Ncert Solutions For Class 7 Maths Ex 7.1

Ncert Solutions For Class 7 Maths Chapter 7 Ex 7.1

 

1. In quadrilateral ACBD, AC = AD and AB bisects A . Show that ΔABC ΔABD. What can you say about BC and BD?

1

Sol.  

Given that,
AC = AD and AB bisects A
To prove that,
ΔABC ΔABD
Proof,
In ΔABC and ΔABD,
AB = AB
AC = AD
CAB =DAB (AB is bisector)
Therefore, ΔABC   ΔABD by SAS congruence condition.
BC and BD are of equal length.

 

2. ABCD is a quadrilateral in which AD = BC and DAB = CBA . Prove that

(i) ΔABD ΔBAC
(ii) BD = AC
(iii) ABD = BAC.

2

Sol.

Given that ,
AD = BC and DAB = CBA

(i) In ΔABD and ΔBAC,
AB = BA (Common)
DAB = CBA (Given)
AD = BC (Given)
Therefore, ΔABD ΔBAC by SAS congruence condition.
(ii) Since, ΔABD ΔBAC
Therefore BD = AC by CPCT
(iii) Since, ΔABD ΔBAC
Therefore ABD BAC by CPCT

 

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

3    

Sol. 

Given that,
AD and BC are equal perpendiculars to AB.
To prove,
CD bisects AB
Proof,
In ΔAOD and ΔBOC,
A = B (Perpendicular)
AOD = BOC (Vertically opposite angles)
AD = BC (Given)
Therefore, ΔAOD ΔBOC by AAS congruence condition.
Now,
AO = OB (CPCT). CD bisects AB.

 

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

4

Sol. 

Given that,
l m and p q
To prove,
ΔABC ≅ ΔCDA
Then Proof that,
In ΔABC and ΔCDA,
BCA =DAC (Alternate interior angles)
AC = CA (Common)
BAC = DCA (Alternate interior angles)
Therefore, ΔABC ΔCDA by ASA congruence condition.

 

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 . Show that:

(i) ΔAPB ΔAQB

(ii) BP = BQ or B is equidistant from the arms of A.              5 

Sol.

Given,
l is the bisector of an angle A.
BP and BQ are perpendiculars.

(i) In ΔAPB and ΔAQB,
P = Q (Right angles)
BAP =BAQ (l is bisector)
AB = AB (Common)
Therefore, ΔAPB ΔAQB by AAS congruence condition.
(ii) BP = BQ by CPCT. Therefore, B is equidistant from the arms of A.

 

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

6 

Sol.

Given that,
AC = AE, AB = AD and BAD = EAC
To show,
BC = DE
then Proof that,
BAD = EAC (Adding DAC both sides)
BAD + DAC = EAC + DAC
BAC = EAD
In ΔABC and ΔADE,
AC = AE (Given)
BAC = EAD
AB = AD (Given)
Therefore, ΔABC  ΔADE by SAS congruence condition.
BC = DE by 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 . Show that
(i) ΔDAP ΔEBP
(ii) AD = BE

7

Sol.

Given,
P is mid-point of AB.
BAD = ABE and EPA = DPB

(i) EPA =DPB (Adding DPE both sides)
EPA + DPE = DPB + DPE
DPA = EPB
In ΔDAP ΔEBP,
DPA = EPB
AP = BP (P is mid-point of AB)
BAD =ABE (Given)
Therefore, ΔDAP ΔEBP by ASA congruence condition.
(ii) AD = BE by CPCT.

 

8. A 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. Show that:

(i) ΔAMC ΔBMD

(ii) DBC is a right angle.

(iii) ΔDBC ΔACB

(iv) CM = 1/2 AB

8

Sol.

Given,
C = 90°, M is the mid-point of AB and DM = CM

(i) In ΔAMC and ΔBMD,
AM = BM (M is the mid-point)
CMA = DMB (Vertically opposite angles)
CM = DM (Given)
Therefore, ΔAMC ΔBMD by SAS congruence condition.

(ii) ACM = BDM (by CPCT)
Therefore, AC || BD as alternate interior angles are equal.
Now,
ACB + DBC = 180° (co-interiors angles)
90° + B = 180°
DBC = 90°

(iii) In ΔDBC and ΔACB,
BC = CB (Common)
ACB = DBC (Right angles)
DB = AC
Therefore, ΔDBC ΔACB by SAS congruence condition.

(iv)  DC = AB (ΔDBC ΔACB)
DM = CM = AM = BM (M is mid-point)
DM + CM = AM + BM
CM + CM = AB
CM = 1/2AB

 

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