Telangana SCERT Class 9 Math Solution Chapter 7 Triangles Exercise 7.1

Telangana SCERT Solution Class IX (9) Math Chapter 7 Triangles Exercise 7.1

1.)

Given,

AB = AD

AB bisects ∠A

∴ ∠BAC = ∠BAD

NOW,

FROM △ ABC and △ABD

i) Sides AC = AD

ii) ∠BAC = ∠BAD

iii) AB IS COMMON SIDE.

∴ △ABC ≅ △ABD by SAS

∴ BC = BC [∵△ABC ≅ △ABD]

2.)

Given,

AD = BC

∠DAB = ∠CBA

∴ In △ABC and △ABD

i) AD = BC (given)

ii) ∠DAB, ∠CBA (″)

iii) AB is common side.

∴ i) △ ABC ≅ △ABD by SAS.

ii) BD = AC [∵Corresponding side of congruent – triangles are equal]

(iii) ∠ABD = ∠ BAC [∵ Corresponding angels of congruent △ are equal]

3.)

Given,

AD > BC

AB ⊥ AD

AB ⊥ BC

∴ From the given information,

BC is parallel to AD Sine both are ⊥ to the same line segment –AB.

In △ ADO and △BCD

∴ i) ∠ADO and ∠BCD [alternate angels]

ii) ∠OAD = ∠ OBC = 90°

iii) BC = AD [given]

△ADO ≅ △ BCO by RHS.

∴ AO = DB [∵△ADO≅ △BCD]

Here we can conclude that – CD bisect – AB.

4.)

Given,

l ∥ m

P ∥ q

∴ From the given into,

In △ABC and △ACD

i) ∠ACB = ∠CAD [alternate angels]

ii) ∠ ACD = ∠ BAC [‘’ ‘’ ‘’]

iii) AC is common side.

∴ △ABC ≅ △ACD by ASA

5.)

Given,

AC = AE, AB = AD

∠BAD = ∠EAC

Now,

∠BAD + ∠DAC = ∠EAC+∠DAC [∵∠BAD = ∠ EAC

∴ ∠BAC = ∠EAD

∴ ∠In △ ABC and △ EAD

i) ∠BAC = ∠EAD

ii) AB = AD

iii) AE = AC

∴ ABC ≅ △EAD by SAS

∴ BC = DE since corresponding side of congruent – △s are equal.

6.)

Given,

∠ACB = 90°

M is midpoint of AB

∴ AM = BM

DM = CM

i) In △AMC and △BMD

i) DM = CM

ii) AM = BM

iii) ∠BMD = ∠AMC [vertically opposite angels]

∴ △AMC ≅ △BMD by SAS

ii) In △BDC and △ABC

i) ∠BDM= ∠CAM [∵△AMC ≅ △BMD]

ii) AC = BD [∵ ‘’ ‘’ ‘’]

iii) BC is common side…..

 

iii) ∴ △ABC ≅ △BDC by SAS

ii) ∴ ∠DBC = ∠ACB = 90° [∵△ABC ≅ △BDC]

∴ AB = DC [∵△ABC ≅ △BDC]

∴Now,
CM = ½ DC [∵DM = CM]

iv) CM = ½ AB [∵DC = AB]

7.)

Given,

ABCD is a square.

AB = BC = CD = AD

∴∠DAB = ∠ABC = ∠BCD = ∠CDA = 90°

△APB is an equilateral △

∴ AP = PB = AB

∠ABP = ∠BPA = ∠PAB =60°

In △APD and △BPC

i) AB = PB [△APB equilateral △]

ii) AD = BC [∴ ABCD is a Square]

iii) ∠DAP = ∠DAB – ∠PAB

= 90° – 60° = 30°

∠PBC = ∠ABC – ∠ABP

= 90° – 60°

= 30°

∴ ∠DAP = ∠PBC

∴ △APD ≅△BPC by SAS

8.)

 Given,

DE = DF

D midpoint of BC

∴ BD = CD

DE ⊥ AB

∴ DE ⊥ BE

DF ⊥ AC

∴ DP ⊥ CF

∴ In △ BED and △CFD

i) DE = DF

ii) BD = CD

iii) ∠BED = ∠CFD = 90° [∵ DE ⊥ BE, DF ⊥CF]

∴ △ BED ≅ △CFD by RHS

9.)

From the given information,

Draw △ABC where

AD Bisects ∠A and side is C

Now,

Draw a parallel line from point – C parallel to AD

Extend BA so that it meets the parallel line from C at point E.

Now,

∠BAD = ∠DAC

BD = CD

∴∠BAD = ∠ABC [corresponding angel]

∠DAC = ∠ACE [alternate angels]

∴ AE = AC [∴ Sides opposite to equilateral of a △are equal]

Now,

In △ BCE, AD ∥ CE

D is the midpoint of BC

∴ A is the midpoint of BE by converse of midpoint theorem

∴ AB = AE

∴ AB = AC [∵ AE = AC]

∴ △ ABC is an isosceles triangle.

10.)

Given,

∠B right angle.

∠BCA = 2∠BAC.

Produce CB to D so that

CB = BD

∴ In △ABC and △ABD

i) BC = BD

ii) AB common side

iii) ∠ABC = ∠ABD = 90° [∴∠ABD is collinear angle to ∠ABC = 180° – 90° = 90°]

∴ △ABC ≅ △ABD by SAS.

Let,

∴ ∠CAB = ∠DAB = x [∴△ADC ≅ △ACD]

∴ ∠BCA = 2x

∴ ∠ CAD = ∠CAB +∠DAB = 2x

∴ ∠CAD = ∠BAC = 2x

∴ CD = AD [∴ Opposite sides of equal angels are equal]

AD = AC [∴△ABC ≅△ABD]

∴ CD = AC

∴ 2BC = AC [∵B midpoint – of CD]

Proved.


Updated: September 21, 2021 — 4:38 pm

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