# Question-1

**Give two different examples of pair of (i) similar figures, (ii) non-similar figures.****Solution:**

**Examples of similar figures:**

i) Two square of sides 4 cm and 8 cm each.

ii) Two circles with the same centre and diameters 7.5 cm and 10 cm each.

**Examples of non similar figures:**

i) A square and a rhombus.

ii) A circle and a triangle.

# Question-2

**State whether the following quadrilaterals are similar or not**

**Solution:**

The figures are not similar as they are not equiangular.

# Question-3

**In the figure (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).****Solution:**

(i)

** **

In Î”ABC __ __

DE is parallel to BC

By Basic Proportionality Theorem

------------------- (1)

**Given: **AD = 1.5 cm, DB = 3 cm, AE = 1 cm

Let EC = â€˜xâ€™ cm

Applying in (1)

1.5x = 3

x =

x = 2 cm

EC = 2cm

(ii)**Given: **DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm

Since DE || BC, using BPT

â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (1)

Let AD be = x

sub. in (1)

x =

=

âˆ´ AD = 2.4 cm

# Question-4

**E and F are points on the sides PQ and PR respectively of a Î” PQR. For each of the following cases, state whether EF || QR**

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

**Solution:**

(i) PE = 3.9 cm, EQ = 3 cm

PF = 3.6 cm, FR = 2.4 cm

=

=

âˆ´

EF is not parallel to QR by convene of BDT

(ii) PE= 4cm QE = 4.5cm PF = 8 RF = 9cm

=

âˆ´

EF || QR by convene of BPT.

(iii) PQ = 1.28 cm , PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm

EQ = PQ â€“ PE

= 1.28 â€“ 0.18

= 1.10

FR = PR â€“ PF

= 2.56 â€“ 0.36

= 2.20

â‡’

â‡’ EF is parallel to QR

# Question-5

**In the figure, if LM || CB and LN || CD, prove that****Solution:**

**Given:**LM || CB and LN || CD

**To prove:**

**Proof:**

In Î” ABC

LM || CB using basic proportionality Theorem

âˆ´ â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)

Also in Î” ADC

âˆ´ Theorem â€¦â€¦â€¦â€¦â€¦.. (2)

from (1) and (2)

Hence proved .

# Question-6

**In the figure, DE || AC and DF || AE. Prove that****Solution:**

**Given:**ABC is a triangle and DE || AC and DF is parallel to AE

**To prove:**

**Proof**

In Î” ABC,

DE || AC (given)

(By BPT) â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)

In Î” AEB,

âˆ´ DF || AE

(By BPT) â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (2)

Comparing equation (1) and equation (2)

Hence proved.

# Question-7

**In the figure, DE || OQ and DF || OR. Show that**

EF || QR.

EF || QR.

**Solution:**

**Given:**DE || OQ and DF || OR

**To prove:**EF || QR

**Proof:**

In Î” POQ

By using BPT

â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (1)

In Î” POR

â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(2)

By comparing equations (1) and (2)

By using inverse of BPT

EF || QR

Hence proved.

# Question-8

**A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR.**

Show that BC || QR.

Show that BC || QR.

**Solution:**

**Given:**A, B and C are points on OP, OQ and OR respectively such that AB || PQ, AC || PR

**Proof:**In Î” OPQ

AB || PQ (Given)

âˆ´ (By using BPT) â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (1)

In Î” OPR

Since AC || PR

(By using BPT) â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (2)

By comparing (1) and (2)

**âˆ´**BC || QR, By using converse of BPT,

Hence proved.

# Question-9

**Using Theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.****Solution:**

**Given:**ABCD is a trapezium and the diagonals AC and BD intersect at O.

**To prove:**The ratio

**Construction :**Draw OM || AB meeting BC at â€˜Mâ€™

**Proof:**In Î” ACB, OM || AB

âˆ´ By using BPT â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (1)

||

^{ly}In Î” BDC

OM || CD [ âˆ´(OM || AB AND AB || CD â‡’ OM || CD)]

âˆ´ using BPT

Taking the reciprocal

â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(2)

from (1) and (2)

(or)

Hence proved.

# Question-10

**The diagonals of a quadrilateral ABCD intersect each other at the point O such that . Show that ABCD is a trapezium.****Solution:**

**Given:**ABCD is a quadrilateral and the diagonals AC and BD intersect of â€˜Oâ€™ such that

**To prove:**ABCD is a trapezium

**Construction:**Draw OM || AB

**Proof:**

In Î” ACB

âˆ´ OM || AB (By construction)

â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (1)

**Given:**

(or)

sub = in equation (1)

âˆ´ =

or =

In Î” BDC by converse of BPT

OM || DC. But OM || AB

â‡’ AB || CD

â‡’ ABCD is a trapezium.

