Ncert Solutions For Class 7 Maths Ex 7.1

 

1. In quadrilateral ACBD, AC = AD and AB bisects $\angle$A . Show that $\Delta$ABC $\cong$ $\Delta$ABD. What can you say about BC and BD?

Sol.  

Given that,
AC = AD and AB bisects $\angle$A
To prove that,
$\Delta$ABC $\cong$ $\Delta$ABD
Proof,
In $\Delta$ABC and $\Delta$ABD,
AB = AB
AC = AD
$\angle$CAB =$\angle$DAB (AB is bisector)
Therefore, $\Delta$ABC $\cong$  $\Delta$ABD by SAS congruence condition.
BC and BD are of equal length.

 

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

(i) $\Delta$ABD $\cong$ $\Delta$BAC
(ii) BD = AC
(iii) $\angle$ABD = $\angle$BAC.

Sol.

Given that ,
AD = BC and $\angle$ DAB = $\angle$CBA

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

 

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

    

Sol. 

Given that,
AD and BC are equal perpendiculars to AB.
To prove,
CD bisects AB
Proof,
In $\Delta$AOD and $\Delta$BOC,
$\angle$A = $\angle$B (Perpendicular)
$\angle$AOD = $\angle$BOC (Vertically opposite angles)
AD = BC (Given)
Therefore, $\Delta$AOD $\cong$ $\Delta$BOC 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 $\Delta$ABC $\cong$ $\Delta$CDA.

Sol. 

Given that,
l $\parallel$ m and p $\parallel$ q
To prove,
$\Delta$ABC ≅ $\Delta$CDA
Then Proof that,
In $\Delta$ABC and $\Delta$CDA,
$\angle$BCA =$\angle$DAC (Alternate interior angles)
AC = CA (Common)
$\angle$BAC = $\angle$DCA (Alternate interior angles)
Therefore, $\Delta$ABC $\cong$ $\Delta$CDA by ASA congruence condition.

 

5. Line l is the bisector of an angle $\angle$A and B is any point on l. BP and BQ are perpendiculars from B to the arms of $\angle$A . Show that:

(i) $\Delta$APB $\cong$ $\Delta$AQB

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

Sol.

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

(i) In $\Delta$APB and $\Delta$AQB,
$\angle$P = $\angle$Q (Right angles)
$\angle$BAP =$\angle$BAQ (l is bisector)
AB = AB (Common)
Therefore, $\Delta$APB $\cong$ $\Delta$AQB by AAS congruence condition.
(ii) BP = BQ by CPCT. Therefore, B is equidistant from the arms of $\angle$A.

 

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

 

Sol.

Given that,
AC = AE, AB = AD and $\angle$BAD = $\angle$EAC
To show,
BC = DE
then Proof that,
$\angle$BAD = $\angle$EAC (Adding $\angle$DAC both sides)
$\angle$BAD + $\angle$DAC = $\angle$EAC + $\angle$DAC
$\Rightarrow$ $\angle$BAC = $\angle$EAD
In $\Delta$ABC and $\Delta$ADE,
AC = AE (Given)
$\angle$BAC = $\angle$EAD
AB = AD (Given)
Therefore, $\Delta$ABC $\cong$  $\cong$$\Delta$ADE 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 $\angle$BAD = $\angle$ABE and $\angle$EPA = $\angle$DPB . Show that
(i) $\Delta$DAP $\cong$ $\Delta$EBP
(ii) AD = BE

Sol.

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

(i) $\angle$EPA =$\angle$DPB (Adding $\angle$DPE both sides)
$\angle$EPA + $\angle$DPE = $\angle$DPB + $\angle$DPE
$\Rightarrow$ $\angle$DPA = $\angle$EPB
In $\Delta$DAP $\cong$ $\Delta$EBP,
$\angle$DPA = $\angle$EPB
AP = BP (P is mid-point of AB)
$\angle$BAD =$\angle$ABE (Given)
Therefore, $\Delta$DAP $\cong$ $\Delta$EBP 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) $\Delta$AMC $\cong$ $\Delta$BMD

(ii) $\angle$DBC is a right angle.

(iii) $\Delta$DBC $\cong$ $\Delta$ACB

(iv) CM = 1/2 AB

Sol.

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

(i) In $\Delta$AMC and $\Delta$BMD,
AM = BM (M is the mid-point)
$\angle$CMA = $\angle$DMB (Vertically opposite angles)
CM = DM (Given)
Therefore, $\Delta$AMC $\cong$ $\Delta$BMD by SAS congruence condition.

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

(iii) In $\Delta$DBC and $\Delta$ACB,
BC = CB (Common)
$\angle$ACB = $\angle$DBC (Right angles)
DB = AC
Therefore, $\Delta$DBC $\cong$ $\Delta$ACB by SAS congruence condition.

(iv)  DC = AB ($\Delta$DBC $\cong$ $\Delta$ACB)
$\Rightarrow$ DM = CM = AM = BM (M is mid-point)
$\Rightarrow$ DM + CM = AM + BM
$\Rightarrow$ CM + CM = AB
$\Rightarrow$ CM = 1/2AB