Class 9 SELINA Solutions Maths Chapter 10 - Isosceles Triangle

Ex. 10(A)

Ex. 10(B)

Test Yourself

Isosceles Triangle Exercise Ex. 10(A)

Solution 1(a)

Correct option: (i) AB = AC

In ∆BAD and ∆CAD

∠B = ∠C

∠BAD = ∠CAD

AD is common

Hence,

∆BAD ≅ ∆CAD … (A.A.S.)

∴ AB = AC … (c.p.c.t.)

Solution 1(b)

Correct option: (ii) BD = CD

In ∆BAD and ∆CAD

AB = AC … given

AD common

∠ADB = ∠ADC … (each 90^o)

∆BAD ≅ ∆CAD … (R.H.S.)

∴ BD = CD … (c.p.c.t.)

Solution 1(c)

Correct option: (iv) 28°

∆ADB is isosceles triangle

∴ ∠ABD = ∠BAD = 65^o

Now,

∠ADC = ∠ABD + ∠BAD …(external angle theorem)

∴ ∠ADC = 130^o

In ∆ADC,

∠ADC + ∠DAC + ∠ACD = 180 … (sum of angles in a triangle)

130 + 22 + ∠ACD = 180

∠ACD = 28^o

Solution 1(d)

Correct option: (iii) AD = AE

∆ABC is isosceles triangle

∴ ∠ABD = ∠ACE

Now, in ∆ABD and ∆ACE

AB = AC

∠ABD = ∠ACE

Now,

BE = DC

BD + DE = DE + EC

BD = EC

∆ABD ≅ ∆ACE … (S.A.S.)

Solution 1(e)

Correct option: (i) isosceles but not congruent

In ΔABC and ΔPQR,

AB = AC,

∴ ΔABC is isosceles

∠C = ∠B …(1)

Also,

∠C = ∠P …(2)

∠B = ∠Q …(3)

∴∠P = ∠Q … (from (1), (2) and (3))

∴ ΔPQR is isosceles

But, we can't prove ΔABC ≅ ΔPQR using the given data.

Therefore, triangles are isosceles but not congruent.

Solution 2

BAC + ACB + ABC = 180⁰

48⁰ + ACB + ABC = 180⁰

But ACB = ABC[AB = AC]

2ABC = 180⁰ - 48⁰

2ABC = 132⁰

ABC = 66⁰ = ACB ……(i)

ACB = 66⁰

ACD + DCB = 66⁰

18⁰ + DCB = 66⁰

DCB = 48⁰………(ii)

Now, In

DBC = 66⁰[From (i), Since ABC = DBC]

DCB = 48⁰[From (ii)]

BDC = 180⁰ - 48⁰ - 66⁰

BDC = 66⁰

Since BDC = DBC

Therefore,BC = CD

Equal angles have equal sides opposite to them.

Solution 3

Given: ACE = 130⁰; AD = BD = CD

Proof:

(i)

(ii)

_{$begin mathsize 11px style angle ADC space equals space angle ABD space plus space angle DAB space open square brackets Exterior space angle space is space equal space to space sum space of space opp. space interior space angles close square brackets But space AD space equals space BD therefore space angle DAB space equals space angle ABD rightwards double arrow 80 to the power of ring operator space equals space angle ABD space plus space angle ABD rightwards double arrow 2 angle ABD space equals space 80 to the power of ring operator rightwards double arrow angle ABD space equals space 40 to the power of ring operator minus angle DAB space........ left parenthesis ii right parenthesis end style$}

(iii)

Solution 4

(i)

(ii)

$Given space DE vertical line vertical line BC space and space DC space is space the space transversal. rightwards double arrow angle CDE equals angle DCB equals 52 degree....... left parenthesis ii right parenthesis Also comma space angle ECB equals 64 degree......... left square bracket From space left parenthesis straight i right parenthesis right square bracket But comma angle ECB equals angle DCE plus angle DCB rightwards double arrow 64 degree equals angle DCE plus 52 degree rightwards double arrow angle DCE equals 64 degree minus 52 degree rightwards double arrow angle DCE equals 12 degree$

Solution 5

(i) Let the triangle be ABC and the altitude be AD.

(ii) Let triangle be ABC and altitude be AD.

Solution 6

Let ABO =OBC = x and ACO = OCB = y

Now,

Solution 7

Given:

(i) We know that the sum of the measure of all the angles of a quadrilateral is 360^o.

In quad. PQNL,

(ii)

Solution 8

Now,

Solution 9

Let us name the figure as following:

For x:

Solution 10

Therefore,

AD=DC

and AB = BC

Substituting the value of x from (i)

Putting y = 3 in (i)

x = 3 + 1

x = 4

Solution 11

Let P and Q be the points as shown below:

Given:

Solution 12

Now,

Solution 13

Solution 14

Let

Given: AB = AC

[Angles opp. to equal sides are equal]

Solution 15

Now,

BP is the bisector of

Isosceles Triangle Exercise Ex. 10(B)

Solution 1(a)

Correct option: (ii) ΔABD ≅ ΔFEC

In ∆ABD and ∆FEC

AB = EF

BC =DE

∴ BC + CD = DE + CD

∴ BD = EC

∠B = ∠E … (both 90^o)

∆ABD ≅ ∆FEC … (S.A.S.)

Solution 1(b)

Correct option: (iii) PQ = PR

In ∆PQO and ∆PRO

∠PQO = ∠PRO … (each 90^o)

∠POQ = ∠POR … given

PO is common

∆PQO ≅ ∆PRO … (A.A.S.)

PQ = PR … (c.p.c.t.)

Solution 1(c)

Correct option: (iv) x = 16, y = 8

∠A = ∠C

∴ ∆ABC is isosceles

∴ AB = BC

∴ 2x = 3y+8 … (1)

Also, in ∆ABD and ∆CBD

∠ABD = ∠CBD … given

AB = BC … given

BD is common

∆ ABD ≅ ∆ CBD … (S.A.S.)

∴ AD = CD … (c.p.c.t.)

∴ x = 2y …(2)

From (1) and (2)

4y = 3y + 8

y = 8

x = 16

Solution 1(d)

Correct option: (iv) ∆ABX ≅ ∆BAY

In ∆ABX and ∆BAY

∠XAB = ∠YBA … (each 90^o)

AX = BY … given

AB is common

∆ ABX ≅ ∆BAY … (S.A.S.)

Solution 1(e)

Correct option: (ii) ∠PBC = ∠PCB

Assuming that quadrilateral ABCD is a Square

We have

DC = AB

∠CDP = ∠BAP … (each 90^o)

DP = AP … (P is mid-point of side AD)

Hence,

In ∆ CDP and ∆BAP

We can say that

∆ CDP ≅ ∆BAP … (S.A.S.)

CP = BP … (c.p.c.t.)

Now, in ∆BPC

We have

CP = BP

Hence, ∆BPC is isosceles

∴ ∠PBC = ∠PCB