# Question-11

**State which pairs of triangles in Fig. 6.34 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:**

AAA Î” ABC âˆ¼ Î” PQR

(ii) Similar

SSS, Î” ABC âˆ¼ Î” QRP

(iii) Not similar

(iv) Not smilar

(v) Similar, SAS Î” ABC âˆ¼ Î” FGE

(vi)similar, AAA Î” DEF âˆ¼ Î” PQR

# Question-12

**Î”ODC ~ Î” OBA, âˆ BOC = 125Â° and âˆ CDO = 70Â°. Find âˆ DOC, âˆ DCO and âˆ OAB.**

**Solution:**

**Given:**Î” ODC âˆ¼ Î” OBA

**âˆ**BOC= 125Â°

**âˆ**CDO = 70Â°

**To find**

**âˆ**DOC,

**âˆ**DCO and

**âˆ**OAB

**âˆ**DOC= 180 â€“ 125 = 55

**âˆ**DCO = 125 â€“ 70 = 55

**âˆ**OAB = 55 ( Q Î” ODC âˆ¼ Î” OBA).

# Question-13

**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****Solution:**

**Given:**ABCD is a trapezium with AB || CD and the diagonals AC and BD intersect at â€˜Oâ€™.

**To prove:**=

**Proof:**In the figure consider the triangle OAB and OCD

âˆ DOC = âˆ AOB (Vertically opposite angles are equal)

âˆ DCO = âˆ OAB (Alternate angles are equal)

âˆ´ By AA corollary of similar triangles.

âˆ´ Î” OAB âˆ¼ Î” OCD When the two triangle are similar, the side are proportionally.

â‡’ =

Hence proved.

# Question-14

**In the figure, = and âˆ 1 = âˆ 2. Show that Î” PQS ~ Î” TQR.****Solution:**

Given: =

and

**âˆ**1 =

**âˆ**2

To prove: Î” PQS âˆ¼ Î” TQR

Proof

In Î” PQR

**âˆ**1 =

**âˆ**2

**âˆ**PQR =

**âˆ**PRQ

**PQ = PR**

=>

=>

=

=

= (âˆ´ QP = PR proved above )

**âˆ**1

**is a common angle**

By SAS similarity

D PQS ~ D TQR

Hence proved.

# Question-15

**S and T are points on sides PR and QR of Î” PQR such that âˆ P = âˆ RTS. Show that Î” RPQ ~ Î” RTS.****Solution:**

**Given:**S and T are the points on sides. PR and QR of a. Î” PQR such that âˆ P = âˆ RTS

**To prove:**Î” RPQ âˆ¼ Î” RTS

**Proof:**

In Î” PQR and Î” TRS

**âˆ**QRP

**=**

**âˆ**TRS

**(common angle)**

**âˆ**STR

**=**

**âˆ**QPR

**(given)**

âˆ´ By AA similarity

Î” RPQ âˆ¼ Î” RTS

Hence proved.

# Question-16

**In the figure, if Î”ABE â‰… Î” ACD, show that Î” ADE ~ Î” ABC.****Solution:**

**Given:**Î” ABE â‰… Î” ACD

**To prove**

**:**Î” ADE ~ Î” ABC

**Prove:**Î” ABE â‰… Î” ACD

âˆ´ AB = AC â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (1) (Since corresponding parts of triangles are equal)

and AD = AE â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (2)

and

**âˆ**A is a common angle

âˆ´

**âˆ**DAE =

**âˆ**BAC

By SAS similarity condition

Î” ADE âˆ¼ Î” ABC.

# Question-17

**In the Figure, altitudes AD and CE of Î” ABC intersect each other at the point P. Show that:**

(i) Î” AEP ~ Î” CDP

(ii) Î” ABD ~ Î” CBE

(iii) Î” AEP ~ Î” ADB

(iv) Î” PDC ~ Î” BEC.

(i) Î” AEP ~ Î” CDP

(ii) Î” ABD ~ Î” CBE

(iii) Î” AEP ~ Î” ADB

(iv) Î” PDC ~ Î” BEC.

**Solution:**

Given: Î” ABC in which the altitudes AD and CE intersect at P.

To prove:

(i) Î” AEP âˆ¼ Î” CDP (ii) Î” ABD âˆ¼ Î” CBE

(iii) Î” AEP âˆ¼ Î” ADB (iv) Î” PDC âˆ¼ Î” BEC

Proof:

(i) Î” AEP and Î” CDP

**âˆ**APE =

**âˆ**CPD (Vertically opposite angle and equal)

**âˆ**AEP =

**âˆ**CDP (Both are equal to 90Â° since CE and AD are altitudes)

âˆ´ BY AA similarity

Î” AEP âˆ¼ Î” CDP

Hence proved.

