The Two Triangles Below Are Similar. What Is the Similarity Ratio of Þâ´xyz to Þâ´abc?

• Textbook Solutions
• Class 9
• Math
• triangles

Mathematics Part II Solutions Solutions for Class nine Math Chapter 3 Triangles are provided here with elementary pace-by-pace explanations. These solutions for Triangles are extremely popular amidst Class 9 students for Math Triangles Solutions come handy for speedily completing your homework and preparing for exams. All questions and answers from the Mathematics Part II Solutions Book of Class ix Math Chapter 3 are provided hither for y'all for free. You lot will also dear the advertisement-free experience on Meritnation's Mathematics Part Ii Solutions Solutions. All Mathematics Function Ii Solutions Solutions for grade Grade 9 Math are prepared past experts and are 100% accurate.

Page No 27:

Question 1:

In the given effigy, $\angle$ACD is an exterior bending of $∆$ABC.$\angle$B = 40^°, $\angle$A = 70^°.

Find the measure of $\angle$ACD.

Answer:

In $∆$ABC,
∠ACD = ∠A + ∠B   (Exterior angle property)
= lxx^∘ + 40^∘
= 110^∘
Hence, the measure out of $\angle$ACD is 110^∘​.

Page No 27:

Question 2:

In

$∆$

PQR,

$\angle$

P = 70^°,

$\angle$

Q = 65 ° then find

$\angle$

R.

Reply:

In $∆$PQR,
∠P + ∠Q + ∠R = 180^∘   (Angle sum property)
⇒ 70^∘ + 65^∘ + ∠R = 180^∘
⇒ 135^∘ + ∠R = 180^∘
⇒ ∠R = 180^∘ − 135^∘
= 45^∘
Hence, the measure of $\angle$R is 45^∘.

Folio No 27:

Question 3:

The measures of angles of a triangle are x ^°, ( x $-$20)^°, (x$-$ forty)°. Observe the measure of each bending.

Answer:

Let united states of america suppose the angles ∠P, ∠Q, ∠Rof a$∆$PQR bex ^°, (x $-$ twenty)^°, (10 $-$ twoscore)° respectively.
∠P + ∠Q + ∠R = 180^∘   (Bending sum property)
⇒ x ^∘ + (10 $-$ xx)^° + (x $-$ 40) ^° = 180^∘
⇒ 3x $-$ 60 = 180
⇒ 3ten= 240
⇒ x​= 80
Therefore,
∠P = 80^∘
∠R = (eighty$-$ 20)^°
= sixty^∘
∠R = (80$-$ 40)^°
= 40^∘
Hence, the mensurate of each bending is 80^∘, 60^∘ and twoscore^∘respectively.

Page No 27:

Question 4:

The mensurate of i of the angles of a triangle is twice the measure of its smallest bending and the mensurate of the other is thrice the measure out of the smallest angle. Find the measures of the three angles.

Answer:

Let usa suppose the angles of a$∆$PQR such that ∠P < ∠Q < ∠R.
A.T.Q,
∠Q = ii∠P
∠R = ii∠P
Now, ∠P + ∠Q + ∠R = 180^∘   (Angle sum property)
⇒ ∠P + ii∠P + iii∠P = 180^∘
⇒ 6∠P = 180^∘
⇒ ∠P= 30^∘
Therefore,
∠P = 30^∘
∠R = 60^∘
∠R = ninety^∘
Hence, the measure of each angle is 30^∘, threescore^∘ and 90^∘respectively.

Page No 28:

Question 5:

In the given figure, measures of some angles are given. Using the measures detect the values of ten, y, z.

Answer:

∠NEM + ∠Internet =  180^∘      (Linear angle property)
⇒y + 100^∘ = 180^∘
⇒y = 80^∘
Also, ∠NME + ∠EMR =  180^∘      (Linear angle belongings)
⇒z + 140^∘ = 180^∘
⇒z = 40^∘
At present, In △NEM
∠North + ∠E + ∠M = 180^∘   (Angle sum property)
⇒x + y + z = 180^∘
⇒ ten+ 80^∘+ xl^∘= 180^∘
⇒x+ 120^∘= 180^∘
⇒x = lx^∘
Hence, the values ofx,y andz are threescore^∘, 80^∘ and forty^∘respectively.

Page No 28:

Question vi:

In the given figure, line AB || line DE. Find the measures of $\angle$DRE and $\angle$ARE using given measures of some angles.

