Question

class 9th maths ch7 ex 7.1 que:7/8​

Answer 1

Explanation:

(i) In ∆ ABC and ∆ BAC,

AD = BC (Given)

∠DAB = ∠CBA (Given)

AB = AB (Common)

∴ ∆ ABD ≅ ∆BAC (By SAS congruence)

(ii) Since ∆ABD ≅ ∆BAC

⇒ BD = AC [By C.P.C.T.]

(iii) Since ∆ABD ≅ ∆BAC

⇒ ∠ABD = ∠BAC [By C.P.C.T.]

hi

Answer 2

Explanation:

3x

3

y

2

×(2x−3y)

=3x

3

y

2

⋅2x−3x

3

y

2

⋅3y

=6x

4

y

2

−9x

3

y

3

Now, for x=−1,y=2, we have

L.H.S.-

3x

3

y

2

×(2x−3y)

=3(−1)

3

(2)

2

(2⋅(−1)−3⋅(2))

=3⋅(−1)⋅4×(−2−6)

=(−12)×(−8)

=96

R.H.S.-

6x

4

y

2

−9x

3

y

3

=6(−1)

4

(2)

2

−9(−1)

3

(2)

3

=6⋅1⋅4−9⋅(−1)⋅8

=24+72

=96

∵ L.H.S. = R.H.S.

Hence verified.