# Exercise 6.1

**1. Fill in the blanks using the correct word given in brackets:**

#### (i) All circles are _______________. (congruent, similar)

#### (ii) All squares are _______________. (similar, congruent)

#### (iii) All _______________ triangles are similar. (isosceles, equilateral)

#### (iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are _______________ and (b) their corresponding sides are _______________. (equal, proportional)

**Answer:**

#### (i) similar

#### (ii) similar

#### (iii) equilateral

#### (iv) equal, proportional

**2. Give two different examples of pair of:**

#### (i) similar figures

#### (ii) non-similar figures

**Answer:**

#### (i) Two different examples of pair of similar figures are:

#### (a) Any two rectangles

#### (b) Any two squares

#### (ii) Two different examples of pair of non-similar figures are:

#### (a) A scalene and an equilateral triangle

#### (b) An equilateral triangle and a right angled triangle

**3. State whether the following quadrilaterals are similar or not:**

**Answer:**

#### On looking at the given figures of the quadrilaterals, we can say that they are not similar because their angles are not equal.

# EXERCISE 6.2

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

**Answer:**

#### (i) Since DE || BC,

#### Let, EC = x cm

#### It is given that DE || BC

#### By using basic proportionality theorem, we obtain

#### x = 2

#### EC = 2 cm.

#### (ii) Let AD = x cm.

#### It is given that DE || BC.

#### By using basic proportionality theorem, we obtain

#### x = 2.4

#### AD = 2.4cm.

**2. E and F are points on the sides PQ and PR respectively of a ∆ PQR. For each of the following cases, state whether EF II QR:**

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

**Answer:**

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

#### Hence,

#### Therefore, EF is not parallel to QR.

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

#### Therefore,EF is parallel to QR.

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

#### And ER = PR – PF = 2.56 – 0.36 = 2.20 cm

#### Hence,

#### Therefore, EF is parallel to QR.

**3. In figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD.**

**Answer:**

#### In figure, LM || CB

#### By using Basic Proportionality theorem

#### And in ∆ ACD, LN || CD

#### ……i

#### Similalry, in ∆ ACD, LN || CD

#### …….ii

#### By using Basic Proportionality theorem

#### From eq. (i) and (ii), we have

**4. In figure, DE || AC and DF || AE. Prove that BF/GE = BE/EC .**

**Answer:**

#### In ∆ BCA, DE || AC

#### [Basic Proportionality theorem] ……….(i)

#### And in ∆ BEA, DF || AE

#### [Basic Proportionality theorem] ……….(ii)

#### From eq. (i) and (ii), we have

**5. In figure, DE || OQ and DF || OR. Show that EF || QR.**

**Answer:**

#### In ∆ PQO, DE || OQ

#### [Basic Proportionality theorem] ……….(i)

#### And in ∆ POR, DF || OR

#### [Basic Proportionality theorem] ……….(ii)

#### From eq. (i) and (ii), we have

#### Therefore, EF || QR [By the converse of Basic Proportionality Theorem]

**6. In figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.**

**Answer:**

#### And in ∆ POQ, AB || PQ

#### [Basic Proportionality theorem] ……….(i)

#### And in ∆ OPR, AC || PR

#### [Basic Proportionality theorem] ……….(ii)

#### From eq. (i) and (ii), we have

#### Therefore, BC || QR (By the converse of Basic Proportionality Theorem)

**7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).**

**Answer:**

#### Consider the given figure in which l is a line drawn through the mid-point P of line segment AB meeting AC at Q, such that PQ || BC

#### By using Basic Proportionality theorem, we obtain,

#### (P is the midpoint of AB ∴ AP = PB)

#### ⇒ AQ = QC

#### Or, Q is the mid-point of AC.

**8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).**

**Answer:**

#### Given: A triangle ABC, in which P and Q are the mid-points of

#### sides AB and AC respectively.

#### Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively.

