KSEEB Solutions For Class 9 Maths Chapter 7 Triangles

KSEEB Solutions For Class 9 Maths Chapter 7 Triangles

Triangles Points to Remember

• A triangle is formed by three line segments obtained by joining three pairs of points taken from a set of three non-collinear points in a plane. A triangle has three vertices, three sides & three angles.
  A triangle ABC is denoted by ΔABC

• Criteria for congruence of triangles.
  Congruence of triangle.
  The geometrical figures of same shape & size are congruent to each other i.e., two Δles ΔABC & ΔPQR are congruent if and only if their corresponding sides & the corresponding angles are equal.

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• If two triangles ΔABC & ΔPQR are congruent under the correspondence A → P, B → Q & C → R, then symbolically it is expressed as ΔABC ≅ ΔPQR
• SAS congruence Rule:
  Two triangles are congruent if two sides & the included angle of one triangle are equal to the sides & the included angle of the other triangle.
• ASA congruence Rule: Two triangles are congruent if two angles & the included side of one triangle are equal to two angles & the included side of the other triangle.
• AAS congruence Rule :
  Two triangles are congruent if any two pairs of angles & one pair of corresponding sides are equal.
• SSS congruence Rule:
  If three sides of a triangle are equal to the three sides of another triangle, then the two triangles are congruent.
• RHS congruence Rule:
  If in two right triangles, the hypotenuse & one side of a triangle are equal to the hypotenuse & one side of the other triangle then the two triangles are congruent.
  Note: It shall be noted that in SAS criteria the equality of included angles is very essential. If two sides & one angle (not included between the two sides) of one triangle are equal to two sides & one angle of the other triangle, then the triangles need not be congruent. So, the equal angle should be the angle included between the sides

• Some properties of triangles.
  A triangle is isosceles if any two sides are equal. Here, we prove some properties related to isosceles triangle.
  (1) Angles opposite to equal sides of a triangle are equal.
  In figure, \(\lfloor B\) = \(\lfloor C\)
  2) The sides opposite to equal angles of a triangle are equal.
  In figure, AB = AC

• In an isosceles triangle, bisector of the vertical angle of a triangle bisect the base.
• The medians of an equilateral triangle are equal in length.
• A point equidistance from two intersecting lines lies on the bisector of the angles formed by two lines.
  Inequalities of a triangle
• Angle opposite to the longer side is larger (greater)
• Side opposite to the larger (greater) angle is longer.
• Sum of any two sides is greater than the third side
  i. e., in ΔABC, AB + BC > CA
• Difference of any two sides of a triangle is less than
  the third side i.e., in ΔABC, AB – BC <CA

KSEEB Class 9 Maths Chapter 7 Solutions PDF

Triangles Exercise 7.1

1. In quadrilateral ACBD, AC =AD( & AB bisects \(\lfloor A\). Show that ΔABC ≅ ΔABD. What can you say about BC & BD?

Solution:

Given: In quadrilaterial ACBD, AC = BD & AB bisects \(\lfloor A\).

To prove : ΔABC ≅ ΔABD

Proof: In ΔABC & ΔABD

AC = AD (given)

AB =AB (common)

\(\lfloor\mathrm{CAB}\) = \(\lfloor\mathrm{DAB}\) (∵ AB bisects \(\lfloor A\))

∴  ΔABC ≅ ΔABD (SAS rule)

∴ BC = BD (C.P.C.T.)

2. ABCD is a quadrilateral in which AD = BC and [DAB = [CBA prove that
i) ΔABD ≅ ΔBAC
ii) BD =AC
iii) \(\lfloor\mathrm{ABD}\) = \(\lfloor\mathrm{BAC}\)

Solution:

Given : ABCD is a quadrilateral in which AD = BC & \(\lfloor\mathrm{DAB}\) = \(\lfloor\mathrm{CBA}\)

To prove:

1) ΔABD ≅ ΔBAC

2) BD = AC

3) \(\lfloor\mathrm{ABD}\) = \(\lfloor\mathrm{BAC}\)

Proof: 1) In ΔABD & ΔBAC

AD = BC (given)

AB = BA (common)

\(\lfloor\mathrm{DAB}\) = \(\lfloor\mathrm{CBA}\) (given)

∴ ΔABD ≅ ΔBAC (SAS rule)

2) ΔABD ≅ ΔBAC (proved in (1))

∴ BD = AC (CPCT)

3) ΔABD ≅ ΔBAC (proved in (1))

\(\lfloor\mathrm{ABD}\) = \(\lfloor\mathrm{BAC}\) (CPCT)

3. AD & BC are equal perpendicular to a line segment AB. Show that CD bisects AB.

Solution:

Given : AD & BC are equal perpendicular to a line segment AB.

To prove : CD bisects AB.

Proof: In ΔOAD & ΔOBC

AD = BC (given)

\(\lfloor\mathrm{OAD}\) = \(\lfloor\mathrm{OBC}\) (each = 90°)

\(\lfloor\mathrm{AOD}\) = \(\lfloor\mathrm{BOC}\) (vertically opposite angles)

∴ ΔOAD ≅ ΔOBC (AAS rule)

∴ OA = OB (CPCT)

∴ CD bisects AB.

4. l and m are two parallel lines intersected by another pair of parallel lines p & q. Show that ΔABC ≅ ΔCDA.

Solution:

Given : l and m are two parallel lines interested by another pair of parallel lines p & q.

To prove : ΔABC ≅ ΔCDA

Proof: AB || DC

& AD || BC

Quadrilateral ABCD is a parallelogram.

(a quadrilateral ABCD is a parallelogram, if both the pairs of opposite sides are parallel)

∴ BC = AD …………….(1)

(opposite sides of a ||gm are equal)

∴ AB = CD ……………(2)

(opposite sides of a ||gm are equal)

& \(\lfloor\mathrm{ABC}\) = \(\lfloor\mathrm{CDA}\) ……………..(3)

(opposite angles of a ||gm are equal).

In ΔABC & ΔCDA

AB = CD from (2)

BC = DA from(1)

\(\lfloor\mathrm{ABC}\) = \(\lfloor\mathrm{CDA}\)from (3)

∴ ΔABC = ΔCDA (SAS rule)

5. Line l is the bisector of an angle\(\lfloor A\) & B is any point on l. BP & BQ are perpendicular from B to the arms of \(\lfloor A\). Show that 1) ΔAPB ≅ ΔAQB, 2) BP = BQ or B is equidistant from the arms of \(\lfloor A\).

Solution:

Given : Line l is the bisector of an angle A & B is any point on A BP & BQ are perpendicular from B to the arms of \(\lfloor A\).

To prove :

1) ΔAPB ≅ ΔAQB

2) BP = BQ

or

B is equidistance from the arms of \(\lfloor A\).

Proof: 1) In ΔAPB & ΔAQB,

\(\lfloor\mathrm{BAP}\) = \(\lfloor\mathrm{BAQ}\)(∵ l is the bisector of \(\lfloor A\))

AB = AB (common)

\(\lfloor\mathrm{BPA}\) =\(\lfloor\mathrm{BQA}\) =90°

(y BP & BQ are perpendiculars from B to the arms of \(\lfloor A\).)