(ii) To prove: Î” ABD âˆ¼ Î” CBE

Proof: In Î” ABD and Î” CBE

**âˆ**CBE =

**âˆ**ABD (Common angle)

**âˆ**ADB =

**âˆ**CEB (Both are equal to 90Â°)

âˆ´ By AA similarity

Î” ABD âˆ¼ D CBE

(iii) To prove: Î” AEP âˆ¼ ADB

Proof: In Î” AEP and Î” ADB

**âˆ**PAE =

**âˆ**DAB (Common angle)

**âˆ**AEP =

**âˆ**ADB (Both are equal to 90Â°)

â‡’ By AA similarity

Î” AEP âˆ¼ Î” ADB

(iv) To prove: Î” PDC âˆ¼ Î” BEC

Proof: In Î” PDC and Î” BEC

**âˆ**PCD =

**âˆ**ECB (Common angle)

**âˆ**PDC =

**âˆ**BEC (Both are equal to 90Â°)

Î” PDC âˆ¼ D BEC (AA similarity).

# Question-18

**In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:**

(i) Î” ABC ~ Î” AMP

(ii)(i) Î” ABC ~ Î” AMP

(ii)

**Solution:**

(i) To prove: Î” ABC âˆ¼ Î” AMP

In Î” ABC and Î” AMP

**âˆ**ABC =

**âˆ**AMP (Both are equal to 90Â°)

**âˆ**BAC =

**âˆ**PAM(common angle)

âˆ´ BY AA corollary

& Î” ABC âˆ¼ AMP,

(ii) To prove:

Proof:

As proved above

Î” ABC âˆ¼ Î” AMP

Since the corresponding sides of the similar triangles are proportional

â‡’

Hence proved.

# Question-19

**CD and GH are respectively the bisectors of âˆ ACB and âˆ EGF such that D and H lie on sides AB and FE of Î” ABC and Î” EFG respectively. If Î” ABC ~ Î” FEG, show that:**

(i)

(ii) Î” DCB ~ Î” HGE

(iii) Î” DCA ~ Î” HGF

(i)

(ii) Î” DCB ~ Î” HGE

(iii) Î” DCA ~ Î” HGF

**Solution:**

**Given:**Î” ABC âˆ¼ Î” FEG and CD and GH are bisectors of the

**âˆ**ACB and

**âˆ**EGF of the Î” ACB and Î” EGF respectively.

**To prove:**

(i)

(ii) Î” DCB âˆ¼ Î” HGE

(iii) Î” DCA âˆ¼ Î” HGF

**Proof:**(i) Consider Î” ADC and Î” FGH

**âˆ**BAC =

**âˆ**EFG (Q D ABC ~ D FEG) is given)

(or)

**âˆ**DAC =

**âˆ**HFG

Also

**âˆ**BCA =

**âˆ**EGF (Q D ABC ~ D FEG is given)

Ã— by

**âˆ**BCA =

**âˆ**EGF

=>

**âˆ**DCA =

**âˆ**HGF

âˆ´ By AA corollary

Î” DAC âˆ¼ Î” HFG

âˆ´

(ii)

**To prove:**Î” DCB âˆ¼ Î” HGE

In Î” DBC and Î” HEG

**âˆ**ABC =

**âˆ**FEG [QÎ” ABC âˆ¼ Î” FEG]

(or)

**âˆ**DBC =

**âˆ**HEG [ Q D and H lie on the line AB and EF respectively]

**âˆ**BCA =

**âˆ**EGF â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(1) [Q Î” ABC âˆ¼ Î” FEG]

multiplying by

**âˆ**BCA =

**âˆ**EGF

(or)

**âˆ**DCB =

**âˆ**HGE â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(2) [Q CD and GH are the bisectors of the

**âˆ**ACB and

**âˆ**EGF of the D ACB and D EGF respectively]

from (1) and (2)

Î” DCB âˆ¼ Î” HGE [By AA corollary]

(iii) Already proved in (i)

In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD perpendicular to BC and EF perpendicular to AC prove that D ABD ~ D ECF.

**Given:**D ABC is isosceles with AB = AC. E is a point as CB such that EF perpendicular to AC. AD is perpendicular to BC.

