NCERT Solution for Triangles

(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).

Sol. (i) All circles are similar.

(ii) All squares are similar.

(iii) All equilateral triangles are similar.

(iv) Two polygons of the same number of sides are similar if:

(a) Their corresponding angles are equal and

(b) Their corresponding sides are proportional.

(i) similar figures (ii) non-similar figures.

Sol. (i) (a) Any two circles are similar figures.

(b) Any two squares are similar figures.

(ii) (a) A circle and a triangle are non-similar figures.

(b) An isosceles triangle and a scalene triangle are non-similar figures.

Sol. On observing the given figures, we find that:

Their corresponding sides are proportional but their corresponding angles are Dot equal.

∴ The given figures are not similar.

Triangles are a special type of polygons. The study of their similarity is important.

Two triangles are said to be similar if:

(i) Their corresponding sides are proportional, and,

(ii) Their corresponding angles are equal.

If a line is drawn parallel to one of the sides of a triangle to intersect the other two sides in distinct points ten the other two sides are divided in the same ratio. (CBSE 2010)

Given:

A ΔABC in which DE || BC and DE intersects AC and AB at h and D respectively.

To Prove:

[ART]

Join BE and ‘ CD.’

Draw EF ⊥ AB and DC ⊥ AC

EF ⊥ AB

EF is height of the ΔADE, corresponding to AD.

Since, ΔDBE and ΔECD being on the same base DE and between the same parallel DE and BC,

we have

ar(ΔDBE) = ar(ΔECD) ...(3)

From (1), (2) & (3), we have:

Since, AD and DB are parts of AB and whereas AE and EC are parts of AC,

∴ D and E divide the sides AB and AC in the same ratio.

Sol. (i) Since DE || BC

∴ Using the Basic proportionality Theorem,

(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, JR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Sol. (i) We have: PE = 3.9 cm, EQ = 3 cm

PF = 3.6 cm and FR = 2.4 cm

Sol. In ΔABC,

∵LM || CB [Given]

∴Using the Basic Proportionality Theorem, we have:

Sol. In ΔPQO

DE || OQ [Given]

∴ Using the Basic Proportionality Theorem, we have:

∴ E and F are two distinct points on PQ and PR respetively and E and F are dividing the two sides PQ and PR in the same ratio in ΔPQR.

∴ EF || QR

Sol. In ΔPQR,

O is a point and 0Q,- OR and joined. We have A, B and C on OP, OQ. and OR respectively such that AB || PQ and AC || OR.

Now, in OPQ,

AB || PQ [Given]

Sol. We Have a ΔABC in which D is the mid point of AB and E is a point on AC such that

DE. || BC.

DE || BC [Given]

∴ Using the Basic Proportionality Theorem, we get

⇒ E is the mid point of AC: Hence lii is proved tht “a line through the Mid-point of one side of a triangle parallel to another side bisects the third side.”

Sol. Given. A triangle ABE in which a line l intersects AB at D and AC at

E, such that:

Sol. We have, a trapezium ABCD such that AB || DC. The diago as AC and BD intersect each other at O.

Let us draw OE parallel toTeither AB or DC.

In ΔADC,

∵OE || DC [By construction]

∴ Using the Basic -Proportionality theorem, we get

Sol. We have a trapezium ABCD in which diagonals AC end BD intersect Ich other at O such that

∴ Using the Basic Proportionality Theorem, we get

i.e., the points 0 and E on the sides AB and AC (of ΔADB) respectively in the same ratio.

∴ Using the converse of the Basic proportionality Theorem, we have

OE || DC and OE ||AB

⇒ AB || DC

⇒ ABCD is a trapezium.

Sol. (i) In ΔABC and ΔPQR

We have:

∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

∴ The corresponding angles are equal,

∴ Using the AAA similarity rule,

ΔABC ~ ΔPQR

(ii) In ΔABC and ΔQRP

Sol. We have:

∠BOC = 125° and ∠CDO = 70°

since, ∠DOC + ∠BOC = 180° [Linear Pair]

⇒ ∠DOC 180° – 125° = 55° ...(1)

In ΔDOC

Using the angle sum property, we get

∠DOC + ∠ODC + ∠DCO = 180°

⇒ 55° + 70° + ∠DCO = 180°

⇒∠DCO =180° – 55° – 70° = 55° ...(2)

Again,

ΔODC ~ ΔOBA [Given]

∴ Their corresponding angles are equal

And ∠OCD = ∠OAB = 55° ...(3)

Thus, from (1), (2) and (3)

∠DOC = 55°, ∠DCO = 55° and ∠OAB = 55°.

