Geometry

Given DE||BC and AD = 6cm, DB = 9cm and AE = 8 cm

Required: AC = ?

Here, DE||BC, ∴ By Thales theorem

∴

⇒

⇒

⇒

∴ EC = 12cm

∴ AC = AE + EC = 8 + 12 = 20cm

∴ Length of AC = 20cm

Given DE||BC and AD = 8cm, AB = 12cm and AE = 12 cm

Required: CE = ?

Here, DB = AB—AD = 12—8 = 4cm

Also, DE||BC, ∴ By Thales theorem

∴

⇒

⇒ CE

⇒

∴ CE = 6cm

∴ Length of CE = 6cm

Given DE||BC and AD = 4x—3, BD = 3x—1, AE = 8x—7 and EC = 5x—3

Required: x = ?

Here, DE||BC, ∴ By Thales theorem

∴

⇒

⇒ 20x²—12x—15x + 9 = 24x²—8x—21x + 7

⇒24x² – 20x² + 12x + 15x—8x—21x—9 + 7 = 0

⇒ 4x²—2x—2 = 0

⇒ 2(2x²—x—1) = 0

⇒ 2x²—x—1 = 0

⇒ 2x²—2x + x—1 = 0

⇒ 2x(x—1) + 1(x—1) = 0

⇒ (2x + 1)(x—1) = 0

⇒ x = 1 or

∴Values x can have are x = 1 and

Given: AP = 3 cm, AR = 4.5 cm, AQ = 6 cm, AB = 5 cm, and AC = 10 cm

Required: Find the length of AD

Consider the ΔABC

Here, PB = AB—AP = 5—3 = 2cm and QC = AC—AQ = 10—6 = 4cm

Now, and

Here, we can clearly see that

∴ By Converse Thales theorem PQ||BC (∵)

Now Consider Δ ABD

Here, PR|| BD

∴ BY Thales theorem

⇒

⇒

⇒

⇒ RD = 3cm

∴ AD = AR + RD = 4.5 + 3 = 7.5 cm

∴ Length of AD = 7.5cm

Given: E and F are points on the sides PQ and PR respectively, of a PQR. PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Required: To verify if EF||QR

Consider the ΔPQR

Here, and

We can clearly see that

∴ Converse Thales theorem is also not satisfied.

∴ No, EF is not parallel to QR

Given: E and F are points on the sides PQ and PR respectively, of a PQR. PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9cm

Required: To verify if EF||QR

Consider the ΔPQR

Here, and

We can clearly see that

∴ Converse Thales theorem is satisfied and EF||QR.

∴ Yes, EF is parallel to QR(EF||QR)

Given: AC||BD and CE||DF.OA = 12cm, AB = 9 cm, OC = 8 cm and EF = 4.5 cm

Required: Length of FO

Consider Δ BOD

Here, AC||BD

∴ By Thales theorem

⇒

⇒

∴ CD = 6cm

Now, Consider ΔDOF

Here, CE||DF

∴ By Thales theorem

⇒

⇒

∴ OE = 6cm

∴ FO = OE + EF = 6 + 4.5 = 10.5cm

∴ Length of FO = 10.5 cm.

Given: ABCD is a quadrilateral with AB parallel to CD and line drawn parallel to AB meets AD at P and BC at Q

Required: To prove

Construction: draw a diagonal AC, which intersects PQ at R

Now, in ΔABC

Here, QR||AB

∴ By Thales theorem

⇒ --eq(1)

Now, consider ΔACD

Here, RP||DC

∴ By Thales theorem --eq(2)

From --eq(1) and –eq(2)

We have,

∴

Hence Proved

Given: PC||QK and BC||HK, AQ = 6 cm, QH = 4 cm, HP = 5 cm, KC = 18 cm

Required: Length of AK and PB

Consider the ΔAPC

Here, PC||QK

∴ By Thales theorem

Here, QP = QH + HP = 4 + 5 = 9cm

⇒

⇒

∴ AK = 12cm

Now, Consider ΔABC

Here, BC||HK

∴ By Thales theorem

Here, AH = AQ + QH = 6 + 4 = 10cm

⇒

⇒

⇒ HB = 15cm

⇒ PB = HB—HP = 15—5 = 10cm

∴ PB = 10cm

∴ Length of AK and PB are 12 cm and 10 cm respectively.