**To prove:**D ABD ~ D ECF

**Proof:**

D ABC is isosceles with AB = AC (Given)

\

**âˆ**ABC =

**âˆ**ACB â€¦â€¦â€¦â€¦â€¦â€¦. (1) Property of an isosceles triangle

and

**âˆ**ADB =

**âˆ**EFC â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(2) Both are equal to 90Â°

Now, in D ABD and D ECF

**âˆ**ABD =

**âˆ**ECF From equation (1)

and

**âˆ**ADB =

**âˆ**EFC From equation (2)

**\**D

**ABD ~ D ECF.**

# Question-20

**In the figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD perpendicular to BC****and EF perpendicular to AC. Prove that Î” ABD ~ Î” ECF.****Solution:**

**Given:**ABC is an isosceles triangle,

â‡’ AB = AC

â‡’

**âˆ**B =

**âˆ**C

â‡’

**âˆ**ABD =

**âˆ**ECF

In Î” ABD and Î” ECF

**âˆ**ADB =

**âˆ**EFC (Each 90Â°)

**âˆ**ABD =

**âˆ**ECF (Proved above)

â‡’

**âˆ**BAD =

**âˆ**CEF

âˆ´ Î” ABD ~ Î” ECF. (By using AA similarity criterion)

# Question-21

**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:**

In Î” ABC and PQR

**âˆ**ACB =

**âˆ**PRQ ( The shadow is cast of the since from the angle made by the sunâ€™s ray is same )

**âˆ**ABC =

**âˆ**PQR

**(**

**âˆµ**both are equal to 90Â°)

â‡’ Î” ABC âˆ¼ Î” PQR (By AA corollary)

âˆ´ =

or x = = 42

Height of the tower = 42 m.

# Question-22

**If AD and PM are medians of triangles ABC and PQR, respectively where Î” ABC ~ Î” PQR, prove that**

**Solution:**

**Given:**Î” ABC âˆ¼ Î” PQR

**To prove:**

**Proof:**

In Î” ABC âˆ¼ Î” PQR

âˆ´ = â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(1)

and

**âˆ**B =

**âˆ**Q â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(2)

replacing BC = 2BD and QR = 2QM in equation (1)

â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(3)

from (2) and (3)

Î” ABD âˆ¼ Î” PQM By SAS similarly

âˆ´

Hence proved.

# Question-23

**Let Î” ABC ~ Î” DEF and their areas be, respectively, 64 cm**^{2}and 121 cm^{2}. If EF = 15.4cm, find BC.**Solution:**

Since Î” ABC âˆ¼ Î” DEF

= (By Theorem 6.6)

(or) =

=

BC = Ã— 15.4

BC = 11.2 cm.

# Question-24

**Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2 CD, find the ratio of the areas of triangles AOB and COD.**

**Solution:**

In Î” AOB and COD

**âˆ**COD =

**âˆ**AOB

**(vertically opposite angles)**

**âˆ**CDO =

**âˆ**OBA (Alternate interior angles)

By AA similarity

Î” COD âˆ¼ Î” AOB

==

âˆ´ =

# Question-25

**In the figure ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that**

**Solution:**

**Construction:**From A and D draw perpendiculars to BC meeting BC at E and F respectively.

**Proof:**

In Î” AEO and Î” DFO

**âˆ**AEO =

**âˆ**DFO = 90Â°

**âˆ**AOE =

**âˆ**DOF (vertically opposite angle)

â‡’ Î” AOE âˆ¼ Î” DOF (By AA similarity)

âˆ´ = = â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦. (1)

âˆ´ =

Hence proved.

# Question-26

**If the areas of two similar triangles are equal, prove that they are congruent.**

**Solution:**

Let ABC and DEF be two similar triangles where areas are equal.

(ie) area(Î” ABC) = area(Î” DEF)

= = =

since ar. Î” ABC = ar.Î” DEF

= = 1

âˆ´ = 1 or AB

^{2}= DE

^{2}

â‡’ AB = DF

||ly BC = EF and AC = DF

By SSS congruency

Î” ABC â‰… Î” DEF

Here the areas of two similar triangles are equal, those two triangles are congruent.

# Question-27

**D, E and F are respectively the mid-points of sides AB, BC and CA of Î” ABC. Find the ratio of the areas of Î”**

**DEF and**Î”**ABC.****Solution:**

Since D and F are the midpoints of the AB and AC, DE || AF or DF || BE

Similarly EF || AB or EF || DB

AFED is a parallelogram as both pairs of opposite sides are parallel.

âˆ´ By the property of parallelogram

âˆ DBE = âˆ DFE = (In an parallelogram opposite angle are equal)

or âˆ DFE = âˆ ABC -------------------(1)

Similarly âˆ FEB = âˆ ACB â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦..(2)

In Î” DEF and Î” ABC, from

Equation (1) and (2)

Î” DEF âˆ¼ Î” CAB

âˆ´ = =

âˆ´ ar Î” DEF: ar Î” CAB = 1 : 4 .