Answer:

AB || DE and Advert is a transversal line.
∠BAR = ∠RDE = seventy^∘      (Alternating angles)
In △DER
∠D + ∠Due east + ∠DRE = 180^∘   (Bending sum holding)
⇒ 70^∘ + 40^∘ + ∠DRE = 180^∘
⇒ 110^∘ + ∠DRE = 180^∘
⇒ ∠DRE = 70^∘
Now, ∠ARE = ∠DRE + ∠RDE   (Outside bending belongings)
= seventy^∘ + 40^∘
= 110^∘
Hence, the measures of ∠DRE and ∠ARE are lxx^∘and 110^∘respectively.

Page No 28:

Question 7:

In $∆$ABC, bisectors of$\angle$A and $\angle$B intersect at point O. If $\angle$C = 70^° . Discover measure of $\angle$AOB.

Answer:

If the bisectors of ∠X and ∠Y of a △XYZ intersect at point O, then$\angle \mathrm{XOY}=90°+\frac{\mathrm{i}}{2}\angle \mathrm{XZY}$
$$\begin{gathered} \therefore \angle \mathrm{AOB}=\mathrm{ninety}°+\frac{1}{\mathrm{ii}}\angle \mathrm{ACB} \\ =\mathrm{ninety}°+\frac{1}{2}\left(\mathrm{seventy}°\right) \end{gathered}$$
= 90^∘ + 35^∘
= 125^∘
Hence, the measure of ∠AOB is 125^∘.

Page No 28:

Question 8:

In the given Figure, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of $\angle$BPQ and $\angle$PQD respectively.
Prove that grand $\angle$PTQ = 90 °.

Reply:

AB || CD and PQ is a transversal line.
∠BPQ + ∠DQP = 180^∘      (Angles on the aforementioned side of a transversal line are supplementary angles)
$\Rightarrow \frac{1}{2}\angle \mathrm{BPQ}+\frac{1}{2}\angle \mathrm{DQP}=\frac{180°}{2}$
⇒ ∠QPT + ∠PQT = 90^∘
In △PQT
∠QPT + ∠PQT + ∠PTQ = 180^∘   (Angle sum property)
⇒ 90^∘ + ∠PTQ = 180^∘
⇒ ∠PTQ = 90^∘
Hence proved.

Folio No 28:

Question 9:

Using the information shown in effigy, find the measures of$\angle$a, $\angle$b and $\angle$c.

Reply:

c + 100^∘ = 180^∘                (Linear angle property)
⇒c = 80^∘
Now,b = 70^∘   (Vertically opposite angles)
a +b +c = 180^∘       (Angle sum property)
⇒a+ seventy^∘+ 80^∘= 180^∘
⇒a+ 150^∘= 180^∘
⇒a= 30^∘
Hence, the values ofa,b andc are 30^∘, 70^∘ and 80^∘respectively.

Page No 28:

Question 10:

In the given figure, line DE ||   line GF ray EG and ray FG are bisectors of

$\angle$

DEF and

$\angle$

DFM respectively. Testify that,
(i)

$\angle$

DEG =

$\frac{\mathrm{one}}{2}\angle$

EDF    (ii)  EF = FG.

Respond:

(i) Given: DE || GF
At present,

$\angle$

DEF =

$\angle$

GFM              (Respective angles every bit DM is a transversal line)
⇒ 2

$\angle$

DEG =

$\angle$

DFG         (Ray EG and ray FG are bisectors of

$\angle$

DEF and

$\angle$

DFM)
⇒ ii

$\angle$

DEG =

$\angle$

EDF          (∵

$\angle$

EDF =

$\angle$

DFG, alternate angles as DF is a transversal line)
⇒

$\angle$

DEG =

$\frac{\mathrm{i}}{2}\angle$

EDF

(ii) Given: DE || GF

$\angle$

DEG =

$\angle$

EGF              (Alternate angles as EG is a transversal line)
∴

$\angle$

GEF =

$\angle$

EGF            (∵

$\angle$

DEG =

$\angle$

GEF)
∴ EF = FG                         (Sides contrary to equal angles)

Page No 31:

Question ane:

In each of the examples given below, a pair of triangles is shown. Equal parts of triangles in each pair are marked with the same signs. Observe the figures and state the test by which the triangles in each pair are congruent.