Given: DE||AQ and DF||AR

Required: To prove EF||QR.

Consider ΔAPQ

Here, ED||QA

∴ By Thales theorem --eq(1)

Now, Consider ΔPAR

Here, DF||AR

∴ By Thales theorem --eq(2)

From –eq(1) and –eq(2), we can see that

∴ By Converse Thales theorem we can say that EF||QR in ΔPQR (∵)

Hence Proved

Given: DE||AB and DF||AC

Required: To prove EF||BC.

Consider ΔAPB

Here, ED||AB

∴ By Thales theorem --eq(1)

Now, Consider ΔPAC

Here, DF||AC

∴ By Thales theorem --eq(2)

From –eq(1) and –eq(2), we can see that

∴ By Converse Thales theorem we can say that EF||BC in ΔPBC (∵)

Hence Proved

Given: A ΔABC with AD as internal bisector of ∠A, meeting BC at D. and BD = 2 cm, AB = 5 cm, DC = 3 cm

Required: The length of AC

Here, In ΔABC AD is the internal bisector of ∠A

∴ By angle bisector theorem

⇒

⇒

∴ AC = 7.5cm

∴ Length of AC = 7.5 cm

Given: A ABC with AD as internal bisector of A, meeting BC at D. and AB = 5.6 cm, AC = 6 cm, DC = 3 cm

Required: The length of BC

Here, In ΔABC AD is the internal bisector of ∠A

∴ By angle bisector theorem

⇒

⇒

∴ BD = 2.8cm

∴ Length of BD = 2.8 cm

Given: A ΔABC with AD as internal bisector of ∠A, meeting BC at D. and AB = x, AC = x–2, BD = x + 2, DC = x–1

Required: The length of BC

Here, In ΔABC AD is the internal bisector of ∠A

∴ By angle bisector theorem

⇒

⇒ (x + 2)(x—2) = x(x—1)

⇒ x²—4 = x²—x (∵ (a + b)(a—b) = a²—b²)

⇒ x = 4

∴The value of x = 4cm

Given: AB = 4 cm, AC = 6 cm, BD = 1.6 cm, and CD = 2.4 cm

Required: To verify if AD is the bisector of ∠A of ΔABC

Consider the ΔABC,

Here, and

Here, we can clearly see that

∴We can say that AD is the angle bisector of ∠A using Converse angle bisector theorem.

∴ Yes, AD is the angle bisector of ∠A

Given: AB = 6 cm, AC = 8 cm, BD = 1.5 cm and CD = 3 cm

Required: To verify if AD is the bisector of ∠A of ΔABC

Consider the ΔABC,

Here, and

Here, we can clearly see that

∴We can say that AD is not the angle bisector of ∠A using Converse angle bisector theorem.

∴ No, AD is not the angle bisector of ∠A

Given: A MNO with MP as external bisector of M meeting NO produced at P, and MN = 10 cm, MO = 6 cm, NO = 12 cm

Required: Length of OP

In ΔMNP , MP is the external bisector of ∠M meeting NO ant produced at P.

Let OP = x cm , Now by angle bisector theorem, we have

⇒

⇒ (x + 12)×6 = 10x

⇒ 6x + 72 = 10x

⇒ 10x—6x = 72

⇒ 4x = 72

⇒ x = = 18

∴The value of x = 18 cm.