# Question-28

**Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.****Solution:**

**Given:**AM and DN are medians of the Î” ABC and Î” DEF and Î” ABC âˆ¼ Î” DEF

**To prove:**

=

**Proof:**since Î” ABC âˆ¼ Î” DEF

or

and

( QÎ” ABC âˆ¼ Î” DEF)

âˆ´ By SAS similarity

Î” ABM âˆ¼ Î” DEN

or â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦... (1)

=

= (From (1)

Hence proved.

# Question-29

**Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.****Solution:**

**Given:**ABCD is a square and two equilateral triangle are drawn with BC and DB as are units side respectively.

**To Prove:**ar Î” PBC =(ar Î”BDQ)

**Proof:**

Consider Î”PBC and Î” BQD

Since both these triangles are equilateral triangles, by the property of equilateral triangles,

both the triangles are equiangular with each angle equal to 60Â°.

â‡’ D PBC ~ D QBD

**âˆ´**= â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (1)

Consider D BCP

Since it is a right angled triangle

BC

^{2}+ DC

^{2}= BD

^{2}( By using Pythagoras theorem)

But BC = DC

2BC

^{2}= BD

^{2}

Substituting in (1)

=

**âˆ´**ar D PBC = ar D BDQ

Hence the results.

# Question-30

**Sides of triangles are given below. Determine which of them are right triangles.In case of a right triangle, write the length of its hypotenuse.**

(i) 7 cm, 24 cm, 25 cm

(ii) 3 cm, 8 cm, 6 cm

(iii) 50 cm, 80 cm, 100 cm

(iv) 13 cm, 12 cm, 5 cm

(i) 7 cm, 24 cm, 25 cm

(ii) 3 cm, 8 cm, 6 cm

(iii) 50 cm, 80 cm, 100 cm

(iv) 13 cm, 12 cm, 5 cm

**Solution:**

(i) 7

^{2}= 49, 24

^{2}= 576, 25

^{2}= 625

âˆ´ 576 + 49 = 625

Since the sum of the squares of the two sides of the triangle is equal to the square of the third side, it is a right triangle.

(ii) 3

^{2}= 9, 8

^{2}= 64, 6

^{2}= 36

from the above 36 + 9 â‰ 64

âˆ´ It is not right triangle.

(iii) 50

^{2}= 2500, 80

^{2}= 6400, 100

^{2}= 10000

since 2500 + 6400 = 8900

8900 â‰ 10000

It is not a right triangle.

(iv) 13

^{2}= 169, 12

^{2}= 144, 5

^{2}= 25, 144 + 25 = 169

â‡’ It is a right triangle.

# Question-31

**PQR is a triangle right angled at P and M is a point on QR such that PM perpendicular to QR. Show that PM**

^{2}**=**QM.MR**Solution:**

**To prove:** PM^{2} = QM Ã— MR

In Î” PMR

PR^{2} = PM^{2} + MR^{2} â€¦â€¦â€¦â€¦â€¦â€¦â€¦.(1)

Also In Î” PMQ

PQ^{2} = PM^{2} + QM^{2} â€¦â€¦â€¦â€¦â€¦.. (2)

(1) + (2)

PR^{2} + PQ^{2} = 2PM^{2} + MR^{2} + QM^{2}

QR^{2} = 2PM^{2} + MR^{2} + QM^{2}

QR^{2} â€“ MR^{2} = QM^{2} + 2PM^{2}

(QR + MR) (QR - MR) â€“ QM^{2} = 2PM^{2}

(QR + MR) QM â€“ QM^{2} = 2PM^{2}

QM( QR + MR â€“ QM) = 2PM^{2}

QM (MR +MR) = 2PM^{2}

QM Ã— 2MR = 2PM^{2}

âˆ´ PM^{2} = QM Ã— MR

# Question-32

**In the figure, ABD is a triangle right angled at A and AC perpendicular to BD. Show that**

(i) AB(i) AB

(ii) AC

(iii) AD^{2}= BC . BD(ii) AC

^{2}= BC . DC(iii) AD

^{2}= BD . CD**Solution:**

(i)

**To prove:**AB

^{2}= BC. BD

In Î” ABC and Î” ABD

âˆ ABC

**=**âˆ DBA

**(common angle)**

âˆ ACB

**=**âˆ BAD

**(equal to 90Â°)**

**âˆ´**Î” ABC âˆ¼ Î” DBA (By AA similarity)

**âˆ´**

â‡’ AB

^{2}= BC Ã— BD

(ii) In Î” ADC and Î” ABC

AD

^{2}= AC

^{2}+ CD

^{2}..........(1)

