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Find an expression for the area enclosed by quadrilateral ABCD below.
a) 7x+25
b)[tex]35x^{2} +125[/tex]
c)[tex]36x^{2}[/tex]
d) [tex]6x^{2} + 30x[/tex]

PLEASE HELP I WILL GIVE BRAINLIEST AND PLEASE PROVIDE EXPLAINATION TYTY

Find an expression for the area enclosed by quadrilateral ABCD below a 7x25 btex35x2 125tex ctex36x2tex d tex6x2 30xtex PLEASE HELP I WILL GIVE BRAINLIEST AND P class=

Respuesta :

∆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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