(i)

 Past . . . . . . . . . . test

$∆$

ABC

$\cong ∆$

PQR

(ii)

By . . . . . . . . . . test

$∆$

XYZ

$\cong ∆$

LMN

(iii)

By . . . . . . . . . . test

$∆$

LMN

$\cong ∆$

PTR

(iv)

By . . . . . . . . . . test

$∆$

LMN

$\cong ∆$

PTR

Answer:

(i) SSC Test
(ii) SAS Test
(iii) ASA Test
(iv) Hypotenuse Side Examination.

Page No 32:

Question 2:

Notice the information shown in pairs of triangles given below. State the test by which the two triangles are congruent. Write the remaining congruent parts of the triangles.

(i)

From the information shown in the effigy,

$\therefore ∆$

ABC

$\cong$$∆$

PQR .......

$\boxed{3243432}$

 test

$\therefore \angle$

BAC

$\cong$$\boxed{3243432}$

 .......corresponding angles of congruent triangles.

$\left.\begin{array}{l}\mathrm{seg}\mathrm{AB}\cong \boxed{1234}\\ \mathrm{and}\boxed{1234}\cong \mathrm{seg}\mathrm{PR}\end{array}\right\}...\mathrm{corresponding}\mathrm{sides}\mathrm{of}\mathrm{congruent}\mathrm{triangles}$

(ii)

From the data shown in the figure,

$\angle$

PTQ

$\cong$$\angle$

STR...... vertically reverse angles

$\therefore ∆$

PTQ

$\cong$$∆$

STR .......

$\boxed{3243432}$

 exam

$\left.\begin{array}{l}\therefore \angle \mathrm{TPQ}\cong \boxed{1234}\\ \mathrm{and}\boxed{1234}\cong \angle \mathrm{TRS}\end{array}\right\}...\mathrm{corresponding}\mathrm{angles}\mathrm{of}\mathrm{congruent}\mathrm{triangles}$

seg PQ

$\cong$$\boxed{3243432}$

  respective sides of coinciding triangles

Answer:

(i)

From the information shown in the effigy,

$\therefore ∆$

ABC

$\cong$$∆$

PQR .......

$\boxed{\mathrm{ASA}}$

 examination

$\therefore \angle$

BAC

$\cong$$\boxed{\angle \mathrm{QPR}}$

 .......respective angles of congruent triangles.

$\left.\begin{array}{l}\mathrm{seg}\mathrm{AB}\cong \boxed{\mathrm{seg}\mathrm{PQ}}123\mathrm{A}\\ \mathrm{and}\boxed{\mathrm{seg}\mathrm{Air-conditioning}}\cong \mathrm{seg}\mathrm{PR}\end{array}\right\}...\mathrm{corresponding}\mathrm{sides}\mathrm{of}\mathrm{congruent}\mathrm{triangles}$

(ii)

From the information shown in the figure,

$\angle$

PTQ

$\cong$$\angle$

STR...... vertically reverse angles

$\therefore ∆$

PTQ

$\cong$$∆$

STR .......

$\boxed{\mathrm{SAS}}$

 test

$\left.\begin{array}{l}\therefore \angle \mathrm{TPQ}\cong \boxed{\angle \mathrm{TSR}}\\ \mathrm{and}\boxed{\angle \mathrm{TQP}}\cong \angle \mathrm{TRS}\end{array}\right\}...\mathrm{corresponding}\mathrm{angles}\mathrm{of}\mathrm{congruent}\mathrm{triangles}$

seg PQ

$\cong$$\boxed{\mathrm{seg}\mathrm{SR}}$

  corresponding sides of congruent triangles

Folio No 32:

Question 3:

From the information shown in the figure, state the test assuring the congruence of $∆$ ABC and$∆$ PQR. Write the remaining congruent parts of the triangles.

Answer:

​In △ABC and △QPR
AB = PQ                        (Given)
BC = PR                        (Given)
∠A = ∠Q = 90^∘             (Given)
By RHS exam of congruency
△ABC ≅ △QPR
∠B = ∠P                (corresponding angles of congruent triangles)
∠C = ∠R               (corresponding angles of coinciding triangles)
​Ac = QR         (corresponding sides of congruent triangles)

Page No 32:

Question 4:

As shown in the following figure, in $∆$ LMN and$∆$PNM, LM = PN, LN = PM. Write the test which assures the congruence of the two triangles. Write their remaining congruent parts.

