Triangles – Exercise 7.1 – (MATHEMATICS) – 9th Class

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TEXTBOOK  QUESTIONS  SOLVED

Exercise 7.1

 

1. In quadrilateral ABCD, AC = AD and AB bisects ∠A (see figure).
Show that ΔABC ≅ ΔABD.
What can you say about BC and BD?


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Sol. Consider triangles ABC and ABD,
We have          AC = AD                             [Given]

AB = AB                                                       [Common]

and ∠CAB = ∠DAB                                    [  ∴ AB bisects ∠CAD]

ΔABC  ≅  ΔABD                                          [SAS rule]

Therefore, BC = BD.


2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see figure). Prove that
(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.


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Sol. (i) Consider triangles ABD and ABC,
We have AD = BC                      [Given]

AB = BA                                       [ Commo ]

and ∠DAB = ∠CBA                    [ Given ]

∴  ΔABD   ≅   ΔBAC                   [ SAS rule]

(ii) BD = AC.                               [ CPCT ]

(iii) LABD = LBAC,                   [ CPCT ]


3. AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

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Sol. AD and. BC are perpendiculars to AB.
AD\parallel BC  and CD is transversal.

∴  ∠BCD = ∠ADC, i.e.. ∠BCO = ∠ADO                                  [ Alternate agnles ]

Consider triangles BCO and ADO,

We have BC = AD                                                                      [ Given ]

∠ OBC =  ∠OAD                                                                          [90° each ]

and  ∠BCO = ∠ADO                                                                  [ Proved above ]

∴   ΔOBC  ≅  ΔOAD                                                                    [ CPCT ]

Therefore, OB = OA

Hence, CD bisects AB.


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


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Sol. Since l\parallel m and AC is transversal.


So, ∠ 1 = ∠ 2                                                           …..  (i)     [Alternate angles]

Again, p\parallel q and AC is transversal.

So, ∠ 3 = ∠ 4                                                            …(ii)     [Alternate angles]

Now, in ΔADC and ΔABC,

AC = CA                                                                               [ Common ]

∠ 1 = ∠2  and     ∠ 3  =  ∠ 4                                               [ From (i) and (ii) ]

∴    ΔABC  ≅   Δ CDA                                                        [ASA rule ]


5. Line l is the bisector of an angle ∠ A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see figure). Show that:
(i) ΔAPB  ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.


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Sol. Line l is the bisector of ΔQAP.
∴       ∠QAB =  ∠PAB                                                             …(i)

AB = BA                                                                                   …(ii) [Common]

and ∠BQA = ∠BPA.                                                               …(iii) [90° each]

(i) In triangles AQB and APB,

∠QAB = ∠PAB ; AB = BA and ∠BQA = ∠BPA.                      [From (i), (ii), (iii)]

∴    ∠APB  ≅  ∠AQB.                                                                   [AAS rule]

(ii) Therefore,      BP = BQ                                                         [CPCT]

i.e., B is equidistant from the arms of ∠A.


6. In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.


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Sol. As ∠BAD = ∠CAE                                      [Given]

⇒       ∠BAD + ∠DAC

= ∠DAC + ∠CAE

[∠ DAC is added to both sides]

⇒  ∠BAC = ∠DAE

Consider triangles BAC and DAE,

We have     AB = AD                                               [ Given ]

AC =AE                                                                     [ Given ]

and ∠BAC = ∠DAE                                                [ SAS rule ]

⇒ BC = DE.                                                             [ CPCT ]


7. AB is a line segment and P is its mid-point. D and E are points on. the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see figure). Show that
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE.


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Sol. Since    ∠EPA = ∠DPB                                                   [Given]

⇒ ∠EPA + ∠EPD = ∠EPD + ∠DPB                                      [Adding ∠EPD to both sides]

⇒     ∠APD = ∠BPE                                                                  …(i)

Also, AP = BP .                                                                          ..(ii) [Given]

and ∠DAP = ∠EBP                                                                    …(iii) [Given]

(i) Consider triangles DAP and EBP,

∠APD = ∠BPE, AP = BP and ∠DAP = ∠EBP.

[From (i), (ii), (iii)]

ΔDAP ≅  ΔEBP.                                   [SAS rule]

(ii)    AD = BE                                       [ CPCT ]


8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that:

(i)  ΔAMC  ≅  ΔBMD

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Sol. –  Consider triangles AMC and DMB,
We have AM = BM                                                             [Given]

CM = DM                                                                             [Given]

and ∠AMC = ∠BMD                                                          [Vertically opposite angles]

∴       Δ AMC  ≅  Δ BMD                                                                [SAS rule]


(ii) ∠ DBC is a right angle.

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Sol. –  As ∠BAC = ∠DBA                                                    [CPCT, from part (i)]

and AB is the transversal.

So,  DB\parallel AC                                                 …(i)

We have AC \bot BC                                                  …(ii) [Given]

So,  DB \bot BC                                                         [From (1) and (ii)]

i.e., ∠ DBC is a right angle, i.e., ∠ DBC = 90°.


(iii) Δ DBC  ≅ ΔACB

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Sol. –  Consider triangles ABC and DCB.

We have AC = DB                                   [Since ΔAMC  ≅ Δ BMD; result (i)]

BC = CB                                                    [Common]

and ∠ACB = ∠DBC                                 [Each 90° ]

∴    ΔDBC  ≅  ΔACB                                [SAS rule]


(iv) CM = {1 \over 2}AB

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Sol. – As               DC = AB                         [CPCT from part (iii)]

⇒             2CM = AB                                   [ ∴ M is mid-point of DC]

⇒             CM = {1 \over 2}AB

 

 

 

 

 

 


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