Sol. We have a trapezium ABCD in which AB || DC. The diagonals AC and BD intersect at O.

In ΔOAB and ΔOCD

AB || DC [Given]

and BD intersects them

∴∠OBA = ∠ODC ...(1) [Alternate angles]

similarly,

∠OAB = ∠OCD ...(2)

∴Using AA similarity rule,

ΔOAB ~ ΔOCD

Sol. In ΔPQR

∵∠1 = ∠2

[Given]

∴ PR = QP

...(1) [∵In a Δ, sides opposite to equal angles are equal]

Sol. In ΔPQR

T is a point on QR and S is a point on PR such that

∠RTS = ∠P

Now in ΔRPQ and ΔRTS

∠RPQ = ∠RTS

[Given]

∠PRQ = ∠TRS

[Common]

∴ Using AA similarity, we have

ΔRPQ and ΔRTS

Sol. We have ΔABE ≌ ΔACD

∴Their corresponding parts are equal,

i.e., AB = AC

AE = AD

∠DAE = ∠BAC

[Common]

∴Using SAS similarity, we have

ΔADE ~ ΔABC

(i) ΔAEP ~ ΔCDP

(ii) ΔABD ~ ΔCBE

(iii) ΔAEP ~ ΔADB

(iv) ΔPDC ~ ΔBEC

Sol. We have a ΔABC in which altitude AD and CE intersect each other at P.

⇒∠D = ∠E = 90°

...(1)

(i) In ΔEAP and ΔCDP

∠AEP = ∠CDP

[From (1)]

∠EPA = ∠DPC

[Vertically opp. angles]

∴Using AA similarity, we get

ΔEAP and ΔCDP

(ii) In ΔABD and ΔCBE

∠ADB = ∠CEB

[From (1)]

Also ∠ABD = ∠CBE

[Common]

∴Using AA similarity, we have

ΔABD and ΔCBE

(iii) In ΔAEP and ΔADB

∠AEP = ∠ADB

[From (1)]

Also ∠EAP = ∠DAB

[Common]

∴Using AA similarity, we have

ΔAEP and ΔADB

(iv) In ΔPDC and ΔBEC

∠PDC = ∠BEC

[From (1)]

And ∠DCP = ∠ECB

[Common]

∴Using AA similarity, we get

ΔPDC and ΔBEC

Sol. We,have a parallelogram ABCD in which AD is produced to E and BE is joined such that BE intersects CD at F.

Now, in ΔABE and ΔCFB

∠BAE = ∠FCB

[Opp. angles of a || gm are always equal]

∠AEB = ∠CBF

[∵Parallel sides are intersected by the transversal BE]

Now, using AA similarity, we have

ΔABE and ΔCFB

Sol. We have ΔABC, right angled at B and ΔAMP, right angled at M.

∴∠B = ∠M = 90°

...(1)

(i) In ΔABC and ΔAMP

∠ABC = ∠AMP

[From (1)]

And ∠BAC = ∠MAP

[Common]

∴ Using AA similarity, we have

ΔABC ~ ΔAMP

(ii) ΔABC ~ ΔAMP

[As proved above]

∴ Their corresponding sides are proportional.

Sol. We have two similar ΔABC and ΔFEG such that CD and GH are the bisectors of ∠ACB and ∠FGE respectively.