Answer:

​In △LMNand △PNM
LM = PN                         (Given)
LN = PM                        (Given)
MN = NM                      (Common)
By SSS test of congruency
△LMN ≅ △PNM
∠L = ∠P                (respective angles of coinciding triangles)
∠LMN = ∠PNM   (respective angles of congruent triangles)
∠MNL = ∠NMP   (corresponding angles of coinciding triangles)

Folio No 32:

Question 5:

In the given figure, seg AB $\cong$ seg CB and seg Advertising$\cong$seg CD.

Prove that $∆$ ABD$\cong$ $∆$ CBD

Reply:

In △ABD and △CBD
AB = CB                         (Given)
Advertizement = CD                        (Given)
BD = BD                      (Common)
By SSS examination of congruency
△ABD ≅ △CBD

Page No 33:

Question half-dozen:

In the given effigy, $\angle$P $\cong$ $\angle$R seg PQ $\cong$ seg RQ

Prove that,$∆$ PQT $\cong$ $∆$ RQS

Reply:

In △PQT and △RQS
∠P = ∠R                         (Given)
PQ = RQ                        (Given)
∠Q = ∠Q                      (Mutual)
By ASA test of congruency
△PQT ≅ △RQS

Page No 38:

Question 1:

Find the values of x and y using the information shown in figure .

Detect the measure of $\angle$ABD and m$\angle$ACD.

Reply:

In △ABC, AB = AC
∴ ∠ABC = ∠ACB            (Angles opposite to equal sides are equal)
⇒x = 50^∘
In △BDC, DB = DC
∴ ∠DBC= ∠DCB            (Angles opposite to equal sides are equal)
⇒y = 60∘
Now, ∠ABD = ∠ABC + ∠DBC
= fifty^∘ + 60^∘
= 110^∘
Also, ∠ACD = ∠ACB + ∠DCB
= l^∘ + lx^∘
= 110^∘
Hence, the values of x andy are 50^∘and threescore^∘respectively.

Page No 38:

Question 2:

The length of hypotenuse of a right angled triangle is 15. Find the length of median of its hypotenuse.

Answer:

In a right angled triangle, the length of the median on its hypotenuse is one-half the length of the hypotenuse.
Therefore, the length of median of its hypotenuse is 7.5 units.

Page No 38:

Question 3:

In $∆$PQR, $\angle$Q = 90^° , PQ = 12, QR = 5 and QS is a median. Observe l(QS).

Answer:

In △PQR,
PR² = PQ² + QRⁱⁱ           (By Pythagoras theorem)
⇒ PR²= 12² + v²
⇒ PR²= 144 + 25
⇒ PRⁱⁱ= thirteen²
⇒ PR= 13
In a right angled triangle, the length of the median on its hypotenuse is half the length of the hypotenuse.
$$\begin{gathered} \therefore l\left(\mathrm{QS}\right)=\frac{1}{2}\left(\mathrm{PR}\right) \\ =\frac{\mathrm{ane}}{2}\left(\mathrm{xiii}\right) \end{gathered}$$
= 6.five units.
Hence, the length of QR is 6.five units.

Page No 38:

Question four:

In the given figure, P indicate Chiliad is the point of concurrence of the medians of

$∆$

PQR . If GT = ii.v, find the lengths of PG and PT.

Answer:

The point of concurrence of medians of a triangle divides each median in the ratio 2 : i.
Let usa suppose the length of PG and GT be 2x andx.
Therefore, the value ofx is 2.5
Now, PG = 210
= 2(2.5)
= 5 units
Now, PT = PG + GT
= v + 2.v
= seven.5 units
Hence, the lenght of PG and GT is 5 and seven.5units respectively.

Page No 43:

Question 1:

In the given figure, point A is on the bisector of $\angle$XYZ. If AX = two cm and so find AZ.

Answer:

By Part I of the angle bisector theorem, every point on the bisector of an angle is equidistant from the sides of the angle.
∴ AZ = AX
= two cm
Hence, the length of AZ is 2 cm.

Page No 43:

Question 2:

In the given figure,

$\angle$

RST = 56^° , seg PT

$\perp$

ray ST, seg PR

$\perp$

ray SR and seg PR

$\cong$

 seg PT Discover the measure of

$\angle$

RSP. Land the reason for your answer.