∴  ΔAPB ≅ ΔAQB (AASrule)

2) ΔAPB ≅ ΔAQB (proved in (1) above)

∴ BP = BQ (CPCT)

6. In figure, AC = AE, AB = AD & \(\lfloor\mathrm{BAD}\) = \(\lfloor\mathrm{EAC}\). Show that BC =DE.

Solution:

Given : In figure, AC = AE, AB = AD & \(\lfloor\mathrm{BAD}\) = \(\lfloor\mathrm{EAC}\).

To prove: BC = DE

Proof: In ΔABC & ΔADE,

AB = AD (given)

AC = AE (given)

\(\lfloor\mathrm{BAD}\) = \(\lfloor\mathrm{EAC}\) (given)

=> \(\lfloor\mathrm{BAD}\) + \(\lfloor\mathrm{DAC}\) = \(\lfloor\mathrm{DAC}\)+ \(\lfloor\mathrm{EAC}\)

(adding \(\lfloor\mathrm{DAC}\) to both sides)

=>\(\lfloor\mathrm{BAC}\) = \(\lfloor\mathrm{DAE}\)

∴ ΔABC ≅ ΔADE (SAS rule)

∴ BC = DE (CPCT)

7. AB is a line segment & P is its mid-point D & E are points on the same side of AB such that
\(\lfloor\mathrm{BAD}\) = \(\lfloor\mathrm{ABE}\) & \(\lfloor\mathrm{EPA}\)=\(\lfloor\mathrm{DPB}\)
Show that i) ΔDAP ≅ ΔEBP 2) AD = BE

Solution:

Given: AB is a line segment & P is its mid-point D & E are points on the same side of AB such that

\(\lfloor\mathrm{BAD}\) = \(\lfloor\mathrm{ABE}\) & \(\lfloor\mathrm{EPA}\) = \(\lfloor\mathrm{DPB}\)

To prove : (1) ΔDAP = ΔEBP

(2) AD = BE

Proof: (1) In ΔDAP & ΔEBP

AP = BP (∵ P is the midpoint of the line segment AB)

\(\lfloor\mathrm{DAP}\) = \(\lfloor\mathrm{EBP}\) (given)

\(\lfloor\mathrm{EPA}\)=\(\lfloor\mathrm{DPB}\) (given)

=> \(\lfloor\mathrm{EPA}\) + \(\lfloor\mathrm{EPD}\)=\(\lfloor\mathrm{EPD}\) + \(\lfloor\mathrm{DPB}\)

(adding \(\lfloor\mathrm{EPD}\) to both sides)

=> \(\lfloor\mathrm{APD}\) –\(\lfloor\mathrm{BPE}\)

ΔDAP ≅ ΔEBP (ASA rule)

2) ΔDAP ≅ ΔEBP (from(1) above)

∴ AD = BE (CPCT)

Karnataka 9th Standard Maths Chapter 7 Notes

8. In right angle triangle ABC, right-angled at C, M is the midpoint of hypotenuse AB, C is joined to M & produced to a point D. Such that DM = CM. Point D is joined to point B show that
1) ΔAMC ≅ ΔBMD
2) \(\lfloor\mathrm{DBC}\) is a right-angled
3) ΔDBC ≅ ΔACB
4) CM = 1/2AB.

Solution:

Given: In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M & produced to a point D. Such that DM = CM, Point D is joined to Point B.

To prove : 1) ΔAMC ≅ ΔBMD

2) \(\lfloor\mathrm{DBC}\) is a right angle

3) ADBC ≅ AACB

4) CM = 1/2AB

Proof: 1) In ΔAMC & ΔBMD

AM = BM (M is the midpoint of the hypotenuse AB)

CM = DM (given)

\(\lfloor\mathrm{AMC}\) = \(\lfloor\mathrm{BMD}\) (vertically opposite angle) (SAS rule)

∴ ΔAMC ≅ ΔBMD

2) ΔAMC ≅ ΔBMD (from(1) above)

∴\(\lfloor\mathrm{ACM}\) = \(\lfloor\mathrm{BDM}\) (CPCT)

but these are alternate interior angles and they are equal.

AC || BD

Now, AC || BD & a transversal BC intersects them

∴ \(\lfloor\mathrm{DBC}\)+\(\lfloor\mathrm{ACB}\) = 180°

(The sum of the consecutive interior angles on the same side of a transversal is 180°)

=> \(\lfloor\mathrm{DBC}\) + 90° = 180°

(∵ \(\lfloor\mathrm{ACB}\) = 90° given)

=> \(\lfloor\mathrm{DBC}\)-180 – 90 – 90°

=> \(\lfloor\mathrm{DBC}\) is a right angle.

3) In ΔDBC & ΔACB

\(\lfloor\mathrm{DBC}\) = \(\lfloor\mathrm{ACB}\)(each = 90°)

(proved in (2) above)

BC = CB (common)

ΔAMC ≅ ΔBMD (proved in (1) above)

∴ AC = BD (CPCT)

ΔDBC ≅ ΔACB (SAS rule)

4) ΔDBC ≅ ΔACD (proved in (3) above)

∴ DC = AB  (CPCT)

=> 2CM = AB (∵ DM = CM = 1/2 DC)

=> CM = 1/2AB

Triangles Exercise 7.2

1. In an isosceles triangle ABC, with AB =AC, the bisectors of \(\lfloor b\) & \(\lfloor c\) intersect each other at O. Join A to O. Show that
1) OB = OC
2) AO bisects \(\lfloor A\)

Solution:

Given: In an isosceles triangle ABC, with AB =AC, the bisectors of \(\lfloor B\) & \(\lfloor C\) intersectAach other at O. Join A to O.

To prove : 1) OB = OC

2) AO bisects \(\lfloor A\)

Proof: 1) AB = AC (given)

\(\lfloor B\) = \(\lfloor C\) (Angles opposite to equal sides of a triangle are equal)

1/2\(\lfloor B\)=1/2\(\lfloor C\)

\(\lfloor\mathrm{OBC}\) = \(\lfloor\mathrm{OCB}\) (∵ BO & CO are the

bisectors of \(\lfloor B\) & \(\lfloor C\) respectively)

∴ OB = OC

(sides opposite to equal angles of triangles are equal)

2) In ΔOAB & ΔOAC

AB = AC (given)

OB = OC (proved in (1) above)

OA = OA (common)

\(\lfloor B\) = \(\lfloor C\)(Angles opposite to equal sides of a Δle are equal)

1/2\(\lfloor B\) = 1/2\(\lfloor C\)

∴ \(\lfloor\mathrm{ABO}\) = \(\lfloor\mathrm{ACO}\) (∵ BO & CO are the

bisectors of \(\lfloor B\) & \(\lfloor C\) respectively)

\(\lfloor\mathrm{OAB}\) = \(\lfloor\mathrm{OAC}\) (CPCT)

∴ AO bisects \(\lfloor C\)

2. In ΔABC,AD is the perpendicular bisector of BC. Show that ΔABC is an isosceles triangle in which AB= AC

Solution:

Given: In AABC, AD is the perpendicular bisector of BC.