AB

^{2}= BC

^{2}+ AC

^{2 }............(2)

(1) + (2) â‡’

AD

^{2 }+ AB

^{2}= 2AC

^{2}+CD

^{2}+ BC

^{2}

BD

^{2 }=

^{ }2AC

^{2}+ CD

^{2}+BC

^{2}

âˆ´ BD

^{2 }- CD

^{2}- BC

^{2}= 2AC

^{2}

(BD + CD)(BD â€“ CD) â€“ BC

^{2}= 2AC

^{2}

(BD + CD) BC â€“ BC

^{2}= 2AC

^{2}

(BC)(BD + CD â€“ BC) = 2AC

^{2}

(BC)(CD + CD) = 2AC

^{2}(

**âˆµ**BD â€“ BC = CD)

**âˆ´**AC

^{2}= BCÃ—CD

(iii) In Î” ACD and D ABD

**âˆ**ADB

**=**âˆ ADC

**(common angle)**

âˆ BAD

**=**âˆ DAC

**(equal to 90Â°)**

Î” ABD âˆ¼ Î” ACD

âˆ´

â‡’ AD

^{2}= BD Ã— DC

# Question-33

**ABC is an isosceles triangle right angled at C. Prove that AB**

^{2}= 2AC^{2}.**Solution:**

In Î” ABC since âˆ ACB= 90Â°

AC^{2} + BC^{2} = AB^{2} (By using Pythagoras theorem)

AC^{2} + AC^{2} = AB^{2} ( Q AC = BC )

âˆ´ AB^{2} = 2AC^{2} (proved)

# Question-34

**ABC is an isosceles triangle with AC = BC. If AB**

^{2}= 2 AC^{2}, prove that ABC is a right triangle.**Solution:**

**Given:**AB

^{2}= 2AC

^{2}and AC = BC

**To prove:**ABC is a right triangle

**Proof:**

Given: AB

^{2}= 2AC

^{2}

(or) AB

^{2}= AC

^{2}+AC

^{2}â€¦â€¦â€¦â€¦â€¦â€¦(1)

since AC = BC given, equation (1) becomes

AB

^{2}= AC

^{2}+ BC

^{2}

âˆ´ By the converse of Pythagoras theorem Î” ABC is a right triangle with âˆ C = 90Â°.

# Question-35

**ABC is an equilateral triangle of side 2**

*a*. Find each of its altitudes.**Solution:**

**Given:**ABC is an equilateral triangle with each side equal to 2a

**Construction:**Draw AD âŠ¥ r to BC

**To find:**Length of AD

In Î” ADB and D ADC

âˆ ADB = âˆ ADC

**= 90Â°**

AC = AB (Q ABC is isosceles)

AD = AD (common side)

âˆ´ Î” ADB â‰… Î” ADC (By RHS property)

â‡’ BD = DC (CPCT)

BC = BD + DC

= BD + BD

2a = 2BD

BD =

*a*

In Î” ABD, AB

^{2}= AD

^{2}+ BD

^{2}

(2

*a*)

^{2}= AD

^{2}+

*a*

^{2}

4

*a*

^{2}= AD

^{2}+

*a*

^{2}

*a*

^{2}â€“ a

^{2}= AB

^{2}

3

*a*

^{2}= AD

^{2}

AD =

*a*

Similarly the other altitudes will also be

*a*only.

# Question-36

**Prove that the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.**

**Solution:**

**Given:**ABCD is a rhombus and the diagonals AC and BD intersect at O.

**To prove:**AB

^{2}+ BC

^{2}+ DC

^{2}+ AD

^{2}= AC

^{2}+ BD

^{2}

**Proof:**The diagonals of rhombus intersect at right angle.

âˆ´ âˆ AOB = 90Â°

In Î” AOB, AO

^{2}+ OB

^{2}= AB

^{2}

+ = AB

^{2}

Hence AC

^{2}+ BD

^{2}= 4AB

^{2}

Since 4AB

^{2}can be taken as sum of the square of the 4 sides of the rhombus as each side of the rhombus equal,

**âˆ´**The sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.

# Question-37

**In the figure O is a point in the interior of a triangle ABC, OD perpendicular to BC, OE perpendicular to AC and OF perpendicular to AB. Show that**

(i) OA

(ii) AF

(i) OA

^{2}+ OB^{2}+ OC^{2}â€“ OD^{2}â€“ OE^{2}â€“ OF^{2}= AF^{2}+ BD^{2}+ CE^{2},(ii) AF