Reply:

Past Part Ii of the angle bisector theorem, any point equidistant from sides of an angle is on the bisector of the angle
∠TSP = ∠RSP
Now, ∠TSP + ∠RSP = ∠TSR
⇒ ∠RSP + ∠RSP = 56^∘
⇒ ii∠RSP = 56^∘
⇒ ∠RSP = 28^∘
Hence, the mensurate of ∠RSP is 28^∘.

Page No 43:

Question iii:

In $∆$PQR, PQ = 10 cm, QR = 12 cm, PR = eight cm. Find out the greatest and the smallest angle of the triangle.

Reply:

In △PQR, QR and PR are the longest and smallest side.
The angle contrary to the longest side is the largest angle and the angle opposite to the smallest side is the smallest bending .
Hence, the greatest and the smallest angle of the triangle are ∠QPR and ∠PQR respectively.

Page No 43:

Question iv:

In $∆$ FAN, $\angle$F = lxxx^° ,$\angle$A = xl^° . Discover out the greatest and the smallest side of the triangle. State the reason.

Answer:

In ∆FAN,
∠F + ∠A + ∠N = 180^∘   (Bending sum belongings)
⇒ fourscore^∘ + 40^∘ + ∠N = 180^∘
⇒ 120^∘ + ∠N = 180^∘
⇒ ∠N = 180^∘ − 120^∘
= 60^∘
Now, in △FAN,  ∠F and ∠A are the largest and smallest angle.
The side reverse to the largest angle is the greatest side and the side contrary to the smallest angle is the smallest side .
Hence, the greatest and the smallest side of the triangle are AN and ​FN respectively.

Page No 43:

Question five:

Prove that an equilateral triangle is equiangular.

Answer:

Consider an equilateral triangle ABC.

In △ABC, AB = BC
∴ ∠C = ∠A               ...(1)          (Angles reverse to equal sides)
In △ABC, AB = CA
∴ ∠C = ∠B               ...(2)          (Angles reverse to equal sides)
From (i) and (two), we go
∠A = ∠B =  ∠C

Hence, an equilateral triangle is equiangular.

Page No 44:

Question six:

Prove that, if the bisector of

$\angle$

BAC of

$∆$

ABC is perpendicular to side BC, and so

$∆$

ABC is an isosceles triangle.

Respond:

In △ADB and △ADC
∠BAD = ∠CAD                         (AD is the bisector of  ∠BAC)
∠BDA = ∠CDA = 90^∘                (Given)
AD = DA                                     (Common)
By ASA exam of congruency
△ADB ≅ △ADC
∴ ∠B = ∠C        (respective angles of coinciding triangles)
If two angles of a traingle are equal, so the traingle is said to exist a isosceles triangle.

Page No 44:

Question 7:

In the given figure, if seg PR

$\cong$

seg PQ, show that seg PS > seg PQ.

Answer:

In △PRS, ∠PRS is an obtuse angle.
Since, a traingle can accept maximum i birdbrained angle.
Hence, ∠PRS is the greatest angle.
Therefore, the side contrary to ∠PRS i.e., PS is the longest side.
Now, seg PS > seg PR > seg RS
∴ seg PS > seg PQ                  [∵seg PR

$\cong$

seg PQ]

Page No 44:

Question 8:

In the given figure, in $∆$ABC, seg AD and seg Exist are altitudes and AE = BD. Prove that seg AD $\cong$seg Be

Respond:

In △ADB and △BEA
∠ADB = ∠BEA = 90^∘                (Given)
BD = AE                                     (Given)
AB = BA                                     (Common)
By RHS exam of congruency
△ADB ≅ △BEA
∴ AD = BE        (respective angles of coinciding triangles)
Hence proved.

Folio No 47:

Question 1:

If $∆$ XYZ ~$∆$ LMN, write the corresponding angles of the two triangles and also write the ratios of respective sides.

Answer:

Consider,$∆$ XYZ ~$∆$ LMN
$$\begin{gathered} \therefore \frac{\mathrm{XY}}{\mathrm{LM}}=\frac{\mathrm{YZ}}{\mathrm{MN}}=\frac{\mathrm{XZ}}{\mathrm{LN}} \\ \angle \mathrm{10}\cong \angle \mathrm{L},\angle \mathrm{Y}\cong \angle \mathrm{K},\mathrm{and}\angle \mathrm{Z}\cong \angle \mathrm{N} \end{gathered}$$

Page No 47:

Question two:

In $∆$ XYZ, XY = 4 cm, YZ = 6 cm, XZ = 5 cm, If$∆$ XYZ ~$∆$​ PQR  and PQ = eight cm so find the lengths of remaining sides of $∆$PQR.