Given: quadrilateral ABCD with bisectors at B and D intersecting at e on AC

Required: To Prove

Consider ΔABC, here BE is the angle bisector of ∠B

∴ By Angle bisector theorem we have, --eq(1)

Now, Consider ΔADC, here DE is the angle bisector of ∠D

∴ By Angle bisector theorem we have, --eq(2)

From –eq(1) and –eq(2)

We have,

Hence proved

Given: The internal bisector of A of ABC meets BC at D and the external bisector of A meets BC produced at E

Required: To prove

Consider ΔABC, here AD is the internal angle bisector of ∠A

∴ By Angle bisector theorem we have, --eq(1)

Again Consider ΔABC, Now AE is the External angle bisector of ∠A

∴ By Angle bisector theorem we have, --eq(2)

From –eq(1) and –eq(2)

We have,

⇒

∴

Hence Proved

Given: ABCD is a quadrilateral with AB = AD. If AE and AF are internal bisectors of ∠BAC and ∠DAC respectively

Required: To prove EF||BD

Consider the ΔABC,

Here, AE is the angle bisector of ∠A

∴ By angle bisector theorem --eq(1)

Now, In ΔACD,

Here, AF is the angle bisector of ∠A

∴ By angle bisector theorem

⇒ (∵ AD = AB) --eq(2)

From –eq(1) and –eq(2)

We have,

Now, Consider ΔBCD

Here,

∴We can say that EF||BD by Converse Thales theorem.

Hence Proved

In this question we must find the value of x and y, i.e. AC and AG respectively.

In such questions where we are given with some sides or angles of one or more triangles and we need find the other remaining sides and angles of triangles. The concept generally used is of congruency and similarity.

In this question we will be dealing only with sides.

Here,

In ΔADE and ΔABC

∠A is common

∠ACB = ∠AED by correspondence(BC∥DE),

∠ABC = ∠ADE by correspondence(FC∥DA)

∴ ΔADE ∼ ΔABC

Similarly, ΔEGA∼ΔEFC

(∵ ΔADE∼ΔABC)

⇒ 24x = 8x + 64

⇒ 16x = 64

⇒ x = 4 cm

Also,

(∵ΔEGA∼ΔEFC)

(∵EA = 8 + x, GA = y)

⇒ y = 9 cm

⇒ y = 4 cm

We need to find x, y and z. i.e. FG and EF respectively.

Now,

In ΔHFG and ΔHBC

∠HFG = ∠HBC by corresponding angles (DG∥BC)

∠HGF = ∠HCB by corresponding angles (FG∥BC)

∴ ΔHFG ∼ ΔHBC

(∵ ΔHFG ∼ ΔHBC)

⇒ x = 3.6 cm

Now,

In ΔFHG and ΔFBD

∠DFB = ∠HFG by vertically opposite angles

∠FDB = ∠FGH by alternate angles (DB∥AC)

∴ ΔFHG ∼ ΔFBD

⇒ 6×3.6 = 12 + 4y (∵ x=3.6)

⇒ y = 2.4 cm

Now,

In ΔAEG and ΔABC

∠AEG = ∠ABC by corresponding angles (EG∥BC)

∠AGE = ∠ACB by corresponding angles (EG∥BC)

∴ ΔAEG ∼ ΔABC

(∵ ΔAEG ∼ ΔABC)

(∵ x=3.6 and y=2.4)

⇒ z = 10 cm

To find x and y i.e. AF and GD respectively.

Now,

In ΔFBC and ΔGBD

∠FBC = ∠GBD (common)

∠BFC = ∠BGD by corresponding angles (DE∥CF)

∴ ΔFBC ∼ ΔGBD

(∵ ΔFBC ∼ ΔGBD)

⇒ y = 2.5 cm

EF = DC (∵ EFCD is a parallelogram)

⇒ EF = 7 cm

Now,

In ΔAEF and ΔABC

∠AEF = ∠ABC by corresponding angles (EF∥BC)

∠AFE = ∠ACB by corresponding angles (EF∥BC)

∴ ΔAEF ∼ ΔABC

(∵ ΔAEF ∼ ΔABC)

⇒ x = 8.4 cm

The figure is given below. Here, FE is the man whose height is 180 cm.