(i) In ΔACD and ΔFGH

∠CAD = ∠GFH

...(1)[ΔABC ~ ΔFEG ∴∠A =∠F]

since ΔABC ~ ΔFEG

[Given]

∴∠C = ∠G

∴∠ACD ~ ∠FGH

...(2)

From (1) and (2),

ΔACD ~ ΔFGH

∴ Their corresponding sides are proportional,

(ii) In ΔDCB and ΔHGE

∠DBC = ∠HEG

...(1)[ΔABC ~ ΔFEG ∴∠B = E]

Again, ΔABC ~ ΔFEG ∴∠ACB ~ ∠FGE

∴∠DBC = ∠HEG

...(2)

From (1) and (2), we get

ΔDCB ~ ΔHGE

[AA similarity]

(iii) In ΔDCA and ΔHGF

∠DAC = ∠HFG

...(1) [ΔABC ~ ΔFEG ∴∠CAB = ∠GFE ⇒ ∠CAD = ∠GFF ∴∠DAC = ∠HFG]

Also ΔABC ~ ΔFEG ∴ΔACB ~ ΔFGE

Sol. We have an isosceles ΔABC in which AB = AC.

In ΔABD and ΔECF

AB = AC

[Given]

⇒Angles opposite to them are equal

∴∠ACB = ∠ABC

⇒∠ECF = ∠ABD

...(1)

Again AD ⊥ BC and EF ⊥ AC

⇒∠ADB = ∠EFC = 90°

...(2)

From (1) and (2), we have

ΔABD ~ ΔECF

[AA criteria of similarity]

Sol. We have ΔABC and ΔPQR in which AD and PM are medial sfeorfesponding to sides BC and QR respectively such, that

∴Using SSS similarity, we have:

Their corresponding q es are equal

⇒∠ABD = ∠PQM

∴∠ABC = ∠PQR

Now, in ∠ABC and ∠PQR

Sol. We have a ΔABC and point D on its side BC such that

∠ADC = ∠BAC

In ΔABC and ΔADC

∠BAC = ∠ADC

[Given]

And ∠BCA = ∠DCA

[Common]

∴ Using AA similarity, we have

∠BAC ~ ∠ADC

∴Their corresponding sides are proportional

Sol. We have two As ABC and PQR such that AD and PM are medians corresponding to BC and QR respectively. Also

From (1), we have:

∴ΔABD ~ ΔPQM.

[using SSS similarity]

since, the corresponding angles of similar triangles are equal.

∴∠ABD = ∠PQM

⇒ ∠ABC = ∠PQR

...(2)

Now, in ΔABC ~ ΔPQR

∠ABC = ∠PQR

[From (2)]

Sol. Let AB = 6 m be the pole and BC = 4 m be its shadow (in right ΔABC), whereas DE and EF denote the tower and its shadow respectively.

EF = Length of the shadow of the tower = 28 m

And DE = h = Height of the tower

In ΔABC and ΔDEF we have

∠B = ∠E = 90°

∠A = ∠D [Angular elevation of the sun at the same time.]

∴Using AA criteria of similarity, we have

ΔABC ~ ΔDEF

∴Their sides are proportional

Thus, the required height of the tower is 42 m.

Sol. We have ΔABC ~ ΔPQR such that AD and PM are the medians corresponding to the sides BC and QR respectively.

∵ΔABC ~ ΔPQR

And the corresponding sides of similar triangles are proportional.

∵Corresponding angles are also equal in two similar triangles

∴∠A = ∠P, ∠B = ∠Q and ∠C = ∠R ...(2)

Since AD and PM are medians

∴BC = 2 BD and QR = 2 QM

∴From (1),

Sol. We have

ar (ΔABC) = 64 cm^{2}

ar (ΔDEF) = 121 cm^{2} and EF = 15.4 cm

ΔABC ~ ΔDEF [Given]

Sol. We have in trap. ABCD, AB || DC.

Diagonals AC and BD intersect at O.

In ΔAOB and ΔCOD

∠AOB = ∠COD, [Vertically opposite angles]

∠OAB = ∠OCD, [Alternate angles]

∴Using AA criterion of similarity, we have:

ΔAOB ~ ΔCOD

Sol. We have:

ΔABC and ΔDBC are on the same base BC. Also BC and

AD intersect at O.

Let us draw AE ⊥ BC and DF ⊥ BC.

In ΔAOE, ΔAEO = 90° and

In ΔDOF, ΔDFO = 90°

∴∠AEO = ∠DFO ...(1)

Also, ΔAOE = ΔDOF ...(2) [Vertically Opposite Angles]

∴ From (1) and (2),

ΔAOE ~ ΔDOF [By AA similarity]

∴ Their corresponding sides are proportional

Sol. We have ΔABC and ΔDEF, such that ΔABC ~ ΔDEF and ar(ΔABC) = ar (ΔDEF).