To prove: AABC is an isosceles triangle in which AB = AC

Proof: In ΔADB & ΔADC

\(\lfloor\mathrm{ADB}\) = \(\lfloor\mathrm{ADC}\) = 90°

DB = DC (AD is the perpendicular bisector of BC)

AD=AD (common)

ΔADB ≅ ΔADC (by SAS rule)

∴ AB = AC (CPCT)

ΔABC is an isosceles triangle in which AB = AC.

3. ABC is an isosceles triangle in which altitudes BE & CF are drawn to sides AC & AB respectively. Show that these altitudes are equal.

Solution:

Given: ABC is an isosceles triangle in which altitudes BE & CF are drawn to sides AC & AB respectively.

To prove : BE = CF

Proof: ABC is an isosceles triangle

∴ AB = AC

& \(\lfloor\mathrm{ABC}\) = \(\lfloor\mathrm{ACB}\) (angles opposite to …………(1) equal sides of a Δle are equal)

In ΔBEC&ΔCFB \(\lfloor\mathrm{BEC}\) = \(\lfloor\mathrm{CFB}\) = 90°

BC = CB (common)

\(\lfloor\mathrm{ECB}\) = \(\lfloor\mathrm{FBC}\) (from(1))

∴ ΔBEC & ΔCFB (by AAS rule)

∴ BE = CF (CPCT)

4. ABC is a Δle in which altitudes BE & CF to sides AC & AB are equal Show that

1) ΔABE ≅ ΔACF
2) AB=AC ie, ΔABC is an isosceles Δle

Solution:

Given : ABC is a Δe in which altitudes BE & CF to sides AC & AB are equal

To prove : 1) ΔABE ≅ ΔACF

2) AB = AC i.e. ABC is an isosceles Δle.

Proof: 1) In ΔABE & ΔACF

BE = CF (Given)

\(\lfloor\mathrm{ABE}\) = \(\lfloor\mathrm{CAF}\) (common)

\(\lfloor\mathrm{AEB}\)= \(\lfloor\mathrm{AFC}\) = 90°

ΔABE = ΔACF (byAAS rule)

2) ΔABE = ΔACF (proved in (1) above)

AB = AC (C.P.C.T.)

ΔABC is an isosceles Δle.

5. ABC & DBC are two isosceles Δle on the same base BC (see figure) show that \(\lfloor\mathrm{ABD}\) = \(\lfloor\mathrm{ACD}\)

Solution:

Given : ABC & DBC are two isosceles Δle on the same base BC.

To prove : \(\lfloor\mathrm{ABD}\) = \(\lfloor\mathrm{ACD}\)

Proof: ABC is an isoscles Δle on the base BC.

∴ \(\lfloor\mathrm{ABC}\) = \(\lfloor\mathrm{ACB}\) …………..(1)

DBC is an isosceles Δle on the base BC.

\(\lfloor\mathrm{DBC}\) – \(\lfloor\mathrm{DCB}\) ……………..(2)

adding the corresponding sides of (1) & (2) we get

\(\lfloor\mathrm{ABC}\) + \(\lfloor\mathrm{DBC}\)= \(\lfloor\mathrm{ACB}\)+ \(\lfloor\mathrm{DCB}\)

=> \(\lfloor\mathrm{ABD}\) = \(\lfloor\mathrm{ACD}\)

Hence proved.

6. ΔABC is an isosceles Δle triangle in which AB = AC. side BA is produced to D such that AD = AB (see fig). Show that \(\lfloor\mathrm{BCD}\) is a right angle.

Solution:

Given : ΔABC is an isosceles Δle in which AB = AC. side BA is produced to D such that AD =AB.

To prove: \(\lfloor\mathrm{BCD}\)is a right angle.

Proof: ABC is an isosceles Δle

∴\(\lfloor\mathrm{ABC}\) = \(\lfloor\mathrm{ACB}\) ………………(1)

AB = AC &

AD =AB

=> AC = AD In ΔACD

\(\lfloor\mathrm{CDA}\) = \(\lfloor\mathrm{ACD}\) (angles opposite to equal sides of a Δle are equal).

\(\lfloor\mathrm{CDB}\) – \(\lfloor\mathrm{ACD}\) …………..(2)

adding the corresponding sides of (1) & (2) we get.

\(\lfloor\mathrm{ABC}\) + \(\lfloor\mathrm{CDB}\) = \(\lfloor\mathrm{ACB}\) + \(\lfloor\mathrm{ACD}\)

=> \(\lfloor\mathrm{ABC}\)+\(\lfloor\mathrm{CDB}\) = \(\lfloor\mathrm{BCD}\) ……………….(3)

In ΔBCD

\(\lfloor\mathrm{BCD}\) + \(\lfloor\mathrm{DBC}\)+ \(\lfloor\mathrm{CDB}\) = 180° (sum of all the angles of a is 180°)

=> \(\lfloor\mathrm{BCD}\) + \(\lfloor\mathrm{ABC}\) + \(\lfloor\mathrm{CDB}\) = 180°

=> \(\lfloor\mathrm{BCD}\)+ \(\lfloor\mathrm{BCD}\) =180° (using(3))

=> 2\(\lfloor\mathrm{BCD}\) =180°

=> \(\lfloor\mathrm{BCD}\) =180°/2= 90°

=> \(\lfloor\mathrm{BCD}\) is a right angle.

7.  ABC is a right angled Δle in which \(\lfloor A\) = 90° & AB =AC. Find \(\lfloor B\) & \(\lfloor C\).

Solution: In ΔABC

AB = AC

∴ \(\lfloor B\) = \(\lfloor C\) (angles opposite to equal sides of a Δle are equal)……………..(1)

In ΔABC

\(\lfloor A\)+\(\lfloor B\)+ \(\lfloor C\) = 180° (angle sum property)

=> 90° +\(\lfloor B\) + \(\lfloor C\) = 180° (∵ \(\lfloor A\)= 90° given)

=> \(\lfloor B\)+ \(\lfloor C\) = 180°- 90° = 90° ……………..(2)

from (1) & (2) we get.

\(\lfloor B\) = \(\lfloor C\) = 45°

8. Show that the angles of an equilateral Δle are 60° each.

Solution:

Given : An equilateral Δle ABC.

To prove : \(\lfloor A\) = \(\lfloor B\)= \(\lfloor C\)= 60°

Proof: ABC is an equilateral Ale.

∴ AB = AC = CA ……………..(1)

AB = BC

=> \(\lfloor A\) = \(\lfloor C\) (Angles opposite to equal sides of a triangle are equal) …………….(2)

BC = CA

=> \(\lfloor A\) = \(\lfloor B\) (Angles opposite to equal sides ……………(3) of a triangle are equal)

from (2) & (3) we get

\(\lfloor A\) = \(\lfloor B\) = \(\lfloor C\) …………………(4)

In ΔABC

\(\lfloor A\)+\(\lfloor B\) + \(\lfloor C\) = 180° (angle sum property)

=> \(\lfloor A\)=\(\lfloor B\) = \(\lfloor C\) = 60°

Triangles Exercise 7.3

1. ΔABC & ΔDbC are two isosceles triangles on the same base BC & vertices A & D are on the same side of BC (see figure). If AD is extended to intersect BC at P. Show that

1) ΔABD ≅ ΔACD
2) ΔABP ≅ ΔACP
3) AP bisects \(\lfloor A\) as well as \(\lfloor D\)
4) AP is the perpendicular bisector of BC

Solution:

Given: ΔABC & ΔDBC are two isosceles Δle s on the same base BC & vertices A & D are on the same side of BC. AD is extended to intersect BC at P.