HG is the image of the man, height of image = 1.5 cm

AB is the camera lens.

We need to find the length of EO.

In ΔFEO and ΔHGO

∠GOH = ∠FOE (Vertically Opposite)

∠GHO = ∠FEO by alternate angles (FE∥GH)

∴ ΔFEO ∼ ΔHGO

(∵ ΔFEO ∼ Δ HGO)

⇒ EO = 3.6 m

Here,

In the figure above,

FE is the height of the girl.

DC is the height of the lamp.

CE is the distance traveled by the girl in 4 seconds.

GE is the shadow of the girl after 4 seconds.

We need to find GE.

CE = 4 × 0.6 m

CE = 240 cm

In ΔGFE and ΔGDC

∠EFG = ∠CDG by corresponding angles (DC∥FE)

∠GFE = ∠GDC by corresponding angles (DC∥FE)

∴ ΔGFE ∼ ΔGDC

(∵ ΔGFE ∼ ΔGDC)

⇒ 3×GE = 240 + GE

⇒ 2×GE = 240

⇒ GE = 120 cm

⇒ GE = 1.2 m

Here,

In the above figure,

B is the girl.

A is her father.

F is the boat.

D is the Drowning Man.

E is the man driving water craft.

We need to find the distance ED.

Now,

In ΔFAB and ΔFED

∠AFB = ∠EFD (vertically opposite angle)

∠FAB = ∠FED by alternate angles (AB∥DE)

∴ ΔFAB ∼ ΔFED

(∵ ΔFAB ∼ ΔFED)

⇒ x = 45 m

(∵ ΔFAB ∼ ΔFED)

⇒ ED = 140 m (∵ x = 45 m)

Now,

In ΔAPQ and ΔABC,

∠QAP = ∠CAB (common)

(given)

∴ ΔACB ∼ ΔAQP

(∵ ΔACB ∼ ΔAQP)

⇒ BC = 3PQ

Here,

DC = (9 - 5) cm

DC = 4 cm

Now,

In ΔBCD and ΔACB

∠DCB = ∠ACB (common)

∴ ΔBCD ∼ ΔACB

(∵ ΔBCD ∼ ΔACB)

⇒ BD = 6 cm

Here,

In ΔADE and ΔABC

∠ADE = ∠ABC by corresponding angles (DE∥BC)

∠DEA = ∠BCA by corresponding angles (DE∥BC)

∴ ΔAED ∼ ΔACB

Similarly,

ΔAGD ∼ ΔAFB (where AF ⊥ BC)

⇒ AF = 3AG (∵ AB = 3AD which is given) ----(1)

Similarly,

⇒ BC = 3×DE ----(2)

{by (1) and (2)}

⇒ Area of ΔABC = 9 × Area of ΔADE

(∵ Area of ΔABC = 72 cm²)

⇒ Area of ΔADE = 8 cm²

⇒ Area of DCEB = Area of ΔABC - Area of ΔADE

⇒ Area of DCEB = 72 – 8 = 64cm²

One of the length should be 35cm.

So, for perimeter to be greatest the smallest length of PQR should be 35 cm so that the other 2 sides could be greater than 35 cm hence greatest Perimeter achieved in that case.

ΔPQR ∼ ΔABC (given)

⇒ PQ = 35 cm (∵ It corresponds AB which is the smaller side of ΔABC)

⇒ PR = 52.5 cm

⇒ QR = 78.75 cm

⇒ Perimeter = PQ + PR + QR

⇒ Perimeter = 35 + 52.5 + 78.75

⇒ Perimeter = 166.25 cm

i) In ΔADG and ΔABF

∠ADE = ∠ABC by corresponding angles (DG∥BF)

∠AED = ∠ACB by corresponding angles (DG∥BF)

∴ ΔADE ∼ ΔABC

Let AD = 3x, so BD = 5x.