Reply:

Consider,$∆$ XYZ ~$∆$ PQR
$$\begin{gathered} \therefore \frac{\mathrm{XY}}{\mathrm{PQ}}=\frac{\mathrm{YZ}}{\mathrm{QR}}=\frac{\mathrm{XZ}}{\mathrm{PR}} \\ \Rightarrow \frac{\mathrm{four}}{8}=\frac{6}{\mathrm{QR}}=\frac{\mathrm{five}}{\mathrm{PR}} \\ \Rightarrow \frac{4}{8}=\frac{6}{\mathrm{QR}}\mathrm{and}\frac{4}{\mathrm{viii}}=\frac{5}{\mathrm{PR}} \end{gathered}$$
$\Rightarrow \mathrm{QR}=12\mathrm{cm}\mathrm{and}\mathrm{PR}=\mathrm{ten}\mathrm{cm}$
Hence, the lengths of remaining sides of$∆$PQR are 12 and 10 cm.

Folio No 47:

Question 3:

Draw a sketch of a pair of similar triangles. Characterization them. Prove their corresponding angles past the aforementioned signs. Show the lengths of corresponding sides by numbers in proportion.

Answer:

The pair of like triangles is given below:

Page No 49:

Question 1:

Choose the correct alternative reply for the post-obit questions.

(i) If two sides of a triangle are 5 cm and 1.5 cm, the lenght of its third side cannot be . . . . . . . .

(A) 3.7 cm (B) 4.i cm (C) 3.eight cm (D) 3.4 cm

(2)  In

$∆$

PQR, If

$\angle$

R >

$\angle$

Q then . . . . . . . .

(A) QR > PR

(B) PQ > PR

(C) PQ < PR
(D) QR < PR

(iii) In

$∆$

TPQ,

$\angle$

T = 65^° ,

$\angle$

P = 95^° which of the following is a true statement ?

(A) PQ < TP

(B) PQ < TQ
(C) TQ < TP < PQ

(D) PQ < TP < TQ

Answer:

(i)
Co-ordinate to the triangle inequality "the sum of the lengths of any 2 sides of a triangle must be greater than or equal to the length of the tertiary side".
Therefore, the lenght of the tertiary side cannot exist greater than 3.iv cm.
Hence, the right option is (D).

(two)  In

$∆$

PQR, If

$\angle$

 R >

$\angle$

Q
∴ PQ > PR                       (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence, the correct option is (B).

(three) In

$∆$

TPQ,

$\angle$

 T = 65^° ,

$\angle$

 P = 95^°
A triangle tin accept maximum one obtuse bending.

$\angle$

T <

$\angle$

 P
PQ < TQ                         (The sides opposite to the greater angle is greater than side opposite to the smaller bending)
Hence, the correct option is (B).

Page No 49:

Question ii:

$∆$ ABC is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that BD = CE.

Answer:

​In △ABC, AB = AC
∴ ∠B = ∠C                    (Angles contrary to equal sides)
Dividing both sides by ii, we get
$\frac{1}{\mathrm{ii}}\angle \mathrm{B}=\frac{{\displaystyle 1}}{{\displaystyle 2}}\angle \mathrm{C}$
⇒ ∠EBC = ∠DCB               ...(ii)
In △BEC and △CDB
∠EBC = ∠DCB                         [From (two)]
Exist = CD                                   (E and D are the mid points of AB & AC respectively and AB = AC)
BC = CB                                     (Common)
By SAS examination of congruency
△BEC ≅ △CDB
∴ BD = CE        (corresponding sides of congruent triangles)

Folio No 49:

Question 3:

In

$∆$

PQR, If PQ > PR and bisectors of

$\angle$

Q and

$\angle$

R intersect at S. Show that SQ > SR.

Answer:

In △PQR, PQ > PR
At present, the angle opposite to the greater side is greater than bending opposite to the smaller side.
∴ ∠R > ∠Q
Dividing both sides by 2, nosotros become

$\frac{\mathrm{ane}}{\mathrm{two}}\angle \mathrm{R}>\frac{{\displaystyle 1}}{{\displaystyle \mathrm{ii}}}\angle \mathrm{Q}$

⇒ ∠SRQ > ∠SQR
⇒ SQ > SR       (The sides opposite to the greater bending is greater than side opposite to the smaller angle)
Hence proved.