Since, the ratio of areas of two sinular triangles is equal to the square of, the ratio of their corresponding sides.

Sol. We have a ΔABC in which D, E and F are mid points of AB, AC and BC respectively. D, E and F are joined to form ΔDEF.

Now, D is mid-point of AB

∴ Using the converse of the Basic Proportionality Theorem, we have

DE || BC

⇒∠ADE = ∠ABC ...(3) [Corresponding angles]

Also ∠AED = ∠ACE ...(4) [Corresponding angles]

Now from (3) and (4), we have

ΔABC ~ ∠DEF [Using AA similarity]

Sol. We have two triangles ABC and DEF such that

ΔABC ~ ΔDEF

AM and DN are medians corresponding to BC and EF respectively.

ΔABC ~ ΔDEF

∴ The ratio of their areas is equal to the square of the ratio of their corresponding sides.

]
Sol. We have a square ABCD, whose diagonal AC. Equilateral ΔBQC is described on the side BC and another equilateral ΔAPC is described on the diagonal AC.

All equilateral triangles are similar.

∴ΔAPC ~ ΔBQC

∴The ratio of they areas is equal to the square of the ratio of their corresponding sides.

(A) 2 : 1

(B) 1 : 2

(C) 4 : 1

(D) 1 : 4

Sol. We have an equilateral ΔABC and D is the mirpoint of BC. DE is drawn such that BDE is also an equilateral triangle.

Since, all equilateral triangles-are sirhilar,

∴ΔABC ~ ΔBDE

⇒The ratio of their areas is, equal to the square of the ratio of their corresponding sides.

From (1) and (2), we have:

(A) 2 :13

(B) 4 : 9

(C) 81 : 16

(D) 16 : 81

Sol. We have two similar triangles such that the ratio of their corresponding sides is 4 : 9

∴ The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

(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

Sol. (i) The sides are:

Here, (7 cm)^{2} = 49 cm^{2}

(24 cm)^{2} = 576 cm^{2}

(25 cm)^{2} = 625 cm^{2}

(49 + 576) cm^{2} = 625 cm^{2}

The given Δ is a right triangle. Hypotenuse = 25 cm.

(ii) The sides, are:

Here, (3 cm)^{2} = 9 cm^{2}

(8 cm)^{2} = 64 cm^{2}

(6 cm)^{2} = 36 cm^{2}

(9 + 36) ≠ 64 cm

It is not a right triangle.

(iii) The sides are:

50 cm, 80 cm, 100 cm

Here, (50 cm)^{2} = 2500 cm^{2}

(80 cm)^{2} = 6400 cm^{2}

(100 cm)^{2} = 10000 cm^{2}

(2500 + 6400) cm^{2} ≠ 10000 cm^{2}

∴ It is not a right triangle.

(iv) The sides are:

13 cm, 12 cm, 5cm

Here, (13 cm)^{2} = 169 cm^{2}

(12 cm)^{2} = 144 cm^{2}

(5 cm)^{2} = 25 cm^{2}

(144 + 25) cm^{2} = 169 cm^{2}

∴ The given triangle is a right triangle.

Sol. In ΔQMP and ΔQPR,

∠QMP = ∠QPR [Each = 90°]

∠Q = LQ [Common]

⇒ΔQMP ~ ∠QPR ...(1) [AA similarity]

Again in ΔPMR and ΔQPR,

∠PMR = ∠QPR [Each = 90°]

∠R = ∠R [Common]

⇒ΔRMP ~ ∠QPR ...(2) [AA similarity]

(i) AB^{2} = BC . BD

(ii) AC^{2} = BC. DC

(iii) AD^{2} = BD CD

Sol. (i) In ΔBAC and ΔBDA

∠ACB = ∠BAD [Each = 90°]

∠B = ∠B [Common]

∠BAC ~ ∠BDA [AA similarity]

⇒AB^{2} = BC . BD

(ii) In ΔACB and ΔDCA

∠ACB = ∠DCA [Each = 90°]

∠C = ∠C [Common]

⇒ΔACB ~ ΔDCA [AA similarity]

∴ Their corresponding sides are proportional

Sol. We have right ΔABC such that∠C = 90° and AC =.BC.