To prove : 1) ΔABD ≅ ΔACD

2) ΔABP ≅ ΔACP

3) AP bisects \(\lfloor A\) as well as \(\lfloor D\)

4) AP is the perpendicular bisector of BC.

Proof: (1) In ΔABD & ΔACD

AB =AC ……….(1) (AABC is an isosceles A/e)

BD = CD …………(2)( ADBC is an isosceles A/e)

AD = AD …………..(3) (common)

ΔABD ≅ ΔACD (SSSrule)

2) In ΔABP & ΔACP

AB=AC ………………(4) (from 1)

\(\lfloor\mathrm{ABP}\) = \(\lfloor\mathrm{ACP}\) ……………(5)

{(∵ AB = AC from (1)

∴\(\lfloor\mathrm{ABP}\) = \(\lfloor\mathrm{ACP}\) angles opposite to equal sides of a Δle are equal)}

∴ ΔABD ≅ ΔACD (proved in (1) above)

∴ \(\lfloor\mathrm{BAP}\) – \(\lfloor\mathrm{CAP}\) …………….(6) (CPCT)

In view (4), (5) & (6)

ΔABP ≅ ΔACP (ASA rule)

(3) ∵ ΔABP ≅ ΔACP (proved in (2) above)

=> AP bisects \(\lfloor A\)

In \(\lfloor\mathrm{BDP}\) & \(\lfloor\mathrm{CDP}\)

BD = CD ………….(7) (from (2))

DP = DP ………….(8) (common)

∴ ΔABP ≅ ΔACP (proved in (2) above)

BP = CP …………..(9) (CPCT)

In view of (7), (8) (9)

ΔBDP ≅ ΔCDP (SSS Rule)

\(\lfloor\mathrm{BDP}\) = \(\lfloor\mathrm{CDP}\) (CPCT)

=> DP bisects \(\lfloor D\)

=> AP bisects \(\lfloor D\)

(4) ∵ ΔBDP ≅ ΔCDP (proved in (3) above)

∴ BP = CP ……………(10) (CPCT)

\(\lfloor\mathrm{BPD}\) = \(\lfloor\mathrm{CPD}\) (CPCT)

but \(\lfloor\mathrm{BPD}\) + \(\lfloor\mathrm{CPD}\)= 180° ……………….(11)

linear pair

In view of (10 & (11)

AP is the perpendicular bisector of BC.

2. AD is an altitude of an isosceles A triangle ABC in which AB = AC.
Show that
1) AD bisects BC
2) AD bisects \(\lfloor A\)

Solution:

Given : AD is an altitude of an isosceles Δle ABC in which AB = AC

To prove : 1) AD bisects BC

2) AD bisects \(\lfloor A\)

Proof: 1) In right ΔADB & right ΔADC

Hyp AB = Hyp. AC (Given)

side AD = side AD (common)

\(\lfloor\mathrm{ADB}\) = \(\lfloor\mathrm{ADC}\)= 90°

ΔADB ≅ ΔADC (RHS rule)

BD = CD (C.P.C.T.)

=> AD bisects BC

ii) ΔADB ≅ ΔADC (proved in (1) above)

\(\lfloor\mathrm{BAD}\) = \(\lfloor\mathrm{CAD}\) (C.P.C.T.)

=> AD bisects \(\lfloor A\)

3. Two sides AB & BC & median AM of one triangle ABC are respectively equal to sides PQ & QR & median PN of triangle PQR (see figure). Show that
1) ΔABM ≅ ΔPQN
2) ΔABC ≅ ΔPQR

Solution:

Given : Two sides AB & BC & median AM of one triangle ABC are respectively equal to sides PQ & QR & median PN of ΔPQR

To prove : 1) ΔABM ≅ ΔPQN

2) ΔABC ≅ ΔPQR

Proof: 1) In ΔABM & ΔPQN

AB = PQ ……………(1) (given)

AM = PN …………….(2) (given)

BC = QR

=> 2BM = 2QN (M & N are the mid-points of BC & QR respectively)

=> BM = QN …………….(3)

In view of(l), (2) & (3)

2) ΔABM = ΔPQN (proved in (1) above)

\(\lfloor\mathrm{ABM}\) = \(\lfloor\mathrm{PQN}\) (C.P.C.T.)

=> \(\lfloor\mathrm{ABC}\) = \(\lfloor\mathrm{PQR}\) (4)

In ΔABC & ΔPQR

AB = PQ (given)

BC = QR (given)

\(\lfloor\mathrm{ABC}\) = \(\lfloor\mathrm{PQR}\) (from (4))

∴ ΔABC = ΔPQR (by SAS rule) 

4. BE & CF are two equal altitudes of a ΔABC using RHS congruence rule, prove that the triangle ABC is isosceles.

Solution:

Given : BE & CF are two equal A altitudes of a ΔABC

To prove : ABC is isosceles. F E

Proof: In right ΔBEC & right

ΔCFB

side BE = side CF (given)

Hyp. BC = Hyp. CB (common)

ΔBEC = ΔCFB (RHS rule)

\(\lfloor\mathrm{BCE}\) = \(\lfloor\mathrm{CBF}\) (C.P.C.T)

AB = AC (sides opposite to equal angles of a Δle are equal)

∴ ΔABC is isosceles.

5. ABC is an isosceles A/e with AB = AC. Draw AP⊥BC show that \(\lfloor B\) = \(\lfloor C\)

Solution:

Given: ABC is an isosceles A/e with AB=AC

To prove : \(\lfloor B\)= \(\lfloor C\)

Construction : Draw AP ⊥ BC

Proof: In right triangle APB & right triangle APC.

Hyp. AB = Hyp. AC (given)

side AP = side AP (common)

∴ ΔAPB ≅ ΔAPC (RHS rule)

\(\lfloor\mathrm{ABP}\) = \(\lfloor\mathrm{ACP}\) (C.P.C.T.)

=> \(\lfloor B\)= \(\lfloor C\)

Triangles Exercise 7.4

1. Show that in a right-angled triangle, the hypotenuse is the longest side.

Solution: Let ABC be a right-angled triangle in which \(\lfloor B\) = 90°

then \(\lfloor A\) + \(\lfloor C\) = 90° (sum of all the angles of a Δle is 180°)

∴ \(\lfloor B\) = \(\lfloor A\) + \(\lfloor C\)

∴ \(\lfloor B\)>\(\lfloor A\) & \(\lfloor B\)>\(\lfloor C\)

∴ AC>BC (∴ side opposite to greater angle is longer)

& AC > AB

∴ AC is the longest side, i.e., the hypotenuse is the longest side.

2. In figure sides AB & AC of AABC are extended to points P & Q respectively. Also \(\lfloor\mathrm{PBC}\) < \(\lfloor\mathrm{QCB}\), Show that AC > AB.

