Question

Two sides of a quadrilateral are 64 cm and 48 cm in length and the angle between them is 90°. The other two sides are 50 cm each. The other three angles of the quadrilateral are not right angles. What is the area of the quadrilateral?​

Answer 1

Answer:

990 $cm^{2}$

Step-by-step explanation:

Consider the diagram attached,

ABC is right angled,

therefore area of ABC = 48 x 64 / 2 cm sq

                                     = 1532 cm sq

In triangle ADC,

AC = $\sqrt{48^{2} + 64^{2} }$

     = 80 cm

s = (50 + 50 + 80)/2 cm

  = 90 cm

Area of ADC = $\sqrt{90 * (90 - 50) * (90 - 50) * (90 - 80)}$ cm sq

                     = $\sqrt{90 * 30 * 30 * 10}$ cm sq

                     = $\sqrt{810000}$ cm sq

                     = 900 cm sq

Total area = area of ABC + area of ADC

                 = (900 + 90) cm sq

                 = 990 cm sq

Hope it helps!!!

Answer 2

Answer:

Consider the diagram attached,

ABC is right angled,

therefore area of ABC = 48 x 64 / 2 cm sq

= 1532 cm sq

In triangle ADC,

AC = \sqrt{48^{2} + 64^{2} }

48

2

+64

2

= 80 cm

s = (50 + 50 + 80)/2 cm

= 90 cm

Area of ADC = \sqrt{90 * (90 - 50) * (90 - 50) * (90 - 80)}

90∗(90−50)∗(90−50)∗(90−80)

cm sq

= \sqrt{90 * 30 * 30 * 10}

90∗30∗30∗10

cm sq

= \sqrt{810000}

810000

cm sq

= 900 cm sq

Total area = area of ABC + area of ADC

= (900 + 90) cm sq

= 990 cm sq

Hope it helps!!!

Step-by-step explanation:

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