#### i.e., AP = PB and AQ = QC

#### It can be observed that

#### and

#### Therefore,

#### Hence, by using basic proportionality theorem, we obtain

#### PQ || BC.

**9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO .**

**Answer:**

#### Given: A trapezium ABCD, in which AB || DC and its diagonals

#### AC and BD intersect each other at O.

#### Draw a line EF through point O, such that EF || CD

#### In ΔADC, EO || CD

#### By using basic proportionality theorem, we obtain

#### …i

#### In ΔABD, OE || AB

#### So, by using basic proportionality theorem, we obtain

#### ⇒ …ii

#### From eq. (i) and (ii), we get

#### ⇒

**10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO Such that ABCD is a trapezium.**

**Answer:**

#### Given: A quadrilateral ABCD, in which its diagonals AC and

#### BD intersect each other at O such that , i.e.

#### Quadrilateral ABCD is a trapezium.

#### Construction: Through O, draw OE || AB meeting AD at E.

#### In ∆ ADB, we have OE || AB [By construction] By Basic Proportionality theorem

#### …..i

#### However, it is given that

#### …..ii

#### From eq. (i) and (ii), we get

#### ⇒ EO || DC [By the converse of basic proportionality theorem]

#### ⇒ AB || OE || DC

#### ⇒ AB || CD

#### ∴ ABCD is a trapezium.

# EXERCISE 6.3

**1. State which pairs of triangles in figure, 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:**

**Answer:**

#### (i) In ΔABC and ΔPQR, we have

∠A = ∠P = 60° (Given)

∠B = ∠Q = 80° (Given)

∠C = ∠R = 40° (Given)

∴ ΔABC ~ ΔPQR (AAA similarity criterion)

(ii) In ΔABC and ΔPQR, we have

AB/QR = BC/RP = CA/PQ

∴ ΔABC ~ ΔQRP (SSS similarity criterion)

(iii) In ΔLMP and ΔDEF, we have

LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6

MP/DE = 2/4 = 1/2

PL/DF = 3/6 = 1/2

LM/EF= 2.7/5 = 27/50

Here, MP/DE = PL/DF ≠ LM/EF

Hence, ΔLMP and ΔDEF are not similar.

(iv) In ΔMNL and ΔQPR, we have

MN/QP = ML/QR = 1/2

∠M = ∠Q = 70°

∴ ΔMNL ~ ΔQPR (SAS similarity criterion)

(v) In ΔABC and ΔDEF, we have

AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°

Here, AB/DF = 2.5/5 = 1/2

And, BC/EF = 3/6 = 1/2

⇒ ∠B ≠ ∠F

Hence, ΔABC and ΔDEF are not similar.

(vi) In ΔDEF,we have

∠D + ∠E + ∠F = 180° (sum of angles of a triangle)

⇒ 70° + 80° + ∠F = 180°

⇒ ∠F = 180° – 70° – 80°

⇒ ∠F = 30°

In PQR, we have

∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)

⇒ ∠P + 80° + 30° = 180°

⇒ ∠P = 180° – 80° -30°

⇒ ∠P = 70°

In ΔDEF and ΔPQR, we have

∠D = ∠P = 70°

∠F = ∠Q = 80°

∠F = ∠R = 30°

Hence, ΔDEF ~ ΔPQR (AAA similarity criterion)

**2. In the figure, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.**

**Answer:**

#### 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°.

**3. 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 AO/OC = OB/OD**

**Answer:**

#### Given: ABCD is a trapezium in which AB DC.

#### In 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]

#### Hence,

**4. In the figure, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.**

**Answer:**

#### We have,

#### In ΔPQR, ∠PQR = ∠PRQ

#### ∴ PQ = PR …..(i)

#### Given,

#### Using (i), we get,

#### In ΔPQS & In ΔTQR

#### ∠Q = ∠Q

#### ∴ ΔPQS ~ ΔTQR [SAS similarity criterion]

**5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.**

**Answer:**

#### In ΔRPQ and ΔRST,

∠RTS = ∠QPS (Given)

∠R = ∠R (Common angle)

∴ ΔRPQ ~ ΔRTS (By AA similarity criterion)

**6. In the figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.**

**Answer:**

#### It is given that ΔABE ≅ ΔACD.

∴ AB = AC [By cpct] …(i)

And, AD = AE [By cpct] …(ii)

In ΔADE and ΔABC,

AD/AB = AE/AC [Dividing equation (ii) by (i)]

∠A = ∠A [Common angle]

∴ ΔADE ~ ΔABC [By SAS similarity criterion]