Solution:

Given: sides AB &AC are extended to points P & Q respectively. Also,\(\lfloor\mathrm{PBC}\)< \(\lfloor\mathrm{QCB}\)

To prove : AC > AB

Proof: \(\lfloor\mathrm{PBC}\) < \(\lfloor\mathrm{QCB}\) (given)

=> –\(\lfloor\mathrm{PBC}\) > –\(\lfloor\mathrm{QCB}\)

=> 180 – \(\lfloor\mathrm{PBC}\) > 180°-\(\lfloor\mathrm{QCB}\)

∴ \(\lfloor\mathrm{ABC}\)>\(\lfloor\mathrm{ACB}\)

∴ AC > AB

∴ sides opposite to greater angle is longer.

3. In figure, \(\lfloor B\) < \(\lfloor A\)& \(\lfloor C\) < \(\lfloor D\) show that AD < BC.

Solution:

Given : In figure, \(\lfloor B\) < \(\lfloor A\) & \(\lfloor C\) < \(\lfloor D\)

To prove :

AD < BC

Proof: \(\lfloor B\)< \(\lfloor A\)(given)

\(\lfloor A\)>\(\lfloor B\)

OB >OA …………….(1)

(side opposite to greater angle is longer)

\(\lfloor C\) < \(\lfloor D\) (given)

∴ \(\lfloor D\) > \(\lfloor C\)

∴ OC>OD ……………..(2)

(side opposite to greater angle is longer)

from (1) & (2), we get

OB + OOOA + OD

=>BC > AD

=> AD < BC

KSEEB Class 9 Maths Triangles Solutions

4. AB & CD are respectively the smallest & longest sides of a quadrilateral ABCD (see figure).
Show that \(\lfloor A\) > \(\lfloor C\) & \(\lfloor B\) > \(\lfloor D\)

Solution:

Given: AB & CD are respectively the smallest and longest sides of a quadrilateral ABCD.

To prove : \(\lfloor A\) > \(\lfloor C\) & \(\lfloor B\) > \(\lfloor D\)

Construction: Join AC

Proof: In ΔABC AB < BC (∵ AB is the smallest side of quadrilateral ABCD)

=> BC > AB

\(\lfloor\mathrm{BAC}\)> \(\lfloor\mathrm{BCA}\) (angle opposite to longer ……………(1)side is greater)

In ΔACD

CD > AD (∵ CD is the longest side of quadrilateral ABCD)

∴ \(\lfloor\mathrm{CAD}\) > \(\lfloor\mathrm{ACD}\) (angle opposite to longer side is greater) from(l) & (2), we obtain

\(\lfloor\mathrm{BAC}\) + \(\lfloor\mathrm{CAD}\) > \(\lfloor\mathrm{BCA}\)+ \(\lfloor\mathrm{ACD}\)

\(\lfloor A\)>\(\lfloor C\)

||ly joining B to D, we can prove that \(\lfloor B\) > \(\lfloor D\)

5. In figure, PR > PQ & PS bisects \(\lfloor\mathrm{QPR}\). Prove that \(\lfloor\mathrm{PSR}\) > \(\lfloor\mathrm{PSQ}\)

Solution:

Given: In figure, PR > PQ & PS bisects \(\lfloor\mathrm{QPR}\)

To prove : \(\lfloor\mathrm{PSR}\) > \(\lfloor\mathrm{PSQ}\)

Proof: In ΔPQR

PR>PQ (given)

∴ \(\lfloor\mathrm{PQR}\) > \(\lfloor\mathrm{PRQ}\) ………………..(1) (angle opposite to the longer side is greater)

PS bisects \(\lfloor\mathrm{QPR}\)

∴ \(\lfloor\mathrm{QPS}\) = \(\lfloor\mathrm{RPS}\) ……………….(2)

In ΔPQS

\(\lfloor\mathrm{PQR}\) + \(\lfloor\mathrm{QPS}\) + \(\lfloor\mathrm{PSQ}\) = 180° ………………..(3)

(angle sum property of a Δle)

In ΔPRS

\(\lfloor\mathrm{PRS}\)+\(\lfloor\mathrm{SPR}\) + \(\lfloor\mathrm{PSR}\) = 180° ……………….(4)

(Angle sum property of a Δle) from (3) & (4) we get

\(\lfloor\mathrm{PQR}\) + \(\lfloor\mathrm{QPS}\)+ \(\lfloor\mathrm{PSQ}\) = \(\lfloor\mathrm{PRS}\) + \(\lfloor\mathrm{SPR}\)+

\(\lfloor\mathrm{PSR}\)

=> \(\lfloor\mathrm{PQR}\) + \(\lfloor\mathrm{PSQ}\) = \(\lfloor\mathrm{PRS}\) + \(\lfloor\mathrm{PSR}\)

(∵ \(\lfloor\mathrm{QPS}\) = \(\lfloor\mathrm{SPR}\))

=> \(\lfloor\mathrm{PRS}\) + \(\lfloor\mathrm{PSR}\) = \(\lfloor\mathrm{PQR}\)+ \(\lfloor\mathrm{PSQ}\)

=> \(\lfloor\mathrm{PRS}\) + \(\lfloor\mathrm{PSR}\) > \(\lfloor\mathrm{PRQ}\)+ \(\lfloor\mathrm{PSQ}\)(from(1))

=> \(\lfloor\mathrm{PRQ}\) + \(\lfloor\mathrm{PSR}\) > \(\lfloor\mathrm{PSR}\)+ \(\lfloor\mathrm{PSQ}\)

(∵ \(\lfloor\mathrm{PSQ}\) = \(\lfloor\mathrm{PSR}\))

=> \(\lfloor\mathrm{PSR}\) > \(\lfloor\mathrm{PSQ}\)

6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

Solution: Let l is a line & P is a point not lying on l , PM ⊥ l N is any point on l other than M.

To prove : PM < PN

Proof: In ΔPMN

\(\lfloor M\) = 90°

∴ \(\lfloor N\) is an acute angle

(angle sum property of a Δle)

\(\lfloor M\) >\(\lfloor N\)

PN > PM (side opposite to greater angle is greater)

=> PM < PN

Additional Problems

Choose the correct answer from the following:

1. In ΔABC = ΔACB then ΔABC is isosceles with

1. AB = AC A
2. AB = BC
3. AC = BC
4. None of these

Solution: 1. AB = AC

2. Which of the following is not a criterion for congruence of triangles?

1. SAS
2. SSA
3. ASA
4. SSS

Solution: 2. SSA

3. In triangle ABC & PQR, if \(\lfloor A\) = \(\lfloor R\) ,\(\lfloor B\)= \(\lfloor P\) & AB = RP, then which one of the following congruence conditions applies.

