QT}{QP}...........(ii)$

$In \Delta PQS and \Delta QTR$

$\frac{QR}{QS}=\frac{QT}{QP}$ using (ii)

$\angle Q=\angle Q$

$\therefore \Delta PQS\sim \Delta TQR$ (SAS similarity)

Comments

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Question 6:

S and T are points on sides PR and QR of $\triangle PQR$ such that $\angle P=\angle RTS$ . Show that $\triangle RPQ\sim\triangle RTS$ .

Solution 6:

In ΔRPQ and ΔRST,

∠RTS = ∠QPS (Given)

∠R = ∠R (Common angle)

∴ ΔRPQ ∼ ΔRTS (By AA similarity criterion)

Comments

Tasneem

Question 7:

In the figure, if $\triangle ABE\cong \triangle ACD$ , show that $\triangle ADE\cong \triangle ABC$ .

Solution 7:

It is given that ΔABE ≅ ΔACD.

∴ AB = AC [By CPCT] (1)

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

In ΔADE and ΔABC,

$\frac{AD}{AB}=\frac{AE}{AC}$ Dividing (2) by (1)

∠A = ∠A [Common angle]

∴ ΔADE ∼ ΔABC [By SAS similarity criterion]

Comments

Tasneem

Question 8:

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $\triangle ABE\sim \triangle CFB$ .

Solution 8:

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)

Comments

Tasneem

Question 9:

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

Prove that:

(i) $\triangle ABC\sim \triangle AMP$

(ii) $\frac{CA}{PA}=\frac{BC}{MP}$

Solution 9:

(i) In ΔABC and ΔAMP,

∠ABC = ∠AMP (Each 90°)

∠A = ∠A (Common)

∴ ΔABC ∼ ΔAMP (By AA similarity criterion)

$\Rightarrow \frac{CA}{PA}=\frac{BC}{MP}$ Corresponding sides of similar triangles are proportional.

Comments

Tasneem

Question 10:

CD and GH are respectively the bisectors of $\angle ACB$ and $\angle EGF$ such that D and H lie on sides AB and FE of $\triangle ABC$ and $\triangle EFG$ respectively. If $\triangle ABC\sim \triangle FEG$ , show that:

(i) $\frac{CD}{GH}=\frac{AC}{FG}$

(ii) $\triangle DCB\sim \triangle HGE$

(iii) $\triangle DCA\sim \triangle HGF$

Solution 10:

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)

$\Rightarrow \frac{CD}{GH}= \frac{AC}{FG}$

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)

Comments

Tasneem

Question 11:

In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If $AD \perp BC$ and $EF\perp AC$ , prove that $\triangle ABD\sim \triangle ECF$ .

Solution 11:

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)

Comments

Tasneem

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 $\triangle PQR$ (see the figure). Show that $\triangle ABC\sim \triangle PQR$

Solution 12:

Median divides the opposite side.

$\therefore BD=\frac{BC}{2}$ and $QM=\frac{QR}{2}$

Given that,

$\frac{AB}{PQ}=\frac{\frac{1}{2}BC}{\frac{1}{2}QR}=\frac{AD}{PM}$

$\Rightarrow \frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$

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)

$\frac{AB}{PQ}=\frac{BC}{QR}$

∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)

Comments

Tasneem

Question 13:

D is a point on the side BC of a triangle ABC such that $\angle ADC=\angle BAC$ . Show that $CA^{2}=CB.CD$ .

Solution 13:

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.

$\therefore \frac{CA}{CB}=\frac{CD}{CA}$

$CA^{2}=CB\times CD$

Comments

Tasneem

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 $\triangle ABC\sim \triangle PQR$ .

Solution 14:

Given that,

$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$

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

$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$

$\frac{AB}{PQ}=\frac{BE}{QL}=\frac{2AD}{2PM}$

$\frac{AB}{PQ}=\frac{BE}{QL}=\frac{AE}{PL}$

∴ 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 $\frac{AB}{PQ}=\frac{AC}{PR}$

∠CAB = ∠RPQ [Using equation (3)]

∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)

Comments

Tasneem

Question 15:

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.

Solution 15:

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)

$\frac{AB}{CD}=\frac{BE}{DF}$

$\frac{AB}{6}=\frac{28}{4}$

$AB=42m$

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

Comments

Tasneem

Question 16:

In the figure, altitudes AD and CE of $\triangle ABC$ 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

Solution 16:

(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

Comments

Tasneem

All Exercises - Chapter 6 - Triangles

Exercise 6.3

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Learn Exercise 6.3 with Free Lessons & Tips

All Exercises - Chapter 6 - Triangles

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