**7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:**

(i) ΔAEP ~ ΔCDP

(ii) ΔABD ~ ΔCBE

(iii) ΔAEP ~ ΔADB

(iv) ΔPDC ~ ΔBEC

**Answer:**

#### (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

**8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.**

**Answer:**

#### In ΔABE and ΔCFB,we have,

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

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

#### ∴By AA-criterion of similarity, we have

#### ∴ ΔABE ~ ΔCFB (By AA similarity criterion)

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

(i) ΔABC ~ ΔAMP

(ii) CA/PA = BC/MP

**Answer:**

#### (i) In Δ ABC and AMP, we have,

#### ∠ABC = ∠AMP =90^{0} [Given]

#### ∠BAC = ∠MAP [Common angles]

#### ∴ Δ ABC ~ Δ AMP [By AA-criterion of similarity, we have]

#### ⇒ CA/PA = BC/MP ….. (Corresponding sides of similar trianlges are proportional)

**10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:**

(i) CD/GH = AC/FG

(ii) ΔDCB ~ ΔHGE

(iii) ΔDCA ~ ΔHGF

**Answer:**

#### We have, Δ ABC ~ Δ FEG

#### ∴ ∠A = ∠F, ∠B = ∠E, & ∠ACB = ∠FGE

#### ∴ ∠ACD = ∠FGH (Angle Bisector)

#### And ∠DCB = ∠HGE (Angle Bisector)

#### In ΔACD & ΔFGH

#### ∠A = ∠F (Proved above)

#### ∠ACD = ∠FGH (Proved above)

#### ∴ Δ ACD ~ Δ FGH [By AA similarity criterion]

#### ⇒ [(CD)/ (GH)] = [(AC) / (FG)]

#### In ΔDCB & ΔHGE,

#### ∠DCB= ∠HGE (Proved above)

#### ∠B= ∠E (Proved above)

#### ∴ Δ DCB~ Δ HGE [By AA similarity criterion]

#### In ΔDCA & ΔHGF,

#### ∠ACD = ∠FGH (Proved above)

#### ∠A = ∠F (Proved above)

#### ∴ Δ DCA~ Δ HGF [By AA similarity criterion]

**11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.**

**Answer:**

#### Here Δ ABC is isosceles with AB = AC

#### ∠B = ∠C

#### In Δ ABD and ECF, we have

#### ∠ABD = ∠ECF[Each 90°]

#### ∠ABD = ∠ECF = [Proved above]

#### By AA-criterion of similarity, we have

#### Δ ABD ~ Δ ECF

**12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see figure). Show that ΔABC ~ ΔPQR.**

**Answer:**

#### Given: AD is the median of Δ ABC and PM is the median of Δ PQR such that

#### Median divides the opposite side.

#### ∴ BD = BC / 2 [Given]

#### And QM = QR / 2 [Given]

#### Given that,

#### AB/PQ = BC/QR = AD/PM

#### ⇒ AB/PQ =[( ½BC) / (½QR) ]= AD/PM

#### ⇒ AB/PQ = BD/QM = AD/PM

#### In ΔABD and ΔPQM,

#### AB/PQ = BD/QM = AD/PM [ Proved above]

#### ∴ ΔABD ∼ ΔPQM (By SSS similarity criterion)

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

#### In ΔABC and ΔPQR,

#### ∠ABD = ∠PQM (Proved above)

#### AB / PQ = BC / QR

#### ∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)

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

^{2}= CB.CD

**Answer:**

#### In triangles ABC and DAC,

∠ADC = ∠BAC (Given)

∠ACD = ∠BCA (Common angle)

∴ ΔADC ~ ΔBAC (By AA similarity criterion)

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

∴ CA/CB =CD/CA

⇒ CA^{2} = CB x CD.

**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 ΔABC ~ ΔPQR.**

**Answer:**

#### Given: AD is the median of Δ ABC and PM is the median of Δ PQR such that

#### ⇒AB / PQ = AC / PR = 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

#### ⇒AB / PQ = AC / PR = AD / PM

#### ⇒AB / PQ = BE / QL = [(2AD) / (2PM)]

#### ⇒AB / PQ = BE / QL = 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,

#### AB / PQ = AC / PR

#### ∠CAB = ∠RPQ [Using equation (3)]

#### ∴ ΔABC ∼ ΔPQR (By SAS similarity criterion)

**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.**

**Answer:**

#### Let AB the vertical pole and AC be its shadow. Also, let DE be the vertical tower and DF be its shadow. Joined BC and EF.

#### Length of the vertical pole = 6m (Given)

#### Shadow of the pole = 4 m (Given)

#### Let Height of tower = h m

#### Length of shadow of the tower = 28 m (Given)

#### In ΔABC and ΔDEF,

#### ∠C = ∠E (angular elevation of sum)

#### ∠B = ∠F = 90°

#### ∴ ΔABC ~ ΔDEF (By AA similarity criterion)

#### ∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)

#### ∴ 6/h = 4/28

#### ⇒ h = 6 x 28/4

#### ⇒ h = 6 x 7

#### ⇒ h = 42 m

#### Hence, the height of the tower is 42 meters.

**16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.**

**Answer:**

#### Given: AD and PM are the medians of triangles

#### ABC and PQR respectively, where

#### It is given that Δ ABC ~ Δ PQR

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

#### ∴ AB / PQ = AC / AD and BC / QR ….(1)

#### Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

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

#### ∴ BD = BC/2 & QM = QR/2 …(3)

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

#### AB/PQ = BD/QM … (4)

#### In ΔABD and ΔPQM,

#### ∠B = ∠Q [Using equation (2)]

#### AB/PQ = BD/QM

#### ∴ ΔABD ∼ ΔPQM (By SAS similarity criterion)

#### AB/PQ = BD/QM = AD/PM

# EXERCISE 6.4

**1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm**^{2} and 121 cm^{2}. If EF = 15.4 cm, find BC.

^{2}and 121 cm

^{2}. If EF = 15.4 cm, find BC.