∴By Pythagoras Theorem, we ave

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

= AC^{2} + AC^{2} [AB = AC (given)]

= 2 AC^{2}

Thus, AB^{2} = 2AC^{2}

Sol. We have a ΔABC such that AB = AC.

Also, AB^{2}, 2 AC^{2}

∴ AB^{2} = AC^{2} + AC^{2}

But AC = BC

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

∴ Using the converse Pythagoras Theorem,

Sol. We have an equilateral ΔABC in which AB = AC = CA = 2a.

Let us draw CD ⊥ AB i.e., CD is an altitude corresponding to AB.

Now, in ΔACD and ΔBCD,

AC = BC [Each = 2a]

CD = CD [Common]

∠ADC = ∠BDC [Each = 90°]

ΔACD ≌ ΔBCD [By AA congruency]

∴ Their corresponding parts are equal.

⇒ AD = BD

i.e., D is the mid point of AB

Now in right ΔADC, we have

AC^{2} = AD^{2} + CD^{2}

⇒CD^{2} = AC^{2} – AD^{2}

= (2a)^{2} – (a)^{2}

= 4a^{2} – a^{2} = 3a^{2}

Similarly, each of the other altitudes are [ Each side of an equilateral Δ is equal]

Sol. Let us have a rhombus ABCD.

Diagonal of a rhombus bisect each otherat right angles.

∴ OA = OC and OB and OD

Also, ∠ADB = ∠BOC [Each = 90°]

And ∠COD = ∠DOA [Each = 90°]

In right, ΔAOB,

We have

AB^{2} = OA^{2} + OB^{2} ...(1) [Using Pythagoras theorem]

Similarly,

BC^{2} = OB^{2} + OC^{2} ...(2)

CD^{2} = OC^{2} + OD^{2} ...(3)

and DA^{2} = OD^{2} + OA^{2} ...(4)

Adding (1), (2), (3) and (4),

AB^{2} + BC^{2} + CD^{2} + DA^{2}

= [OA^{2} + OB^{2}] + [OB^{2} + OC^{2}]+ [OC^{2} + OD^{2}] + [OD^{2} + OA^{2}]

= 2OA^{2} + 2OB^{2} + 2OC^{2} + 2OD^{2}

= 2 [OA^{2} + OB^{2} + OC^{2} + OD^{2}]

= 2[OA^{2} + OB^{2} + OA^{2} + OB^{2}]

[OA = QC and OB = OD]

= 2 [2 OA^{2} + 2 OB^{2}]

Thus, sum of the squares of the sides of a rhombus is equal to theuq of thg,squares of its diagonals.

(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} + CD^{2} + BF^{2}.

Sol. We have a pointlin the interior of a ΔABCsulhat QD ⊥ BC, OE ⊥ AC and On ⊥ AB.

(i) Let us join OA, OB and OC.

In right ΔOAF, we have

OA^{2} = OF^{2}+ AF^{2} [Using Pythagoras Theorem]

Similarly, from right triangles ODB and OEC, we have

OB^{2} = BD^{2} + OD^{2}, and

OC^{2} = CE^{2} + OE^{2}

Adding,

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

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

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

⇒ OA^{2} + OB^{2} + OC^{2} – OD^{2} – OE^{2} – OF^{2} = AF^{2} +BD^{2} + CE^{2}

(ii) In right triangles OBD and OCD:

OB^{2} = OD^{2} + BD^{2} [Using Pythagoras Theorem]

and OC^{2} =OD^{2} + CD^{2}

⇒ OB^{2} – OC^{2} = OD^{2} + BD^{2} – OD^{2} – CD^{2}

⇒ OB^{2} – OC^{2} = BD^{2} – CD^{2}

Similarly, we have

OC^{2} – OA^{2} = CE^{2} – AE^{2} ...(2)

and OA^{2} – OB^{2} = AF+ – BF^{2} ...(3)

Adding (1), (2) and (3) we get:

(OB^{2} – OC^{2}) + (OC^{2} – OA^{2}) + (OA^{2} – OB^{2}) = (BD^{2} – CD^{2}) + (CE^{2} – AE^{2}) + (AF^{2} – BF^{2})

⇒ 0 = BD + CE^{2} + AF^{2} – (CD^{2} + AE^{2} – BF^{2})

⇒ BD^{2} + CE^{2} + AF^{2} = CD^{2} + AE^{2} + BF^{2}

or AF^{2} + BD^{2} + CE^{2} = AE^{2} + BF^{2} + CD^{2}

Sol. Let PQ be the ladder = PQ = 10 m

PR, the wall ⇒PR = 8 m

RQ is the base = RQ = ?

Now, in the right ΔPQR,

PQ^{2} = PR^{2} + QR^{2}

⇒ 10^{2} = 8^{2} + QR^{2}

[using Pythagoras theorem]

⇒ QR^{2} = 10^{2} – 8^{2}

= (10 +8) (10 – 8)

= 18 × 2 = 36

Thus, the distance of the foot of the ladder from’ the base to the all is 6 m.

Sol. Let AB is the wireand BC is the vertical pole. The point A is the stake.

∴ AB = 24 m, BC = 18 m

Now, in the right ΔAAC, using Pythagoilas Theorem, we have:

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

24^{2} = AC^{2} + 18^{2}

AC^{2} = ^{2}4^{2} – 18^{2}

= (24 – 18) (24 + 18)

= 6 × 42 = 252

= 7 × 36

Thus, the stake is required to be taken at from the base of the pole to make the wire taut.

Sol. Let the point A represent the airport.

Let the plane-I fly towards North

∴ Distance of the plane-l from the airport after hours

= speed × time

Let the plane-II flies towards West,

∴ Distance of the plane-lI from the airport after hours

= 1800 km

Now, in right ΔABC, using Pythagoras theorem, we have:

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

⇒BC^{2} = (1500)^{2} + (1800)^{2}

= 2250000 + 3240000

= 5490000

Thus, after hours the two planes are apart from each other.

Sol. Let the two poles AB and CD are such that the distance between their feet AC = 12 m.

∴ Height of poleI AB = 11

Height of pole-II, CD = 6 m

∴ Extra height of pole-I = BE = 11m – 6m = 15m

Let us join the tops of the poles D and B.

Now, in rt. ΔBED, using the Pythagoras Theorem, we have:

DB^{2} = DE^{2} + EB^{2}

⇒ DB^{2} = 12^{2} + 5^{2}

= 144 + 25 = 169

Thus, the required distance between the tops =13 m.

Sol. We have a right ΔABC such that ∠C = 90°. Also D and E are points on CA and CB respcetively. Let us join AE and BD.

In right ΔACB,

Using Pythagoras Theorem, we have

AB^{2} = AC^{2} + BC^{2} ...(1)

Using Pythagoras Theorem,

DE^{2} = CD^{2} + CE^{2} ...(2)

Adding (1) and (2), we get

AB^{2} + DE^{2} = [AC^{2} + BC^{2}] + [CD^{2} + CE^{2}]

= AC^{2} + BC^{2} + CD^{2} + CE^{2}

= [AC^{2} + CE^{2}] + [BC^{2} + CD^{2}]

From the figure, we can have

In right ΔACE,

[AC^{2} + CE^{2}] = AE^{2} and

In right ΔBCD,

[BC^{2} + CD^{2}] = BD^{2}

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

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

Sol. We have, a ΔABC such that AD ⊥ BC. The position of D is such that, BD = 3 CD.

In right ΔABC, we have [using Pythagoras Theorem] ...(1)

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

Similarly from right ΔACD, we have:

AC^{2} = AD^{2} + CD^{2} ...(2)

Subtracting (2) from (1), we get

AB^{2} – AC^{2} = DB^{2} – CD^{2} ...(3)

Now BC = DB = CD

= 3CD + CD = 4 CD [BD = 3 CD]

Now, substituting the values of CD and BD in (3), we get

Sol. We have an equalateral A ABC; in which D is a point on BC such that

Let us draw

In right ΔAPB, we have

AD^{2} = AP^{2} + BP^{2} ...(1) [using Pythagoras Theorem]

In right ΔAPD, we have

AD^{2} = AP^{2} + DP^{2} ...(1) [using Pythagoras Theorem]

AP^{2} = AD^{2} – DP^{2}

∴ From (1), we have

AB^{2} = (AD^{2} – DP^{2}) + BP^{2}

Sol. We hae an equilateral ΔABC, in which AD ⊥ BC.