----(1)

Similarly, ΔADG ∼ ΔABF

----(2)

Now,

(by (1) and (2)) ----(3)

ii) Area of BCED = Area of ΔABC – Area of ΔADE

(by (3))

To find the Area of ΔEBD.

EF = 1.4 Km

In ΔEBD and ΔEAC

∠EBD = ∠EAC by alternate angles (BD∥CA)

∠EDB = ∠ECA by alternate angles (BD∥CA)

∴ ΔEBD ∼ ΔEAC

(∵ BD = 3, AC = 1)

Similarly,

ΔEBG ∼ ΔEAF

---(1)

(by (1))

⇒ Area of ΔEBD = 6.3

To find DB

Let ∠EAD = x ----(1)

⇒ ∠ADE = 90 – x (∵ AED = 90 and sum of angles of triangle is 180)

⇒ ∠EDC = x (∵ ∠ADC = 90) ----(2)

⇒ ∠EAD = ∠EDC (by (1) and (2)) ----(3)

Now,

In ΔAED and ΔDEC

∠AED = ∠DEC (both perpendicular to AC)

∠EAD = ∠EDC (by (3))

∴ ΔAED ∼ ΔDEC

(∵ ΔAED ∼ ΔDEC)

⇒ 16×81 = ED²

⇒ ED = 36 cm

⇒ 2×ED = DB = 72 cm

Here,

In the above Figure,

DE is the distance of student and mirror.

EG is the distance of mirror and flagpole.

CD is the height of student till its eyes.

FG is the height of flagpole which we need to find.

Now,

In ΔCDE and ΔFGE

∠FEG = ∠CED (by mirror property)

∠CDE = ∠FGE (both perpendicular given)

∴ ΔCDE ∼ ΔFGE

(∵ ΔCDE ∼ ΔFGE)

⇒ FG = 9 m

In ΔYZW

Let ∠YZW = x

⇒ ∠ZYW = 90-x (∵ It is a right-angled triangle)

⇒ ∠XYW = x (∵ ∠ZYX is 90)

⇒ ∠YZW = ∠XYW ----(1)

Similarly,

∠XYW = ∠XZY ----(2)

Now,

In ΔYWZ and ΔXYZ

∠WXY = ∠ZXY (common)

∠YWZ = ∠XYZ (by (1))

∴ ΔYWZ ∼ ΔXYZ ----(3)

Now,

In ΔXWY and ΔXYZ

∠YXW = ∠ZXY (common)

∠XYW = ∠XZY (by (2))

∴ ΔXWY ∼ ΔXYZ ----(4)

⇒ ΔXWY ∼ ΔXYZ ∼ ΔYWZ (by (3) and (4))

The alternate segment theorem (also known as the tangent-chord theorem) states that in any circle, the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment.

Therefore by above theorem,

∠BAT = ∠BTP = 72°

Sum of all angles of a triangle = 180°

In ΔABT,

∠ABT + ∠BTA + ∠TAB = 180°

⇒ ∠ABT + 43° + 72° = 180°

⇒ ∠ABT + 115° = 180°

⇒ ∠ABT = 180° – 115°

⇒ ∠ABT = 65°

Hence, ∠ABT = 65°

If two chords of a circle intersect each other internally or externally, then area of rectangle contained by the segment of one chord is equal to the area of rectangle contained by the segment of other chord.

Using above theorem,

PA × PB = PC × PD …(1)

(i) Given: CP = 4 cm

AP = 8 cm

PB = 2 cm

Putting the values in (1),

8 cm × 2 cm = 4 cm × PD

⇒ PD = 4 cm

Hence, PD = 4 cm

(ii) Given: AP = 12 cm

AB = 15 cm

CP = PD

∵ AB = AP + PB

⇒ PB = AB – AP

⇒ PB = 15 cm – 12 cm

⇒ PB = 3 cm

Putting the values in (1),

12 cm × 3 cm = PD × PD

⇒ PD² = 36 cm²
⇒ PD = 6 cm

Hence, PD = 6 cm

If two chords of a circle intersect each other internally or externally, then area of rectangle contained by the segment of one chord is equal to the area of rectangle contained by the segment of other chord.

