Question

Find an expression for the area enclosed by quadrilateral ABCD below.

Answer 1

∆ABD is right angled hence area:-

$\\ \sf\longmapsto \dfrac{1}{2}bh$

$\\ \sf\longmapsto \dfrac{1}{2}(4x)(3x)$

$\\ \sf\longmapsto \dfrac{1}{2}(12x^2)$

$\\ \sf\longmapsto 6x^2$

There is only one option containing 6x^2 i.e Option D.

Hence without calculating further

Option D is correct

Answer 2

Answer:

• Area of a triangle:

${ \boxed{ \rm{area = \frac{1}{2} \times base \times height }}}$

For triangle BDC:

• height = √[(4x)² + (3x)²] = 5x
• base = 12

→ therefore:

${ \tt{area = \frac{1}{2} \times 12 \times 5x }} \\ \\ = { \tt{6 \times 5x}} \\ \\ = { \tt{30x}}$

For triangle ABD:

${ \tt{area = \frac{1}{2} \times 4x \times 3x }} \\ \\ = { \tt{2x \times 3x}} \\ \\ = { \tt{6 {x}^{2} }}$

Total area = area of BDC + area of ABD

${ \rm{area = 6 {x}^{2} + 30x}}$

Answer: Objective D