## RD Sharma Solutions Class 9 Chapter 10 Ex 10.1

Â (1) In Fig. (10).22, the sides BA and CA have been produced such that: BA = AD and CA = AE. Prove that segment DE \(\parallel\)

Â Solution:

Given that, the sides BA and CA have been produced such that BA = AD and CA = AE and given to prove DE \(\parallel\)

We have

BA = AD and CA= AEÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [given in the data]

And also \(\angle BAC\)

So, by SAS congruence criterion, we have

\(\angle BAC\simeq \angle DAE\)

- BC = DE and \(\angle DEA = \angle BCA\)
, \(\angle EDA = \angle CBA\)

Now, DE and BC are two lines intersected by a transversal DB such that \(\angle DEA = \angle BCA\)

(2) In a PQR, if PQ = QR and L, M and N are the mid-points of the sides PQ, QR and RP respectively. Prove that LN = MN.

Solution: Given that,

In PQR, PQ = QR and L, M, N are midpoints of the sides PQ, QP and RP respectively and given to prove that LN = MN

Here we can observe that PQR is an isosceles triangle

Â

- PQ = QR and \(\angle\)
QPR = \(\angle\) QRP _____(i)

And also, L and M are midpoints of PQ and QR respectively

- PL = LQ = QM = MR = \(\frac{PQ}{2}\)
= \(\frac{QR}{2}\)

And also, PQ = QR

Now, consider \(\Delta\)

\(\angle\)

\(\angle\)

PN = NR Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [N is midpoint of PR]

So, by SAS congruence criterion, we have \(\Delta\)

- LN = MN [Corresponding parts of congruent triangles are equal]

(3) In fig. (10).23, PQRS is a square and SRT is an equilateral triangle. Prove that (i) PT = QTÂ (ii) \(\angle\)

Â Solution: Given that PQRS is a square and SRT is an equilateral triangle. And given to prove that

(i) PT =QT and (ii) \(\angle\)

Now, PQRS is a square

- PQ = QR = RS = SP ____(i)
- And \(\angle\)
SPQ = \(\angle\) PQR = \(\angle\) QRS = \(\angle\) RSP = 90 \(^{\circ}\) = right angle

And also, SRT is an equilateral triangle.

- SR = RT = TS ___(ii)

And \(\angle\)

From (i) and (ii)

PQ = QR = SP = SR = RT = TSÂ ___(iii)

And also,

\(\angle\)

SP = RQ Â Â Â Â Â Â Â Â Â Â [From (iii)]

So, by SAS congruence criterion we have

\(\Delta\)

PT = QTÂ Â [Corresponding parts of congruent triangles are equal] Consider \(\Delta\)

QR = TR Â Â Â Â Â Â Â [From (iii)]

\(\Delta\)

\(\angle\)

Now,

Sum of angles in a triangle is equal to 180\(^{\circ}\)

=>Â \(\angle\)

=>Â 2 \(\angle\)

=>Â 2 \(\angle\)

=>Â 2 \(\angle\)

Hence proved

Â

(4) Prove that the medians of an equilateral triangle are equal.

Solution:

Given,

To prove the medians of an equilateral triangle are equal.

Median: The line Joining the vertex and midpoint of opposite side. Â Now, consider an equilateral triangle ABC.

Let D,E,F are midpoints of BC, CA and AB.

Then, AD, BE and CF are medians of ABC.

Â

Now,

D Is midpoint of BC => BD = DC = \(\frac{BC}{2}\)

Similarly, CE = EA = \(\frac{AC}{2}\)

AF = FB = \(\frac{AB}{2}\)

Since \(\Delta\)

=> AB = BC = CAÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â _____(i)

=> Â BD = DC = CE = EA = AF = FB = \(\frac{BC}{2}\)

And also, \(\angle\)

Now, consider \(\Delta\)

BD = CEÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [From (ii)]

Now, in \(\Delta\)

TS = TR Â Â Â [From (iii)]

\(\angle\)

So, from SAS congruence criterion, we have

\(\Delta\)

AD = BEÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â ____(iv)

[Corresponding parts of congruent triangles are equal]Now, consider \(\Delta\)

\(\angle\)

CE = AFÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â [From (ii)]

So, from SAS congruence criterion, we have

\(\Delta\)

- BE = CF (v)

From (iv) and (v), we have

AD = BE = CF

- Median AD = Median BE = Median CF

The medians of an equilateral triangle are equal.