1. SAS
2. ASA
3. SSS
4. RHS

Solution: 2. ASA

4. In the following figure, write the relation between AB & AC.

1. AB > AC
2. AB < AC
3. AB = AC
4. AB = 1/2 AC

Solution: 1. AB > AC

5. In the following figure, in ΔABC, AB = AC, CD = CA & \(\lfloor\mathrm{ADC}\) = 20°, then \(\lfloor\mathrm{ABC}\) =

1. 10°
2. 20°
3. 30°
4. 40°

Solution: 4. 40°

6. In ΔABC, \(\lfloor C\) = \(\lfloor A\) & BC = 6cm & AC = 5cm then the length of AB is

1. 6cm
2. 5cm
3. 3cm
4. 2.5cm

Solution: 1. 6cm

7. ΔABC ≅ ΔPQR, then which of the following is true

1. A↔R
2. AB = QR
3. AC = PQ
4. AB = PQ

Solution: 4. AB = PQ

8. Two sides of a triangle are of length 7cm & 3.5 cm. The length of the third side of the triangle cannot be

1. 3.6cm
2. 4.1cm
3. 3.4cm
4. 3.8cm

Solution: 3. 3.4cm

9. In given figure, if AE || DC & AB = AC, the value of \(\lfloor\mathrm{ABD}\) is

1. 70°
2. 110°
3. 120°
4. 130°

Solution: 2. 110°

10. In given figure, the measure of \(\lfloor\mathrm{PQR}\) is _

1. 50°
2. 60°
3. 70
4. 80°

Solution: 2. 60°

11. In ΔABC AB = AC, BD = EC, then ΔADE is

1. right-angled
2. Scalene
3. Isosceles
4. equilateral

Solution: 3. Isosceles

One Mark Questions

1. In ΔABC & ΔDEF, AB = DE, \(\lfloor A\) = \(\lfloor D\) What will be the condition in which the two triangles will be congruent by SAS axiom?

Solution: Since AB = DE

\(\lfloor A\) = \(\lfloor D\)

& ΔABC ≅ ΔDEF

by SAS

∴ AC = DF

2. What do we call a triangle if the angles are in the ratio 5:3:7?

Solution: Let the angles of triangles are 5x, 3x& 7x, then

5x+3x + 7x=180°

15x= 180

x = 180/15=12°

∴ Angles are 60°, 36°, 84°

∴ Each angle is less than 90°

The Δle is an acute-angled triangle.

3. In the figure below, if AB = AC, then what is the value of x?

Solution: \(\lfloor\mathrm{ABC}\) = 180- 125 = 55°

AB = AC,

=> \(\lfloor\mathrm{ACB}\) = \(\lfloor\mathrm{ABC}\) = 55°

=> x = 180-55-55°

=> x=70°

4. Is it possible to construct a triangle, when its sides are 5.4cm, 2.3cm, 3.1cm?

Solution: No, Because 2.3 + 3.1 = 5.4cm

Not possible to construct a Δle.

5. From the given figure A show that AB > BD

Solution: If D is a point on the side BC of a ΔABC such that

AD bisects \(\lfloor\mathrm{BAC}\) then AB > BD

6. In ΔABC \(\lfloor B\) = 30°, \(\lfloor C\) = 80°&\(\lfloor A\) = 70° then prove that AB > BC > AC

Solution:

Since \(\lfloor C\) > \(\lfloor A\) > \(\lfloor B\)

∴ AB > BC > AC

7. ΔABC is an isosceles right-angled triangle in which \(\lfloor A\) =90°. Calculate \(\lfloor B\).

Solution: AB=AC \(\lfloor B\) = \(\lfloor C\)

By angle sum property.

\(\lfloor A\) + \(\lfloor B\) + \(\lfloor C\) = 180°

=> 90° + \(\lfloor B\) + \(\lfloor B\) = 180°

2\(\lfloor B\) = 90°

\(\lfloor B\) = 9%/2 = 45°

8. In ΔPQR PE is the perpendicular bisector of \(\lfloor\mathrm{QPR}\) , then prove that PQ = PR

Solution: In ΔPEQ & ΔPER

\(\lfloor\mathrm{PEQ}\) = \(\lfloor\mathrm{PER}\) = 90°

PE = PE (common)

ΔQPE = ΔRPE (given)

ΔPEQ ≅ ΔPER

PQ = PR (cpct)

9. ΔPQR ≅ ΔABC if PQ = 5cm, \(\lfloor Q\) = 40° & \(\lfloor P\) 80°. Calculate the value of \(\lfloor C\)

Solution: \(\lfloor R\) = 180-80°-40° = 60°

ΔPQR ≅ ΔABC

\(\lfloor R\) = \(\lfloor C\) = 60°

Karnataka Board Class 9 Maths Chapter 7 MCQs

10. Given ΔOAP =ΔOBP in the figure below. Prove the criteria by which the triangles are congruent.

Solution: OA = OB (given)

OP = OP (common)

\(\lfloor\mathrm{AOP}\) = \(\lfloor\mathrm{BOP}\) (given)

∴ ΔOAP ≅ ΔOBP (bySAS)

Two Mark Questions

1. In the figure below, ABC is a Δle in which AB = AC, x & y are points on AB & AC. Such that AX = AY. Prove that ΔABY ≅ ΔACX

Solution: In ΔABY & ΔACX

AB=AC (given)

AY=AX (given)

\(\lfloor A\) = \(\lfloor A\) (common)

ΔABY = ΔACX (by SAS rule)

2. In the figure below ABCD A is a square & P is the midpoint of AD. BP & CP are joined. Prove that \(\lfloor\mathrm{PCB}\) = \(\lfloor\mathrm{PBC}\)

Solution: In ΔPAB &ΔPDC

PA= PD

(P is the mid-point of AD)

AB = CD

\(\lfloor\mathrm{PAB}\) = \(\lfloor\mathrm{PDC}\)= 90°

ΔPAB ≅ ΔPDC (by RHS rule)

PB = PC (C.P.C.T.)

=> \(\lfloor\mathrm{PCB}\) = \(\lfloor\mathrm{PBC}\)

3. In the figure below, the diagonal AC of quadrilateral ABCD, A bisects \(\lfloor\mathrm{BAD}\)& \(\lfloor\mathrm{BCD}\) .
Prove that BC = CD

Solution: In ΔADC & ΔABC

AC = AC (common)

\(\lfloor\mathrm{DAC}\) = \(\lfloor\mathrm{BAC}\) (given)

\(\lfloor\mathrm{DCA}\) = \(\lfloor\mathrm{BCA}\) (given)

Hence ΔADC ≅ ΔABC (by AAS rule)

=> CD = BC (byCPCT)

4. PS is an altitude of an isosceles Δle PQR in which PQ = PR. Show that PS bisects \(\lfloor P\)

Solution: In ΔPQS & ΔPRS

PQ = PR (given)

PS = PS (common)

\(\lfloor\mathrm{PSQ}\) = \(\lfloor\mathrm{PSR}\) = 90°

ΔPQS ≅ ΔPRS (byR.H.S. rule)

\(\lfloor\mathrm{QPS}\) = \(\lfloor\mathrm{RPS}\) (CPCT)

Hence, PS bisects \(\lfloor P\)

5. In the figure, given AC > AB & AD is the bisector of \(\lfloor A\). Show that \(\lfloor\mathrm{ADC}\) > \(\lfloor\mathrm{ADB}\)

Solution:

In ΔABC

AC>AB (given)

∴ \(\lfloor\mathrm{ABC}\) >\(\lfloor\mathrm{ACB}\) (angles opposite to larger side is greater)

\(\lfloor\mathrm{ABC}\) + \(\lfloor 1\) > \(\lfloor\mathrm{ACB}\) + \(\lfloor 1\) (adding \(\lfloor 1\) on both sides)

