Question

Help me with qs 4b) i) and ii) and Qs 5

Answer 1

Answer:

4 a) In triangle ABD and triangle ADC

angle ADB = angle ADC ( both 90° )

AB = AC ( given )

AD = AD ( common )

triangle ABD congruent triangle ADC ( RHS cong. rule )

So , BD = DC ( C.P.C.T )

BC = 2× BD = 2 × ( 6x - 4 )

Area of triangle = 1/2 × base × alt.

= 1/2 × 2 ( 6x - 4 ) × ( 2x + 7 )

= ( 6x - 4 ) × ( 2x + 7 )

= 12x² + 42x - 8x - 28

= (12x² + 34x - 28 ) units²

Answer 2

Answer:

4 a) In triangle ABD and triangle ADC

angle ADB = angle ADC ( both 90° )

AB = AC ( given )

AD = AD ( common )

triangle ABD congruent triangle ADC ( RHS cong. rule )

So , BD = DC ( C.P.C.T )

BC = 2× BD = 2 × ( 6x - 4 )

Area of triangle = 1/2 × base × alt.

= 1/2 × 2 ( 6x - 4 ) × ( 2x + 7 )

= ( 6x - 4 ) × ( 2x + 7 )

= 12x² + 42x - 8x - 28

= (12x² + 34x - 28 ) units²