Using above theorem,

PA × PB = PC × PD …(1)

(i) Given: AB = 4 cm

BP = 5 cm

PD = 3 cm

∵ AP = AB + BP

⇒ AP = 4 cm + 5 cm

⇒ AP = 9 cm

Putting the values in (1),

9 cm × 5 cm = PC × 3 cm

⇒ PC = 15 cm

∵ PC = CD + DP

⇒ 15 cm = CD + 3 cm

⇒ CD = 15 cm – 3 cm

⇒ CD = 12 cm

Hence, CD = 12 cm

(ii) Given: BP = 3 cm

CP = 6 cm

CD = 2 cm

∵ CP = CD + DP

⇒ DP = CP – CD

⇒ DP = 6 cm – 2 cm

⇒ DP = 4 cm

Putting the values in (1),

PA × 3 cm = PC × PD

⇒ PA × 3 cm = 6 cm × 4 cm

⇒ PA = 8 cm

∵ AP = AB + BP

⇒ 8 cm = AB + 3 cm

⇒ AB = 8 cm – 3 cm

⇒ AB = 5 cm

Hence, AB = 5 cm

Lengths of tangents drawn from an exterior point to the circle are equal.

From the above theorem,

AR = AQ …(1)

BP = BQ …(2)

CP = CR …(3)

Now,

Perimeter of ΔABC = AB + BC + CA

⇒ Perimeter of ΔABC = AB + BP + PC + CA

⇒ Perimeter of ΔABC = (AB + BQ) + (CR + CA) [Using (2) and (3)]

⇒ Perimeter of ΔABC = AQ + AR

⇒ Perimeter of ΔABC = AQ + AQ

⇒ Perimeter of ΔABC = 2AQ [Using (1)]

∴ AQ = AR = 1/2Perimeter of ΔABC

Hence, Proved.

Lengths of tangents drawn from an exterior point to the circle are equal.

From the above theorem,

AP = AS

DS = DR

CR = CQ

BQ = BP

Adding above equations,

AP + DR + CR + BP = AS + DS + CQ + BQ

⇒ (AP + BP) + (DR + CR) = (AS + DS) + (CQ + BQ)

⇒ AB + CD = AD + BC

Also,

AB = CD

AD = BC

[∵ Parallel sides of a parallelogram are equal]

⇒ AB + CD = AD + BC

⇒ AB + AB = BC + BC

⇒ 2AB = 2BC

⇒ AB = BC

⇒ AB = BC = CD = AD

⇒ ABCD is a rhombus

Hence, Proved

Let AB be the length of stem

D is the point of lotus that touch the ground

x cm be the length of stem that lies below the ground

AD = 20 cm

Clearly,

AB = AD + DB

⇒ AB = 20 cm + x cm

⇒ AB = (20 + x)cm

Also,

BC = AB [∵ length of stem]

⇒ BC = (20 + x)cm

Now by Pythagoras theorem,

BC² = BD² + DC²

⇒ (20 + x)² = x² + 40²

⇒ 20² + x² + 2 × 20 × x = x² + 40²

⇒ 400 + x² + 40x = x² + 1600

⇒ x² + 1600 – x² – 400 – 40x = 0

⇒ x² + 1600 – x² – 400 – 40x = 0

⇒ 1200 – 40x = 0

⇒ 40x = 1200

⇒ x = 30

Hence, length of stem below water = 30 cm

Draw EF || AB || DC passing through O.