\(\lfloor\mathrm{ABC}\)+ \(\lfloor 1\) > \(\lfloor\mathrm{ACB}\) +\(\lfloor 2\) (AD bisects \(\lfloor A\),\(\lfloor 1\) =

\(\lfloor 2\))

∴ \(\lfloor\mathrm{ADC}\)>\(\lfloor\mathrm{ADB}\)

(Exterior angle property of Δle)

6. In figure, AD ⊥ CD & BC ⊥ CD & If AQ = BP & DP = CQ. Prove that
\(\lfloor\mathrm{DAQ}\) = \(\lfloor\mathrm{CBP}\)

Solution: DP = CQ

=> DP + PQ = CQ + PQ (adding PQ to both side)

=> DQ = CP

In right Δles ADQ&BCP

\(\lfloor\mathrm{ADQ}\) = \(\lfloor\mathrm{BCP}\) = 90°

Hyp. AQ = Hyp. BP (given)

side DQ = side CP

ΔADQ ≅ ΔBCP (by RHS rule)

∴ \(\lfloor\mathrm{DAQ}\)– \(\lfloor\mathrm{CBP}\) (C.P.C.T.)

Class 9 Maths Triangles KSEEB Guide

7. Angles A, B & C of a triangle ABC are equal to each other, prove that ΔABC is equilateral Solution: \(\lfloor A\)= \(\lfloor B\)

=> BC = CA (sides opp. to equal angles of AABC) ………………(1)

\(\lfloor B\)=\(\lfloor C\)

=> CA=AB ……………(2) (sides opp. to equal angles of ΔABC)

\(\lfloor C\)= \(\lfloor A\)

=> AB = BC ………………(3) (sides opp. to equal angles of ΔABC)

from(l), (2) & (3) we get

AB = BC = CA

=> ΔABC is equilateral.

Three Mark Questions

1. In figure PQRS is a square & SRT is an equilateral triangle. Prove that
1) PT = QT
2) \(\lfloor\mathrm{TQR}\) =15°

Solution: 1) SRT is an equilateral Δle.

\(\lfloor\mathrm{PSR}\) = 90°, \(\lfloor\mathrm{TSR}\) – 60°

=> \(\lfloor\mathrm{PSR}\) + \(\lfloor\mathrm{TSR}\)=150°

||ly \(\lfloor\mathrm{QRT}\) = 150°

In ΔPST & ΔQRT

PS = QR

\(\lfloor\mathrm{PST}\) = \(\lfloor\mathrm{QRT}\) = 150°

ST = RT (sides of equilateral Δle)

ΔPST ≅ ΔQRT (by SAS rule)

=> PT=QT (byCPCT)

2) In ΔTQR

Square & equilateral Δle on same base)

=> \(\lfloor\mathrm{TQR}\) = \(\lfloor\mathrm{QTR}\) = x

∴ x + x+\(\lfloor\mathrm{QRT}\)= 180°

∴ 2x+150 = 180°

=> x=(180-150)/2

=> \(\lfloor\mathrm{TQR}\) = x =30/2=15°

2. In the figure, prove that CD+DA+AB+BC > 2AC.

Solution: In ΔABC

AB + BC > AC …………….(1) (as sum of two sides is greater than the 3rd side)

In ΔACD

CD + DA > AC …………….(2) (as sum of two sides is greater than the 3rd side)

adding (1) & (2), we get

CD + DA +AB + BC > 2AC

3. PQR is a Δle in which PQ = PR. S is any point on the side PQ, through S, a line is drawn parallel to QR intersecting PR at T. Prove that PS = PT.

Solution: PQ = PR

\(\lfloor\mathrm{PQR}\) = \(\lfloor\mathrm{PRQ}\)(angles opp. to p equal sides are equal)

ST || QR

=> \(\lfloor\mathrm{PST}\) = \(\lfloor\mathrm{PQR}\)(corresponding angles)

∴ \(\lfloor\mathrm{PTS}\) = \(\lfloor\mathrm{PRQ}\)(corresponding angles)

\(\lfloor\mathrm{PST}\) = \(\lfloor\mathrm{PTS}\)

=> PS = PT (sides opp. to equal angles are equal)

4. In the given figure, AD = BD, prove that BD < AC

Solution:

AD = BD

\(\lfloor\mathrm{ABD}\) = \(\lfloor\mathrm{DAB}\) = 59° (anglesopp. to equal sides are equal)

In ΔABD

59° + 59° + \(\lfloor\mathrm{ADB}\) = 180°

\(\lfloor\mathrm{ADB}\) = 180 – 118 = 62°

\(\lfloor\mathrm{ACD}\) = 62 – 32 = 30° (exterior angle is equal to the sum of interior opp. angles)

In ΔABD, AB > BD

(sides opp. to greater angle is the longest)

Also in ΔABC AB < AC

BD < AC

5. In figure, OA = OD & \(\lfloor 1\) = \(\lfloor 2\) Prove that ΔOCBis an isosceles Δle.

Solution: \(\lfloor 1\) + \(\lfloor 3\) = 180° (linear pair)

\(\lfloor 2\)+ \(\lfloor 4\) = 180° (linear pair)

\(\lfloor 1\) + \(\lfloor 3\) = \(\lfloor 2\) + \(\lfloor 4\)

but \(\lfloor 1\) = \(\lfloor 2\) (given)

=> 12 = 14

In ΔDAC & ΔODB \(\lfloor 3\) = \(\lfloor 4\)

\(\lfloor\mathrm{AOC}\) = \(\lfloor\mathrm{DOB}\) (verticallyopp. angle)

OA = OD (given)

∴ ΔOAC ≅ ΔODB (by AS A rule)

OC = OB (C.P.C.T.)

=> ΔOCB is an isosceles triangle.

6. AD & BC are equal perpendiculars to a line segment AB
(1) show that CD bisects AB.
(2) What is its value?

Solution: AB & CD intersect at O

\(\lfloor\mathrm{AOB}\) = \(\lfloor\mathrm{BOC}\)

(vertically opp. angles)……..(1)

In ΔAOD & ΔBOC

\(\lfloor\mathrm{AOD}\) = \(\lfloor\mathrm{BOC}\) (V.O.A)

\(\lfloor\mathrm{DAO}\) = \(\lfloor\mathrm{CBO}\) = 90° (given)

AD = BC (given)

∴ ΔAOD ≅ ΔBOC (by AAS rule)

=> OA = OB

ie, O is the mid point of AB Hence CD bisects AB.

2) Equality is the sign of democracy.

7. l & m are two parallel equal lines intersected by another pair of parallel lines p & q. Show that ΔABC ≅ ΔCD A

Solution: l and m are two parallel equal lines intersected by another pair of parallel lines p & q.