Since, an altitude in an equilateral Δ, that bisects the corresponding side.

∴D is the mid point of BC.

⇒ 4 AB^{2} = 4 AD^{2} + BC^{2}

⇒ 4 AB^{2} = 4 AD^{2} + AB^{2}

⇒ 4 AD^{2} = 4 AB^{2} – AB^{2}

⇒ 4 AD^{2} = 3 AB^{2}

⇒ 3 AB^{2} = 4 AD^{2}

(A) 120°

(B) 60°

(C) 90°

(D) 45°

Sol. We have AB = cm, AC = 12 cm and BC = 6 cm

∴AB^{2} = ()^{2} = 36 × 3 = 108

AC^{2} = 12^{2} = 144

BC^{2} = 6^{2} = 36

Since 144 = 108 + 36

i.e., AC^{2} = AB^{2} = BC^{2}

∴ The given sides form a triangle ΔABC right angled at B.

⇒ B = 90°

∴ The correct answer is (C) 90°.

Sol. We have ΔPQR in which PS is the bisector of ∠QPR

∠QPS = ∠RPS

Let us draw RT || PS to meet QP produced at T, such that

∠1 = ∠RPS [Alternate angles]

Also ∠3 = ∠QPS [Corresponding angles]

But ∠RPS = ∠QPS [Given]

∠1 = ∠3

⇒ PT = PR

[Equal sides of a triangle opposite to equal angles]

Now, in ΔQRT,

∠PR || RT [By construction]

Using the Basic Proportionality Theorem, we have:

BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that.

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

Sol. We have AC as the hypotenuse of ΔABC.

Also BD ⊥ AC, DM ⊥ BC and DN ⊥ AB

∠ BMDN is a rectangle.

∴BM = ND [Opp. sides of a rectangle]

(i) In ΔBMD and DMC,

∠DMB = 90° = ∠DMC ...(1)

∵3D ⊥ AC [Given]

∠1 + ∠2 = 90°

In ΔBDM,

∠3 + ∠2 = 90°

⇒∠1 = ∠3 ...(2)

From (1) and (2)

∴ΔBMD ~ ΔDMC [By AA similarity]

∴ Their corresponding sides are proportional.

[∵ DN and BM are opposite sides of a rectangle, ∴ DN= BM ]

⇒ DN×MC = DM×DM

⇒ DN×MC = DM^{2}

or DM^{2} = DN × MC

(ii) In ΔBND and ΔDNA, we have:

∠BND = ∠DNA [Each = 90°]

∠DBN = ∠ADN [As in part (i)]

∴ΔBND ~ ΔDNA [AA similarity]

∴ Their corresponding sides are proportional.

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

Sol. ABC is a right triangle ∠ABC > 90° and AD ⊥ CB

In ΔADB,

∠D = 90°

∴ Using pythagoras Theorem, we have:

AB^{2} = AD^{2} + DB^{2} ...(1)

Again in ΔADC,

∠D = 90°

∴ Using Pythagoras theorem,

AC^{2} = AD^{2} + DC^{2}

=AD^{2} + [BD + BC]^{2}

= AD^{2} + [BD^{2} + BC^{2} + 2BD BC]

⇒ AC^{2} = [AD^{2} + DB^{2}] + BC^{2} + 2BC BD

⇒ AC^{2} = AB^{2} + BC^{2} + 2BC BD [From (1)]

Thus, we have:

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

AC^{2} = AB^{2} + BC^{2} – 2BC.BD.

Sol. We have ΔABC in which ∠ABC < 90° and AD ⊥ BC.