Also AB = EF = DC

∴ ABCD and CDEF are rectangles

Now by Pythagoras theorem,

BO² = BF² + OF² …(1)

And,

OD² = OE² + ED² …(2)

Adding (1) and (2),

BO² + OD² = BF² + OF² + OE² + ED²

⇒ BO² + OD² = AE² + OF² + OE² + CF²

⇒ BO² + OD² = AE² + OE² + OF² + CF²

⇒ BO² + OD² = AO² + OC²

Hence, Proved

Given: DE intersect AB and AC and DE || BC

Using Basic Proportionality theorem which states that if a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the ratio

{Reciprocal of above}

{Adding 1 on both sides}

{Reciprocal of above}

Given: DE intersect AB and AC and DE || BC

AD = 3 cm, DB = 2 cm and AE = 2.7 cm

Using Basic Proportionality theorem which states that if a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the ratio

Putting the values,

Now, AC = AE + EC = 2.7 +1.8 = 4.5 cm

Given: ∠ PRS =∠ SRQ and PQ = 6 cm, QR = 8 cm, RP = 4 cm

Using Angle Bisector Theorem which states that the internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the corresponding sides containing the angle

Here, RS is the bisector of ∠ A so

Let PS be x then SQ will be 6– x.

So, PS = 2 cm

Given: and ∠ B =40°, ∠ C =60°

We know that Converse of Angle Bisector Theorem states that if a straight line through one vertex of a triangle divides the opposite side internally in the ratio of the other two sides, then the line

bisects the angle internally at the vertex.

Here, {Given}

So, AD is the bisector of ∠ A

∵ ∠ A +∠ B +∠ C = 180° {Angle sum property}

⇒ ∠ A = 180° – 40° – 60°

⇒ ∠ A = 80°

Also,

Here, DE intersect AB and AC and DE || BC

AD = x, DB = 8, EC =10 and AE = 4

Using Basic Proportionality theorem which states that if a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the ratio

Putting the values,

Given: ∠ B =∠ E and ∠ C =∠ F

By the criterion of similar triangle AA which says that

if two angles of one triangle are respectively equal to two angles of another triangle then the two triangles are similar

Hence, ABC~DEF (AA similarity criterion)

{Corresponding sides of the similar triangles are in same ratio}

In ΔABC and ΔADB,

∠ A = ∠ A {Common}

∠ B = ∠ D {90° each}

By the criterion of similar triangle AA which says that

if two angles of one triangle are respectively equal to two angles of another triangle then the two triangles are similar

ΔABC ~ ΔADB …(1)

In ΔABC and ΔBDC,

∠ C = ∠ C {Common}

∠ B = ∠ D {90° each}

By the criterion of similar triangle AA which says that

if two angles of one triangle are respectively equal to two angles of another triangle then the two triangles are similar

ΔABC ~ ΔBDC …..(2)

From (1) and (2),

ΔBDC ~ ΔADB

Hence, option (B) is wrong.

Let AB be the stick of length 12 m and BC be the shadow 8 m long. Similarly DE be the tower with shadow 40 m long.

In ΔABC and ΔDEF,

∠ ABC =∠ DEF {90° each}

∠ BCA = ∠ EFD {angular elevation is same at the same instant}

By the criterion of similar triangle AA which says that

if two angles of one triangle are respectively equal to two angles of another triangle then the two triangles are similar

ΔABC ~ ΔDEF

{Corresponding sides of the similar triangles are in same ratio}

Given: Sides of two similar triangles are in the ratio 2:3

We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of

their corresponding sides

= 4:9

Given: ar(ABC) = 100 cm², ar(DEF) = 49 cm² and BC = 8.2 cm

Areas of two similar triangles are in the ratio 100:49

We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of

their corresponding sides

Putting the values

Given: Perimeter (ABC) = 24 cm², Perimeter (DEF) = 18 cm² and BC = 8 cm

Perimeters of two similar triangles are in the ratio 24:18 = 4:3

We know that if two triangles are similar, then the ratio of the corresponding sides is equal to the

ratio of the corresponding perimeters.