Folio No 49:

Question 4:

In the figure, bespeak D and E are on side BC of

$∆$

ABC, such that BD = CE and AD = AE.

Reply:

​In △ADE, AD = AE
∴ ∠ADE = ∠AED                    (Angles opposite to equal sides)
Now,
∠ADE + ∠ADB = 180^∘            ...(1)
∠AED + ∠AEC = 180^∘            ...(2)
Subtracting (2) from (ane), we get
∠ADB − ∠AEC = 0
⇒ ∠ADB = ∠AEC                  ...(3)
In △ABD and △ACE
∠ADE = ∠AEC                         [From (3)]
BD = CE                                   (Given)
AD = AE                                   (Given)
By SAS exam of congruency
△ABD ≅ △ACE

Folio No 49:

Question v:

In the given figure, signal S is any point on side QR of $∆$PQR Prove that : PQ + QR + RP > 2PS

Respond:

In △PQS
PQ + QS > PS                     ...(1)
(Sum of two sides of a traingle is greater than the tertiary side)
In △PRS
RP + RS > PS                     ...(ii)
​                                                      (Sum of two sides of a traingle is greater than the third side)
Calculation (1) and (2), we become
PQ + QS + RP + RS  > PS + PS
⇒ PQ + QR + RP > 2PS

Page No 50:

Question 6:

In the given figure, bisector of$\angle$BAC intersects side BC at point D. Bear witness that AB > BD

Answer:

Given: Advertisement is the bisector of ∠A
∴ ∠DAB = ∠DAC
At present, ∠BDA is the exterior angle of the ∆DAC
∴ ∠BDA > ∠DAC
⇒ ∠BDA > ∠DAB        (∵ ∠DAC = ∠DAB)
⇒ AB > BD                   (The sides opposite to the greater angle is greater than side opposite to the smaller bending)
Hence proved.

Folio No 50:

Question seven:

In the given effigy, seg PT is the bisector of $\angle$QPR. A line through R intersects ray QP at bespeak South. Prove that PS = PR

Answer:

According to the figure, we have
PT || SR
If PT || SR and PS is a transversal line, then
$\angle$QPT =$\angle$PSR   (Corresponding angles)           ....(1)
If PT || SR and PR is a transversal line, then
$\angle$TPR =$\angle$PRS  (Alternate angles)              ....(2)
We are given that seg PT is the bisector of$\angle$QPR.
∴$\angle$QPT =$\angle$TPR         ...(3)
From (1), (2) and (three)
$\angle$PSR =$\angle$PRS
The sides opposite to equal angles are besides equal in a triangle.
∴ PS = PR

Page No l:

Question 8:

In the given effigy, seg AD

$\perp$

seg BC. seg AE is the bisector of

$\angle$

CAB and C - Eastward - D. Prove that

$\angle$

DAE =

$\frac{1}{2}\left(\angle \mathrm{C}-\angle \mathrm{B}\right)$ 

Answer:

In ∆ABC, since AE bisects ∠CAB, then

∠BAE = ∠CAE                         .......(1)
In ∆ADB,
∠ADB + ∠DAB + ∠B = 180°   [Angle sum property]
⇒ 90° + ∠DAB + ∠B = 180°
⇒∠B = 90° − ∠DAB                    .....(2)
In ∆ADC,
∠ADC+∠DAC+∠C = 180°   [Angle sum property]
⇒ 90° + ∠DAC + ∠C = 180°
⇒∠C = 90° − ∠DAC                         .....(three)
Subtracting (2) from (iii), we get
∠C − ∠B = ∠DAC − ∠DAB
⇒ ∠C − ∠B = (∠AEC + ∠DAE) − (∠BAE − ∠DAE)
⇒ ∠C − ∠B = ∠AEC + ∠DAE − ∠BAE + ​∠DAE
​⇒ ∠C − ∠B = 2∠DAE        [From (ane)]
​⇒

$\angle$

DAE =

$\frac{1}{2}\left(\angle \mathrm{C}-\angle \mathrm{B}\right)$ 

View NCERT Solutions for all chapters of Class 9

quickgimber60.blogspot.com

Source: https://www.meritnation.com/maharashtra-class-9/math/mathematics-part-ii-solutions/triangles/textbook-solutions/90_1_3386_24442_27_136040

Share this post