In right ΔADB,

∠D = 90°

∴ Using Pythagoras theorem, we have:

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

Also in right ΔADC,

∠D = 90°

∴ Using Pythagoras Theorem, we have:

AC^{2} = AD^{2} + DC^{2}

= AD^{2} + [BC . BD]^{2}

= AD^{2} + [BC^{2} + BD^{2} – 2BC.BD]

= [AD^{2} + BD^{2}] + BC^{2} – 2BC.BD

= [AB^{2}] + BC^{2} – 2BC.BD, From (1)]

Thus, AC^{2} = AB^{2} + BC^{2} – 2 BC.BD

which is the required relation.

Sol. We have ΔABC in which AD is median and AM ⊥ BC such that

∠ADC > 90° and ∠ADM > 90°

(i) IN ΔAMC,

∠M = 90°

AC^{2} = AM^{2} + MC^{2}

= AM^{2} + MD^{2} + DC^{2} + 2MD.DC

= AD^{2} + DC^{2} + 2MD. DC [MD^{2} + AM^{2} = AD^{2}]

(ii) In ΔAMB,

∠AMC = 90°

∴ Using Pythagoras theorem, we have:

AB^{2} = AM^{2} + BM^{2}

= AM^{2} + (BD – DM)^{2}

= AM^{2} + BD^{2} + DM^{2} + 2BD.DM

= AD^{2} + BDC^{2} – 2BD. DM [DM^{2} + AM^{2} = AD^{2}]

(iii) Adding (1) and (2) we get,

Sol. We have a parallelogram ABCD.

AC and BD are the diagonals of ||gm ABCD.

Diagonals of a || gm bisect each other.

∴ O is the mid-point of AC and BD.

Now, in ΔABC,

BO is a median

Also, in ΔADC,

Do is a median,

(i) ΔAPC ~ ΔDPB (ii) AP.PB = CP. DP

Sol. We have two chords, AB and CD of a circle. AB and CD intersect at P.

∴∠AOC = ∠DPB [Vertically opp. angles] ...(1)

Let us join AC and BD.

(i) In ΔAPC and ΔDPB,

∠CDP = ∠BDP [Angles in the same segment]...(2)

From (1) and (2) and using AA similarity we have

ΔAPC ~ ΔDPB

(ii) Since ΔAPC ! ΔDPB [As proved above]

∴ Their corresponding sides are proportional,

(i) ΔPAC ~ ΔPDB (ii) PA.PB =PC.PD

Sol. We have two chords AB and CD, when produced meet outside the circle at P.

(i) Since, in cyclic quadrilateral, the exterior angle is equal to the interiror opposite angle,

∠PAC = ∠PDB ...(1)

and ∠PCA = ∠PBD ...(2)

∴ From (1) and (2) and using the AA similarity, we have

ΔPAC ~ ΔPDB

(ii) Since, ΔPAC ~ ΔPDB

∴ Their corresponding sides are proportional.

Sol. Let us produce BA to E such that

AE = AC

Join EC.

And BE is a transversal,

∠BAD = ∠AEC [Corresponding angles] ...(1)

Also ∠CAD = ∠ACE [Alternate angles] ...(2)

Since, AC = AE

∴ Their opposite sides are equal

⇒∠AEC = ∠ACE ...(3)

From (1) and (3), we have

∠BAD = ∠ACE ...(4)

From (2) and (4), we have

∠BAD = ∠CAD

⇒ AD is bisector of ∠BAC.

Sol. Let us find the lenght of the string that she has out.

In right ΔABC,

AC^{2} = AB^{2} + CB^{2} [using Pythagoras Theorem.]

AC^{2} = (2.4)^{2} + (1.8)^{2}

⇒ AC^{2} = 5.76 + 3.24 = 9.00

i.e. Length of string she has out = 3m

Since, the string is pulled out at the rate of 5 cm/ sec,

Length of the string pulled out in 12 seconds.

= 5 cm × 12 = 60 cm

∴ Remaining string let out

= (3 – 0.60) m

= 2.4 m

In the right ΔPBC, let PB be the required horizonal distance of fly.

Since, PB^{2} = PC^{2} – BC^{2}

PB^{2} = (2.4)^{2} – (1.8)^{2}

= 5.76 – 3.24 = 2.52

Thus, the horizontal distance of the fly from Nazima after 12 seconds.

= (1.59 + 1.2) m (approximately)

= 2.76 m (approximately)