Hence proved

(5) In a \(\Delta\)

Solution:

Consider a \(\Delta\)

Given Mat \(\angle\)

We can observe that \(\Delta\)

- \(\angle\)
B = \(\angle\) C (i)

We know that sum of angles in a triangle is equal to 180Â°

=> \(\angle\)

=> \(\angle\)

=> 120\(^{\circ}\)

=> 2\(\angle\)

=>Â \(\angle\)

(6) In a \(\Delta\)

Solution:

Consider a \(\Delta\)

Since, AB = AC \(\Delta\)

- \(\angle\)
B = \(\angle\) C [Angles opposite to equal sides are equal] - \(\angle\)
B = \(\angle\) C = 70\(^{\circ}\)

And also,

Sum of angles in a triangle = 180\(^{\circ}\)

- \(\angle\)
A + \(\angle\) B +\(\angle\) C = 180 \(^{\circ}\) - \(\angle\)
A + 70 \(^{\circ}\) + 70 \(^{\circ}\) = 180 \(^{\circ}\) - \(\angle\)
A = 180\(^{\circ}\) â€“ 140 \(^{\circ}\) - \(\angle\)
A = 40 \(^{\circ}\)

(7) The vertical angle of an isosceles triangle is (10)0 \(^{\circ}\)

Solution:

Consider an isosceles \(\Delta\)

Given that vertical angle A is (10)0\(^{\circ}\)

To find the base angles

Since \(\Delta\)

- \(\angle\)
B = \(\angle\) C [Angles opposite to equal sides are equal]

And also,

Sum of interior angles of a triangle = 180 \(^{\circ}\)

- \(\angle\)
A + \(\angle\) B +\(\angle\) C = 180 \(^{\circ}\) - (10)0 \(^{\circ}\)
+ \(\angle\) B \(\angle\) B = 180 \(^{\circ}\) - 2\(\angle\)
B = 180\(^{\circ}\) â€“ (10)0 \(^{\circ}\) - \(\angle\)
B = 40 \(^{\circ}\) - \(\angle\)
B = \(\angle\) C = 40 \(^{\circ}\)

Â

(8) In Fig. (10).24, AB = AC and \(\angle\)

Solution:

Consider the given figure

We have,

AB = AC and \(\angle\)

Since, \(\angle\)

- \(\angle\)
BCA + \(\angle\) ACD =180Â° - \(\angle\)
BCA + (10)5Â° = 180Â° - \(\angle\)
BCA = l80Â° -(10)5Â° - \(\angle\)
BCA= 75Â°

And also,

\(\Delta ABC\)

- \(\angle\)
ABC = \(\angle\) ACB [Angles opposite to equal sides are equal]

From (i), we have

\(\angle\)

- \(\angle\)
ABC = \(\angle\) ACB = 75Â°

And also,

Sum of Interior angles of a triangle = 180 \(^{\circ}\)

- \(\angle\)
ABC = \(\angle\) BCA + \(\angle\) CAB =180Â° - 75Â° + 75Â° + \(\angle\)
CAB =180Â° - 150Â° + \(\angle\)
BAC = 180Â° - \(\angle\)
BAC = 180Â° -150Â° = 30Â° - \(\angle\)
BAC = 30Â°

Â

(9) Find the measure of each exterior angle of an equilateral triangle.

Solution:

Given to find the measure of each exterior angle of an equilateral triangle consider an equilateral triangle ABC.

We know that for an equilateral triangle

AB = BC = CA and \(\angle\)

Now,

Extend side BC to D, CA to E and AB to F.