**Answer:**

#### It is given that,

Area of ΔABC = 64 cm^{2}

Area of ΔDEF = 121 cm^{2}

EF = 15.4 cm

and, ΔABC ~ ΔDEF

∴ Area of ΔABC/Area of ΔDEF = AB^{2}/DE^{2}

= AC2/DF^{2} = BC^{2}/EF^{2} …(i)

[If two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides]

∴ 64/121 = BC^{2}/EF^{2}

⇒ (8/11)^{2} = (BC/15.4)^{2}

⇒ 8/11 = BC/15.4

⇒ BC = 8×15.4/11

⇒ BC = 8 × 1.4

⇒ BC = 11.2 cm

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

**Answer:**

#### ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.

In ΔAOB and ΔCOD, we have

∠1 = ∠2 (Alternate angles)

∠3 = ∠4 (Alternate angles)

∠5 = ∠6 (Vertically opposite angle)

∴ ΔAOB ~ ΔCOD [By AAA similarity criterion]

Now, Area of (ΔAOB)/Area of (ΔCOD)

= AB^{2}/CD^{2} [If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides]

= (2CD)^{2}/CD^{2} [∴ AB = CD]

∴ Area of (ΔAOB)/Area of (ΔCOD)

= 4CD^{2}/CD = 4/1

Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1

**3. In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.**

**Answer:**

#### Let us draw two perpendiculars AP and DM on line BC.

#### We know that area of a triangle = 1/2 x Base x Height

#### ∴

#### In ΔAPO and ΔDMO,

#### ∠APO = ∠DMO (Each = 90°)

#### ∠AOP = ∠DOM (Vertically opposite angles)

#### ∴ ΔAPO ∼ ΔDMO (By AA similarity criterion)

#### ∴ AP/DM = AO/DO

#### ⇒

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

**Answer:**

#### Let us assume two similar triangle as ΔABC ~ ΔPQR

#### Given that, ar(ΔABC) = ar(ΔABC)

#### Putting this value in equation (1) we obtain

#### 1 =

#### ⇒ AB = PQ, BC = QR and AC = PR

#### ∴ ΔABC ≅ ΔPQR (By SSS congruence criterion)

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

**Answer:**

#### D and E are the mid-points of ΔABC

#### :.DE || AC and DE = 1/2 AC

#### In ΔBED and ΔBCA

#### ∠BED = ∠BCA (Corresponding angle)

#### ∠BDE = ∠BAC (Corresponding angle)

#### ∠EBD = ∠CBA (Common angles)

#### ∴ΔBED ~ ΔBCA (AAA similarity criterion)

#### ⇒

#### Similary

#### ⇒

#### Also ar(ΔDEF) = ar(ΔABC) – [ar(ΔBED) + ar(ΔCFE) + ar(ΔADF)]

#### ⇒

#### ⇒

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

**Answer:**

#### Let us assume two similar triangles as ΔABC ∼ ΔPQR. Let AD and PS be the medians of these triangles.

#### ∵ ΔABC ∼ ΔPQR

#### :.(AB)/(PQ) = (BC)/(QR) = (AC)/(PR)…(1)

#### ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)

#### Since AD and PS are medians,

#### ∴ BD = DC = BC/2

#### And, QS = SR = QR/2

#### Equation (1) becomes

#### (AB)/(PQ) = (BD)/(QS) = (AC)/(PR) ….(3)

#### In ΔABD and ΔPQS,

#### ∠B = ∠Q [Using equation (2)]

#### and (AB)/(PQ) = (BD)/(QS) [Using equation (3)]

#### ∴ ΔABD ∼ ΔPQS (SAS similarity criterion)

#### Therefore, it can be said that

#### AB/PQ = BD/QS = AD/PS ….(4)

#### From equations (1) and (4), we may find that

#### AB/PQ = BC/QR = AC/PR = AD/PS

#### And hence,

**7. 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 the diagonals.**

Tick the correct answer and justify:

Tick the correct answer and justify:

**Answer:**

#### Let ABCD be a square of side a.

#### Therefore, its diagonal = √2a

#### Two desired equilateral triangles are formed as ΔABE and ΔDBF.

#### Side of an equilateral triangle, ΔABE, described on one of its sides = a

#### Side of an equilateral triangle, ΔDBF, described on one of its diagonals =√2a

#### We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

**8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is:**

#### (A) 2: 1

#### (B) 1: 2

#### (C) 4: 1

#### (D) 1: 4

**Answer:**

#### We know that equilateral triangles have all its angles as 60º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.

#### Let side of ΔABC = x

#### Therefore, side of ΔBDE = x/2

#### Hence, (C) is the correct answer.

**9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio:**

#### (A) 2: 3

#### (B) 4: 9

#### (C) 81: 16

#### (D) 16: 81

**Answer:**

#### If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides of these triangles.

#### It is given that the sides are in the ratio 4:9.

#### Therefore, ratio between areas of these triangles = (4/9)² = 16/81

#### Hence,(D) is the correct answer.

# Exercise 6.5

**1. 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

**Answer:**

(i) Given that the sides of the triangle are 7 cm, 24 cm, and 25 cm.

Squaring the lengths of these sides, we will get 49, 576, and 625.

49 + 576 = 625

(7)^{2} + (24)^{2} = (25)^{2}

The sides of the given triangle are satisfying Pythagoras theorem.Hence, it is right angled triangle.

Length of Hypotenuse = 25 cm

(ii) Given that the sides of the triangle are 3 cm, 8 cm, and 6 cm.

Squaring the lengths of these sides, we will get 9, 64, and 36.

However, 9 + 36 ≠ 64

Or, 3^{2} + 6^{2} ≠ 8^{2}

Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle is not satisfying Pythagoras theorem.

(iii) Given that sides are 50 cm, 80 cm, and 100 cm.

Squaring the lengths of these sides, we will get 2500, 6400, and 10000.

However, 2500 + 6400 ≠ 10000

Or, 50^{2 }+ 80^{2} ≠ 100^{2}

Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle is not satisfying Pythagoras theorem.

Hence, it is not a right triangle.

(iv) Given that sides are 13 cm, 12 cm, and 5 cm.

Squaring the lengths of these sides, we will get 169, 144, and 25.

Clearly, 144 +25 = 169

Or, 12^{2} + 5^{2 }= 13^{2}

The sides of the given triangle are satisfying Pythagoras theorem.

Therefore, it is a right triangle.

Length of the hypotenuse of this triangle is 13 cm.

**2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM ^{2} = QM × MR.**

**Answer:**

#### Given: ΔPQR is right angled at P is a point on QR such that PM ⊥QR.

**To prove:**. PM^{2} = QM × MR

**Proof:** In ΔPQM, we have

PQ^{2} = PM^{2} + QM^{2 }[By Pythagoras theorem]

Or, PM^{2 }= PQ^{2 }– QM^{2} …(i)

In ΔPMR, we have

PR^{2} = PM^{2} + MR^{2} [By Pythagoras theorem]

Or, PM^{2} = PR^{2} – MR^{2} …(ii)

Adding (i) and (ii), we get

2PM^{2} = (PQ^{2} + PM2) – (QM^{2} + MR^{2})

= QR^{2 }– QM^{2 }– MR^{2 } [∴ QR^{2} = PQ^{2} + PR^{2}]

