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Sagot :
∆ABD is right angled hence area:-
[tex]\\ \sf\longmapsto \dfrac{1}{2}bh[/tex]
[tex]\\ \sf\longmapsto \dfrac{1}{2}(4x)(3x)[/tex]
[tex]\\ \sf\longmapsto \dfrac{1}{2}(12x^2)[/tex]
[tex]\\ \sf\longmapsto 6x^2[/tex]
There is only one option containing 6x^2 i.e Option D.
Hence without calculating further
Option D is correct
Answer:
• Area of a triangle:
[tex]{ \boxed{ \rm{area = \frac{1}{2} \times base \times height }}} \\ [/tex]
For triangle BDC:
- height = √[(4x)² + (3x)²] = 5x
- base = 12
→ therefore:
[tex]{ \tt{area = \frac{1}{2} \times 12 \times 5x }} \\ \\ = { \tt{6 \times 5x}} \\ \\ = { \tt{30x}}[/tex]
For triangle ABD:
[tex]{ \tt{area = \frac{1}{2} \times 4x \times 3x }} \\ \\ = { \tt{2x \times 3x}} \\ \\ = { \tt{6 {x}^{2} }}[/tex]
Total area = area of BDC + area of ABD
[tex]{ \rm{area = 6 {x}^{2} + 30x}}[/tex]
Answer: Objective D
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