^{2}+ BD^{2}+ CE^{2}= AE^{2}+ CD^{2}+ BF^{2}.**Solution:**

(i)

**Given :**In Î” ABC OD âŠ¥ BC, OE âŠ¥ AC and OF âŠ¥ AB

**Construction:**Join OC, OA and OB.

In Î” OBD, OB

^{2}= OD

^{2}+ BD

^{2}

In Î” OFA, OA

^{2}= OF

^{2}+ AF

^{2}

In Î” OCE, OC

^{2}= OE

^{2}+ CE

^{2}

âˆ´ OB

^{2}+ OA

^{2}+ OC

^{2}= OD

^{2}+ OF

^{2}+ OE

^{2}+ BD

^{2}+ AF

^{2}+ CE

^{2}

(or)

OB

^{2}+ OA

^{2}+ OC

^{2}â€“ OD

^{2}â€“ OF

^{2}â€“ OE

^{2}= BD

^{2}+ AF

^{2}+ CF

^{2}

(ii) From the result above

AF

^{2}+ BD

^{2}+ CF

^{2}= OB

^{2}+ OA

^{2}+ OC

^{2}

**â€“**OD

^{2}â€“ OF

^{2}â€“ OE

^{2}

= (OA

^{2}â€“ OE

^{2}) + (OC

^{2}â€“ OD

^{2}) +(OB

^{2}â€“ OF

^{2})

= AE

^{2}+ CD

^{2}+ BF

^{2}

# Question-38

**A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.**

**Solution:**

In the figure AC = ladder = 10 m

AB = Height of the window = 8 m

BC = Distance from the foot of the wall to the foot of the ladder = x m

ABC is a right angle triangle

AB

^{2}+ BC

^{2}= AC

^{2}

(or) 8

^{2}+ x

^{2}= 10

^{2}

x

^{2}+ 10

^{2}- 8

^{2}

x

^{2}= (10 + 8) (10 â€“ 8)

= (18)(2)

= 36 x = = 6 m

âˆ´ The distance of the foot of the ladder from the wall = 6 m

# Question-39

**A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?**

**Solution:**

In Î” ACB, AC = length of the wire = 24m and the largest of the tower = 18 m

18

^{2}+ x

^{2}= 24

^{2}(By using Pythagoras theorem)

x

^{2}+ 24

^{2}= 18

^{2}

x

^{2 }= 576 â€“ 324

= 252

x =

= 6m

Therefore, the stake should be driven to a distance of 6m so that the wire will be taut.

# Question-40

**An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. How far apart will be the two planes after 1hours?****Solution:**

Distance traveled by the aeroplane in the North directions is

AB = 1000 Â´ = 1500 km. (Distance = speed Ã— time)

And the Distance traveled by the aeroplane in the west direction is AC = 1200 Ã— = 1800 km

BC is the distance between the two aeroplanes

**âˆ´**AB

^{2}+ AC

^{2}= BC

^{2}

(1500)

^{2}+ (1800)

^{2}= BC

^{2}

2250000 + 3240000 = BC

^{2}

5490000 = BC

^{2}

BC = 100km.

# Question-41

**Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.****Solution:**

Let AD be a pole of height 11m and BC be a pole of height 6m

**Construction:**Join CD and draw âŠ¥ r from C to AD meeting AD at E.

**Proof:**ABCD is rectangle

âˆ´ AB = CE = 12 m

and BC = AE = 6 m

âˆ´ DE = AD â€“ AE = 11 â€“ 6 = 5 m

In Î” DEC

DE

^{2}+ EC

^{2}= DC

^{2}

or 5

^{2}+ 12

^{2}= DC

^{2}

25 + 144 = DC

^{2}

DC = = 13 m

**âˆ´**The distance between the two poles = 13 m.

# Question-42

**D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE**

^{2}+ BD^{2}= AB^{2}+ DE^{2}.**Solution:**

**To prove:**AE

^{2}+ BD

^{2}= AB

^{2}+ DE

^{2}

**Proof:**

In Î” AEC

AE

^{2}= EC

^{2}+ AC

^{2}â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)

In Î” BDC

BD

^{2}= BC

^{2}+ CD

^{2}â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (2)

(1) + (2)

AE

^{2}

_{ }+ BD

^{2}= EC

^{2}+ BC

^{2}+ AC

^{2}+ CD

^{2}

= (EC

^{2}+ CD

^{2}) + (AC

^{2}+ BC

^{2})

= DE

^{2}+ AB

^{2}

Hence proved.