= (QM + MR)^{2} – QM^{2} – MR^{2}

= 2QM × MR

∴ PM^{2} = QM × MR

**3. In figure, ABD is a triangle right angled at A and AC **⊥** BD. Show that**

(i) AB^{2 }= BC × BD

(ii) AC^{2 }= BC × DC

(iii) AD^{2} = BD × CD

**Answer:**

#### Given: ABD is a triangle right angled at A and AC **⊥** BD.

#### (i) In ΔADB and ΔCAB, we have

#### ∠DAB = ∠ACB (Each equals to 90°)

#### ∠ABD = ∠CBA (Common angle)

#### ∴ ΔADB ~ ΔCAB [AA similarity criterion]

#### ⇒ AB/CB = BD/AB

#### ⇒ AB² = CB × BD

#### (ii) Let ∠CAB = x

#### In ΔCBA,

#### ∠CBA = 180° – 90° – x

#### ∠CBA = 90° – x

#### Similarly, in ΔCAD

#### ∠CAD = 90° – ∠CBA

#### = 90° – x

#### ∠CDA = 180° – 90° – (90° – x)

#### ∠CDA = x

#### In ΔCBA and ΔCAD, we have

#### ∠CBA = ∠CAD

#### ∠CAB = ∠CDA

#### ∠ACB = ∠DCA (Each equals to 90°)

#### ∴ ΔCBA ~ ΔCAD [By AAA similarity criterion]

#### ⇒ AC/DC = BC/AC

#### ⇒ AC² = DC × BC

#### (iii) In ΔDCA and ΔDAB, we have

#### ∠DCA = ∠DAB (Each equals to 90°)

#### ∠CDA = ∠ADB (common angle)

#### ∴ ΔDCA ~ ΔDAB [By AA similarity criterion]

#### ⇒ DC/DA = DA/DA

#### ⇒ AD² = BD × CD

**4. ABC is an isosceles triangle right angled at C. Prove that AB**^{2} = 2AC^{2}.

^{2}= 2AC

^{2}.

**Answer:**

#### Since ABC is an isosceles right triangle, right angled at C.

#### ∴ AC = CB

#### Applying Pythagoras theorem in ΔABC (i.e., right-angled at point C), we obtain

#### AC² + CB² = AB²

#### ⇒ AC² + AC² = AB² ( AC = CB)

#### ⇒ 2 AC² = AB²

**5. ABC is an isosceles triangle with AC = BC. If AB**^{2} = 2AC^{2}, prove that ABC is a right triangle.

^{2}= 2AC

^{2}, prove that ABC is a right triangle.

**Answer:**

Given that ΔABC is an isosceles triangle having AC = BC and AB^{2} = 2AC^{2}

In ΔACB,

AC = BC (Given)

AB^{2} = 2AC^{2} (Given)

AB2 = AC^{2} + AC^{2}

= AC^{2} + BC^{2} [Since, AC = BC]

Hence, By Pythagoras theorem ΔABC is right angle triangle.

**6. ABC is an equilateral triangle of side 2a. Find each of its altitudes.**

**Answer:**

#### ABC is an equilateral triangle of side 2a.

Draw, AD ⊥ BC

In ΔADB and ΔADC, we have

AB = AC [Given]

AD = AD [Given]

∠ADB = ∠ADC [equal to 90°]

Therefore, ΔADB ≅ ΔADC by RHS congruence.

Hence, BD = DC [by CPCT]

In right angled ΔADB,

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

(2a)^{2} = AD^{2 }+ a^{2}

⇒ AD^{2} = 4a^{2} – a^{2}

⇒ AD^{2} = 3a^{2}

⇒ AD = √3a

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

**Answer:**

#### Let the diagonals AC and BD of rhombus ABCD intersect each other at O. Since the diagonals of a rhombus bisect each other at right angles.

#### In ΔAOB, ΔBOC, ΔCOD, ΔAOD,

#### Applying Pythagoras theorem, we obtain

#### AB² = AO² + OB² ….(1)

#### BC² = BO² + OC² …. (2)

#### CD² = CO² + OD² …..(3)

#### AD² = AO² + OD² ……(4)

#### Adding all these equation we obtain

#### AB² + BC² + CD² + AD² = 2(AO² + OB² + OC² + OD²)

#### =

#### (Diagonals bisect each other)

#### =

#### = (AC)² + (BD)²

**8. In Fig. O is a point in the interior of a triangle**

ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

(i) OA^{2} + OB^{2 }+ OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} + BD^{2} + CE^{2} ,

(ii) AF^{2} + BD^{2} + CE^{2} = AE^{2 }+ CD² + BF².

ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that

**Answer:**

#### Applying Pythagoras theorem in ΔAOF, we obtain

#### OA² = OF² + AF²

#### Similarly, in ΔBOD

#### OB² = OD² + BD²

#### Similarly, in ΔCOE

#### OC² = OE² + EC²

#### Adding these equations,

#### OA² + OB² + OC² = OF² + AF² + OD² + BD² + OE² + EC²

#### OA² + OB² + OC² – OF² – OD² – OE² = AF² + BD² + EC²

#### From above result

#### AF² + BD² + EC² = (OA²- OE² ) + (OC² – OD²) + (OB² – OF² )

#### ∴ AF² + BD² + EC² = AE² + CD² + BF².

**9. 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.**

**Answer:**

#### Let OA be the wall and AB be the ladder.

#### Therefore, by Pythagoras theorem,

#### AB² = OA² + BO²

#### (10 m)² =( 8 m)² + OB²

#### 100m² = 64m² + OB²

#### OB² = 36m²

#### OB = 6m

#### Therefore the distance of the foot of the ladder from the base of the wall is 6 m.

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

**Answer:**

#### Let AB (= 24m) be a guy wire attached to a vertical pole. BC of height 18 m. To keep the wire taut, let it be fixed to a stake at A. Then, ABC is a right triangle, right angled at O.