AD || BC

& AB || CD

=> ABCD is a parallelogram

i.e.,AB = CD

& BC = AD

Now in ΔABC & ΔCDA

AB = CD

BC =AD

& AC = AC

∴ ΔABC ≅ ΔCDA (by SSS rule)

Four Mark Questions

1. If two parallel lines are intersected by a transversal then prove that bisectors of the interior angles form a rectangle

Solution: \(\lfloor\mathrm{AGH}\) = \(\lfloor\mathrm{HGD}\) (Alternate angle)

 

=> 1/2\(\lfloor\mathrm{AGH}\) = 1/2\(\lfloor\mathrm{GHD}\)

=> \(\lfloor 1\) =\(\lfloor 2\)

GM II LH

|| ly GL || MH

=> GMHL is a parallelogram

\(\lfloor\mathrm{BGH}\) + \(\lfloor\mathrm{GHD}\) = 180°

=> 1/2\(\lfloor\mathrm{BGH}\) + 1/2\(\lfloor\mathrm{GHD}\) = 90°

=> \(\lfloor 3\) + \(\lfloor 2\) = 90°

In ΔGLH

\(\lfloor\mathrm{GLH}\) = 180 — (\(\lfloor 2\)+ \(\lfloor 3\)) =180-90

\(\lfloor\mathrm{GLH}\) = 90°

=> \(\lfloor\mathrm{GMH}\) = 90°

So \(\lfloor\mathrm{MGL}\) + \(\lfloor\mathrm{GLH}\) = 180°

=>\(\lfloor\mathrm{MGL}\) +90°= 180°

=> \(\lfloor\mathrm{MGL}\) = 90°

∴ MGLH is a rectangle.

KSEEB Class 9 Maths Chapter 7 Exercise Solutions 

2. In figure, the sides AB & AC of AABC are produced to points E&D respectively. If bisectors BO & CO of \(\lfloor\mathrm{CBE}\) & \(\lfloor\mathrm{BCD}\) respectively meet at a point O, then prove that
\(\lfloor\mathrm{BOC}\) = 90°-1/2\(\lfloor\mathrm{BAC}\)

Solution: As \(\lfloor\mathrm{ABC}\) & \(\lfloor\mathrm{CBE}\) form a linear pair.

\(\lfloor\mathrm{ABC}\) + \(\lfloor\mathrm{CBE}\) = 180°

BO is the bisector of \(\lfloor\mathrm{CBE}\)

\(\lfloor\mathrm{CBE}\)= 2\(\lfloor 1\)\(\lfloor\mathrm{ABC}\)+ 2\(\lfloor 1\) = 180°

=> 2\(\lfloor 1\) = 180°-\(\lfloor\mathrm{ABC}\)

=> \(\lfloor 1\) = 90°-1/2\(\lfloor\mathrm{ABC}\) ……………..(1)

Again, \(\lfloor\mathrm{ACB}\)& \(\lfloor\mathrm{BCD}\) form a linear pair.

\(\lfloor\mathrm{ACB}\) + \(\lfloor\mathrm{BCD}\) = 180°

As CO if the bisector of \(\lfloor\mathrm{BCD}\)

\(\lfloor\mathrm{BCD}\) = 2\(\lfloor 2\)

\(\lfloor\mathrm{ACB}\) + 2\(\lfloor 2\) = 180°

2\(\lfloor 2\) = 180°-\(\lfloor\mathrm{ACB}\)

\(\lfloor 2\) = 90°-1/2\(\lfloor\mathrm{ACB}\) ……………(2)

In ΔOBC, we have

\(\lfloor 1\) + \(\lfloor 2\) + \(\lfloor\mathrm{BOC}\) = 180° …………….(3)

from (1), (2) & (3) we have

90- 1/2\(\lfloor\mathrm{ABC}\) + 90-1/2\(\lfloor\mathrm{ACB}\) + \(\lfloor\mathrm{BOC}\) = 180° ……………….(4)

Now in AABC we have \(\lfloor A\)+\(\lfloor B\) + \(\lfloor C\)=180°

\(\lfloor B\)+ \(\lfloor C\)=180°-\(\lfloor A\) …………………(5)

from equations (4) & (5), we have

180 – 1/2(180°-\(\lfloor A\)) + \(\lfloor\mathrm{BOC}\) = 180°

=> \(\lfloor\mathrm{BOC}\) = 180-180 +1/2(180-\(\lfloor A\))

\(\lfloor\mathrm{BOC}\) = 1/2(180-\(\lfloor A\))

=> \(\lfloor\mathrm{BOC}\)=90-1/2\(\lfloor A\)

3. In figure, ABCD is a square & EF is parallel to diagonal BD & EM = FM. Prove that
1) DF = BE
2) AM bisects \(\lfloor\mathrm{BAD}\)

Solution:

1) EF || BD

=> \(\lfloor 1\) = \(\lfloor 2\) [corresponding angles ]

\(\lfloor 3\) = \(\lfloor 4\)

but\(\lfloor 2\)= \(\lfloor 4\)

∴ \(\lfloor 3\) = \(\lfloor 1\)

=> FC = EC (sides opp. to equal angle)

CD – FC = CB – CE

=>  DF = BE

AD =AB

\(\lfloor D\) = \(\lfloor B\) = 90°

∴ ΔADF = ΔABE(by SAS rule)

=> AF=AE (C.P.C.T.)

\(\lfloor 5\) = \(\lfloor 6\)

2) In ΔAMF & ΔAME

AF = AE

AM = AM (common)

FM = EM (given)

ΔAMF = ΔAME (by SSS rule)

∴  \(\lfloor 7\)= \(\lfloor 8\) (by CPCT)

\(\lfloor 7\) + \(\lfloor 5\) = \(\lfloor 8\)+ \(\lfloor 6\)

=> \(\lfloor\mathrm{MAD}\) = \(\lfloor\mathrm{MAB}\)

=> AM bisects \(\lfloor\mathrm{BAD}\)

4. Show that the difference of any two sides of a triangle is less than the third side.

Solution:

Construction: Take BD = AB, join AD

Proof: In ΔABD

AB = BD

=> \(\lfloor 1\) = \(\lfloor 3\) & \(\lfloor 4\)>\(\lfloor 1\)

=> \(\lfloor 4\) > \(\lfloor 3\)

In ΔADC

\(\lfloor 3\)>\(\lfloor 2\)

∴ \(\lfloor 4\)>\(\lfloor 3\)>\(\lfloor 2\)

=> \(\lfloor 4\)>\(\lfloor 2\)

=> AC > DC

=> AC > BC – BD (BC = BD + CD)

AC> BC – AB  (BD = AB)

BC – AB < AC || ly,  AC – AB < BC

& BC – AC < AB

5. In the figure, BL ⊥ AC, MC⊥LN, AL= CN & BL = CM, prove that ΔABC ≅ ΔNML

Solution: AL = CN (given)

AL + LC = CN + LC

ie AC = LN …………..(1)

In ΔALB & ΔNCM

AL = CN (given)

\(\lfloor\mathrm{ALB}\) = \(\lfloor\mathrm{NCM}\) = 90°

BL = CM (given)

∴ ΔALB & ΔNCM (by SAS rule)

=> AB=NM(CPCT) ……………(2)

In ΔBLC & ΔMCL

BL = MC (given)

\(\lfloor\mathrm{BLC}\)= \(\lfloor\mathrm{MCL}\) = 90°

LC = CL ……………(3)

∴ ΔBLC = ΔMCL (by SAS rule)

=> BC = ML(CPCT)

In ΔABC & ΔNML

AB=NM (proved in (1))

AC = NL (proved in (2))

BC = ML (proved in (3)

∴  ΔABC ≅ ΔNML (by SSS rule)