Putting the values

Given: AB = 5 cm, AP = 8 cm, and CD = 2 cm

We know that two chords AB and CD intersect at P inside the circle with centre at O.

Then PA× PB = PC× PD.

PB =AP – AB = 8 – 5 =3 cm

Let PD = x then PC = PD + CD = 2 + x

⇒ 8 × 3 = x (2+x)

⇒ 2x +x² =24

⇒ x² + (6–4)x – 24 = 0

⇒ (x+6)(x –4) =0

⇒ x = 4 and –6

x = –6 is not possible because length is always positive

⇒ PD = 4 cm

Given: AB = 16 cm, PD = 8 cm, and PC = 6 cm

We know that two chords AB and CD intersect at P inside the circle with centre at O.

Then PA× PB = PC× PD.

Let PA = x then PB = AB – AP = 16 – x

⇒ x × (16–x) = 6× 8

⇒ 16x – x² =48

⇒ x² – (12+4)x + 48 = 0

⇒ (x–12)(x –4) =0

⇒ x = 4 and 12

x = 4 is not possible because AP >PB

⇒ AP = 12 cm

Given: OP = 26 cm, PT = 10 cm

We know that a tangent at any point on a circle is perpendicular to the radius through the point of contact

⇒ ∠ OTP = 90°

Using Pythagoras theorem,

OP² = PT² + OT²

⇒ OT² = OP² – PT²

⇒ OT² = 26² – 10²

⇒ OT = 24 cm

From the figure, we find that ABCP is a cyclic quadrilateral

∵Opposite angles are supplementary

∴∠ BAP + ∠ PCB = 180°

⇒ ∠ PCB = 180° – ∠ BAP

⇒ ∠ PCB = 180° – 120°

⇒ ∠ PCB = 60°

Now, using Tangent–Chord theorem

∠ PCB = ∠ BPT

⇒ ∠ BPT = 60°

In ΔAOP and ΔBOP,

OP = OP {Common}

PB = PA {Tangents from the same external point are equal}

OB =OA {Radii}

⇒ ΔAOP ΔBOP

⇒ ∠ BPO = ∠ APO {Corresponding parts of congruent triangles}

Given: ∠ BPA = 40°

⇒ ∠ APO=20°

Also, ∠ OAP = 90° {a tangent at any point on a circle is perpendicular to the radius through the point of contact}

So, By angle sum property

∠ POA = 180° – 90° – 20°

⇒ ∠ POA = 70°

Given: PA = 8 cm and CQ = 3 cm

We know that the lengths of the two tangents from an exterior point to a circle

are equal.

DA = DQ, PA = PB and CQ = BC

PB = 8 cm BC = 3 cm

PC = PB – BC = 8 – 3 = 5 cm

In ΔABC and ΔADB,

∠ A = ∠ A {Common}

∠ B = ∠ D {90° each}

By the criterion of similar triangle AA which says that

if two angles of one triangle are respectively equal to two angles of another triangle then the two triangles are similar

ΔABC ~ ΔADB …(1)

In ΔABC and ΔBDC,

∠ C = ∠ C {Common}

∠ B = ∠ D {90° each}

By the criterion of similar triangle AA which says that

if two angles of one triangle are respectively equal to two angles of another triangle then the two triangles are similar

ΔABC ~ ΔBDC …..(2)

From (1) and (2),

ΔBDC ~ ΔADB

Putting the values,

⇒ CD = 16 cm

Given: ar(ABC) = 16 cm², ar(DEF) = 36 cm² and altitude of first triangle = 3 cm

Areas of two similar triangles are in the ratio 16:36 = 4: 9

We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of

their corresponding sides or altitudes

Putting the values

Given: Perimeter (ABC) = 36 cm², Perimeter (DEF) = 24 cm² and DE = 10 cm

Perimeters of two similar triangles are in the ratio 36:24 = 3:2

We know that if two triangles are similar, then the ratio of the corresponding sides is equal to the

ratio of the corresponding perimeters.

Putting the values