#### Let OB be the pole and AB be the wire.

#### By Pythagoras theorem,

#### AB² = OB² + OA²

#### (24 m)² = ( 18 m)² + OA²

#### 576m² = 324m² + OA²

#### OA² = (576 – 324)m² = 25m²

#### OA² = √252 m = √6 x 6 x 7 = 6√7 m.

#### OA =6√7 m.

#### Hence, the stake may be placed at distance of 6√7 m from the base of the pole.

**11. An aeroplane leaves an airport and flies due north at a speed of 1000 km pwe 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 hours?**

**Answer:**

#### Let the first aeroplane starts from O and goes upto A towards north where

#### Distance travelled by the plane flying towards north in hrs

#### Similarly, distance travelled by the plane flying towards west inhrs

#### Let these distances be represented by OA and OB respectively.

#### Applying Pythagoras theorem,

#### Distance between these planes after

#### = √[(1500)² + (1800)²]

#### = √(5490000)

#### Therefore, the distance between these planes will be 300√61 km.

**12. 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.**

**Answer:**

#### Let CD and AB be the poles of height 11 m and 6 m.

#### Therefore, CP = 11 − 6 = 5 m.

#### From the figure, it can be observed that AP = 12m

#### Applying Pythagoras theorem for ΔAPC, we obtain

#### AP² + PC² = AC²

#### (12m)² + (5m)² = AC²

#### AC² = (144 + 25)m² = 169m²

#### AC = 13m

#### Therefore, the distance between their tops is 13 m.

**13. 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}.

^{2}+ BD

^{2 }= AB

^{2}+ DE

^{2}.

**Answer:**

#### Applying Pythagoras theorem in ΔACE, we get

AC^{2 }+ CE^{2} = AE^{2} ….(i)

Applying Pythagoras theorem in ΔBCD, we get

BC^{2} + CD^{2} = BD^{2} ….(ii)

Using equations (i) and (ii), we get

AC^{2} + CE^{2} + BC^{2} + CD^{2} = AE^{2} + BD^{2} …(iii)

Applying Pythagoras theorem in ΔCDE, we get

DE^{2} = CD^{2} + CE^{2}

Applying Pythagoras theorem in ΔABC, we get

AB^{2} = AC^{2} + CB^{2}

Putting these values in equation (iii), we get

DE^{2} + AB^{2} = AE^{2} + BD^{2}.

**14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Fig.). Prove that 2AB**^{2} = 2AC^{2} + BC^{2}.

^{2}= 2AC

^{2}+ BC

^{2}.

**Answer:**

#### Given that in ΔABC, we have

AD ⊥BC and BD = 3CD

In right angle triangles ADB and ADC, we have

AB^{2} = AD^{2} + BD^{2} …(i)

AC^{2} = AD^{2} + DC^{2} …(ii) [By Pythagoras theorem]

Subtracting equation (ii) from equation (i), we get

AB^{2} – AC^{2} = BD^{2} – DC^{2}

= 9CD^{2} – CD^{2} [∴ BD = 3CD]

= 9CD^{2} = 8(BC/4)^{2} [Since, BC = DB + CD = 3CD + CD = 4CD]

Therefore, AB^{2} – AC^{2} = BC^{2}/2

⇒ 2(AB^{2} – AC^{2}) = BC^{2}

⇒ 2AB^{2} – 2AC^{2} = BC^{2}

∴ 2AB^{2 }= 2AC^{2 }+ BC^{2}.

**15. In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD**^{2} = 7AB^{2}.

^{2}= 7AB

^{2}.

**Answer:**

#### Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

#### ∴ BE = EC = BC/2 = a/2

#### and AE= a√3/2

#### Given that, BD = 1/3 BC

#### ∴ BD = a/3

#### DE = BE – BD = a/2 – a/3 = a/6

#### Applying Pythagoras theorem in ΔADE, we get

#### AD^{2} = AE^{2} + DE^{2}

#### AD² = (a√3/2)² + (a/6)²

#### AD² = (3a²/4) + (a²/36)

#### ⇒ AD² = (28a²/36)

#### ⇒ AD² = 7/9AB²

#### ⇒ 9AD² = 7AB²

**16. 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.**

**Answer:**

#### Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.

#### ∴ BE = EC = BC/2 = a/2

#### Applying Pythagoras theorem in ΔABE, we obtain

#### AB² = AE² + BE²

#### a² = AE² + (a/2)²

#### AE² = a² – a²/4

#### AE² = 3a²/4

#### 4 AE² = 3a²

#### ⇒ 4 × (Square of altitude) = 3 × (Square of one side).

**17. Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.**

The angle B is:

(A) 120°

(B) 60°

(C) 90°

(D) 45°

**Answer:**

#### Given that, AB = 6√3 cm, AC = 12 cm, and BC = 6 cm

We can observe that

AB^{2} = 108

AC^{2} = 144

And, BC^{2} = 36

AB^{2} + BC^{2} = AC^{2}

The given triangle, ΔABC, is satisfying Pythagoras theorem.