# Question-43

**The perpendicular from A on side BC of a Î” ABC intersects BC at D such that DB = 3 CD. Prove that 2 AB**

^{2}**=**2 AC^{2}+ BC^{2}.**Solution:**

From Î”ABD,

AD

^{2}= AB

^{2}â€“ BD

^{2}...................(1)

From Î”ADC,

AD

^{2}= AC

^{2 }â€“ DC

^{2}......................(2)

From (1) and (2), AB

^{2}â€“ BD

^{2}= AC

^{2 }â€“ DC

^{2}.................(3)

BD = 3 DC (given)

âˆ´ BC = BD + DC = 4 DC

BC

^{2}= 16 DC

^{2}......................(4)

Now AB

^{2}= AC

^{2}+ BD

^{2}â€“ DC

^{2}

AB

^{2}= AC

^{2 }+ (3DC)

^{2}â€“ DC

^{2}

AB

^{2}= AC

^{2}+ 9DC

^{2}â€“ DC

^{2}

= AC

^{2}+ 8DC

^{2}

AB

^{2}= AC

^{2}+

= AC

^{2}+ BC

^{2}/2 [ from (4) ]

2AB

^{2}= 2AC

^{2}+ BC

^{2}

# Question-44

**In an equilateral triangle ABC, D is a point on side BC such that BD = BC. Prove that 9 AD**

^{2}= 7 AB^{2}.**Solution:**

**Given:**Î” ABC is an equilateral triangle. â€˜Dâ€™ is a point on BC such that BD = BC.

**To prove:**9AD

^{2}= 7AB

^{2}

**Construction:**Draw AE âŠ¥ r to BC meeting BC at E

**Proof:**

In an equilateral triangle the altitude and median are same line regret. Therefore AE is the median of the Î” ABC. E is the midpoint of BC

â‡’ BE + EC = BC

2EC = BC

BE = EC = BC â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)

BD = BC (Given)â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦(2)

DC = BC â€“ BD

= BC - BC = BC â€¦â€¦â€¦â€¦â€¦â€¦â€¦ (3)

DE = BE â€“ BD

= BC -

= BC

= = BC

In Î” ADE

Î” ADE

AD

^{2}= AE

^{2}+ ED

^{2}â€¦â€¦â€¦â€¦â€¦â€¦â€¦(1)

But in Î” AEC

AC

^{2}= AE

^{2}+ EC

^{2}

(or) AE

^{2}= AC

^{2}- EC

^{2}

substituting in (1)

AD

^{2}= AC

^{2}â€“ EC

^{2}+ ED

^{2}

= AC

^{2}â€“ DE

^{2}â€“ CE

^{2}

= AC

^{2}+ (DC + CE)(DE â€“ CE)

= AC

^{2}+ (DC)(DE â€“ CE)

= AC

^{2}+ DC.DE â€“ DC.CE

= AC

^{2}+

= AC

^{2}+

= AC

^{2}+

= AC

^{2}+

= AC

^{2}â€“

=

AD

^{2}= AB

^{2}

**âˆ´**9 AD

^{2}= 7 AB

^{2}.

# Question-45

**In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.**

**Solution:**

Let the side of the equilateral Î” ABC be â€˜aâ€™

Let AE be the altitude from A to BC meeting

BC at E.

E is the midpoint of C

Therefore BE = BC = a

In Î” ABE, AB

^{2}= AE

^{2}+ BE

^{2}

AE

^{2}= AB

^{2}â€“ BE

^{2}

= a

^{2}-

= a

^{2}-

= =

AE = â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦â€¦.. (1)

Now,

Three times, square of one of its side = 3a

^{2}

Four times the square of y

Its one of the altitude = 4 = 3a

^{2}

Therefore it is proved that in an equilateral triangle three times the square of one side is equal to four times the square of its altitude.

# Question-46

**Tick the correct answer and justify: In Î” ABC, AB = 6âˆš3 cm, AC = 12 cm and BC = 6 cm. The angle B is:**

(A) 120Â° (B) 60Â°

(C) 90Â° (D) 45Â°

(A) 120Â° (B) 60Â°

(C) 90Â° (D) 45Â°

**Solution:**

In Î” ABC, AB = 6cm, AC = 12 cm, BC = 6 cm

AB

^{2}= (6)

^{2}= 36 Ã— 3 = 108

BC

^{2}= 6

^{2}= 36

AC

^{2}= 12

^{2}= 144

AB

^{2}+ BC

^{2}= 108 + 36 = 144 = AC

^{2}âˆ´ AB

^{2}+ BC

^{2}= AC

^{2}

â‡’ By converse of pythagoras theorem, âˆ B = 90Â°

Answer = (c) 90Â°.