Here BCD is a straight line segment

- BCD = Straight angle =180Â°
- \(\angle\)
BCA + \(\angle\) ACD = 180Â° [From (i)] - 60Â° + \(\angle\)
ACD = 180Â° - \(\angle\)
ACD = 120Â°

Similarly, we can find \(\angle\)

\(\angle\)

Hence, the median of each exterior angle of an equilateral triangle is 120Â°

Â

(10) If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.

Solution:

ED is a straight line segment and B and C an points on it.

- \(\angle\)
EBC = \(\angle\) BCD = straight angle = 180Â° - \(\angle\)
EBA+ \(\angle\) ABC = \(\angle\) ACB + \(\angle\) ACD - \(\angle\)
EBA = \(\angle\) ACD + \(\angle\) ACB – \(\angle\) ABC - \(\angle\)
EBA = \(\angle\) ACD [From (i) ABC = ACD]

\(\angle\)

Hence proved

Â

(11) In Fig. (10).2(5) AB = AC and DB = DC, find the ratio \(\angle\)

Solution:

Consider the figure

Given,

AB = AC, DB = DC and given to find the ratio

\(\angle\)

Now, \(\Delta\)

- \(\angle\)
ABC = \(\angle\) ACB and \(\angle\) DBC = \(\angle\) DCB [Angles opposite to equal sides are equal]

Now consider,

\(\angle\)

- (\(\angle\)
ABC – \(\angle\) DBC) : (\(\angle\) ACB – \(\angle\) DCB) - (\(\angle\)
ABC – \(\angle\) DBC) : (\(\angle\) ABC – \(\angle\) DBC) [\(\angle\) ABC = \(\angle\) ACB and \(\angle\) DBC = \(\angle\) DCB] - 1:1

ABD: ACD = 1:1

Â

(12) Determine the measure of each of the equal angles of a right-angled isosceles triangle.

OR

Â ABC is a right-angled triangle in which \(\angle\)

Solution:

ABC is a right angled triangle

Consider on a right – angled isosceles triangle ABC such that

\(\angle\)

AB = AC => \(\angle\)

Now, Sum of angles in a triangle = 180 \(^{\circ}\)

\(\angle\)

=> 90Â° + \(\angle\)

=>Â 2\(\angle\)

=> \(\angle\)

\(\angle\)

Hence, the measure of each of the equal angles of a right-angled Isosceles triangle Is 45Â°

(13) AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See Fig. (10).26). Show that the line PQ is perpendicular bisector of AB.

Solution:

Consider the figure.

We have

AB is a line segment and P, Q are points on opposite sides of AB such that

AP = BP Â Â Â Â Â Â Â Â Â __________(i)

AQ = BQÂ Â Â Â Â Â Â ___________(ii)

We have to prove that PQ is perpendicular bisector of AB.

Now consider \(\Delta\)

We have

AP = BPÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [From (i)]

AQ = BQÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [From (ii)]

And PQ – PQ Â Â Â Â Â Â Â Â Â Â Â Â [Common site]

\(\Delta\)

Now, we can observe that APB and ABQ are isosceles triangles. [From (i) and (ii)]

- \(\angle\)
PAB = \(\angle\) ABQ and \(\angle\) QAB = \(\angle\) QBA

Now consider \(\Delta\)

C is the point of intersection of AB and PQ

PA = PBÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [From (i)]

\(\angle\)

PC = PCÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [common side]

So, from SAS congruency of triangle \(\Delta PAC \cong \Delta PBC\)

- AC = CB and \(\angle\)
PCA = \(\angle\) PBC _________(iv)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [ Corresponding parts of congruent triangles are equal]Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â And also, ACB is line segment - \(\angle\)
ACP + \(\angle\) BCP = 180 \(^{\circ}\) - \(\angle\)
ACP = \(\angle\) PCB - \(\angle\)
ACP = \(\angle\) PCB = 90\(^{\circ}\) <

We have AC = CB => C is the midpoint of AB

From (iv) and (v)

We can conclude that PC is the perpendicular bisector of AB

Since C is a point on the line PQ, we can say that PQ is the perpendicular bisector of AB.