Therefore, the triangle is a right triangle, right-angled at B.

∴ ∠B = 90°

Hence, the correct option is (C).

# EXERCISE 6.6

**1. In figure, PS is the bisector of ∠QPR of Δ PQR. Prove that QS/SR = PQ/PR .**

**Answer:**

**Given:**

#### Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.

#### Given that, PS is the angle bisector of ∠QPR.

#### ∠QPS = ∠SPR … (1)

#### By construction,

#### ∠SPR = ∠PRT (As PS || TR) … (2)

#### ∠QPS = ∠QTR (As PS || TR) … (3)

#### Using these equations, we obtain

#### ∠PRT = ∠QTR

#### ∴ PT = PR

#### By construction,

#### PS || TR

#### By using basic proportionality theorem for ΔQTR,

#### ⇒ QS/SR = QP/PT

#### ⇒ QS/SR = PQ/PR (∴PT = PR)

**2. In figure, D is a point on hypotenuse AC of Δ ABC, BD ⊥ AC, DM ⊥ BC and DN ⊥AB. Prove that:**

#### (i) DM^{2}= DN.MC

#### (ii) DN^{2}= DM.AN

**Answer: **(i) Let us join DB

#### We have, DN || CB, DM || AB, and ∠B = 90°

#### ∴ DMBN is a rectangle.

#### ∴ DN = MB and DM = NB

#### The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC.

#### ∴ ∠CDB = 90°

#### ⇒ ∠2 + ∠3 = 90° … (1)

#### In ΔCDM,

#### ∠1 + ∠2 + ∠DMC = 180°

#### ⇒ ∠1 + ∠2 = 90° … (2)

#### In ΔDMB,

#### ∠3 + ∠DMB + ∠4 = 180°

#### ⇒ ∠3 + ∠4 = 90° … (3)

#### From equation (1) and (2), we obtain

#### ∠1 = ∠3

#### From equation (1) and (3), we obtain

#### ∠2 = ∠4

#### In ΔDCM and ΔBDM,

#### ∠1 = ∠3 (Proved above)

#### ∠2 = ∠4 (Proved above)

#### ∴ ΔDCM ∼ ΔBDM (AA similarity criterion)

#### ⇒ (BM)/(DM) = (DM)/(MC)

#### ⇒ (DN)/(DM) = (DM)/(MC) ∴ (BM = DN)

#### ⇒ DM² = DN × MC

#### (ii) In right triangle DBN,

#### ∠5 + ∠7 = 90° … (4)

#### In right triangle DAN,

#### ∠6 + ∠8 = 90° … (5)

#### D is the foot of the perpendicular drawn from B to AC.

#### ∴ ∠ADB = 90°

#### ⇒ ∠5 + ∠6 = 90° … (6)

#### From equation (4) and (6), we obtain

#### ∠6 = ∠7

#### From equation (5) and (6), we obtain

#### ∠8 = ∠5

#### In ΔDNA and ΔBND,

#### ∠6 = ∠7 (Proved above)

#### ∠8 = ∠5 (Proved above)

#### ∴ ΔDNA ∼ ΔBND (AA similarity criterion)

#### => AN/DN = DN/NB

#### ⇒ DN² = AN × NB

#### ⇒DN² = AN × DM (As NB = DM)

**3. In figure, ABC is a triangle in which ∠ABC > 90**^{0} and AD ⊥ CB produced. Prove that:

^{0}and AD ⊥ CB produced. Prove that:

#### AC^{2} = AB^{2} + BC.BD

**Answer**

#### Applying Pythagoras theorem in ΔADB, we obtain

#### AB² = AD² + DB² …(1)

#### Applying Pythagoras theorem in ΔACD, we obtain

#### AC² = AD² + DC²

#### AC² = AD² + (DB + BC)²

#### AC² = AD² + DB² + BC² + 2DB x BC

#### AC² = AB² + BC² + 2DB x BC [Using equation (1)]

**4. In figure, ABC is a triangle in which ∠ABC < 90**^{0} and AD ⊥ BC produced. Prove that: AC^{2} =AB^{2} + BC^{2} – 2BC.BD

^{0}and AD ⊥ BC produced. Prove that: AC

^{2}=AB

^{2}+ BC

^{2}– 2BC.BD