Question

Please answer this question it is urgent. ​

Answer 1

Step-by-step explanation:

Since the diagnolas of a rhombus bisect each other and at the same time, also at an angle of 90°, then I can say that, OD= 8 cm,

And side of the rhombus, DC= 10 cm.

Now, I can see the ∆ OCD, where I will find the value of OC, by Pythagoras Theorem.

So, Pythagoras Theorem States that,

• (DC)²-(OD)²= (OC)².
• THEN, (10)²-(8)²= (OC)².
• SO, OC= 6 cm.

Now, I have the values of the diagnolas of this rhombus, which are 12 cm, and 16 cm.

So I use the formula for solving the area of any rhombus, which is= 1/2× product of diagnolas.

then, area of this rhombus= 1/2× 6× 8,

=24 cm².

Now, I know that, CD= BC, since they are the sides of a rhombus.

So, I focus on the trapezium BCQP, where the parallel sides are 10 cm and 20 cm.

So, I use the formula for the area of any trapezium, that is,

1/2×(sum of the parallel sides)× perpendicular height.

Then, Area of this trapezium = 1/2× (10+20)× 8,

= 120 cm².

Now, I also can say that, BC= CD, because these are the sides of a trapezium.

So, now I focus on the trap. DCYX, where the parallel sides are 10 cm and 16 cm.

Here also, I use the same formula for the area of any trapezium,

Then, area of this trapezium= 1/2× (10+16)× 6 cm,

= 84 cm².

So now, if a add all the areas, I will get the area of the whole figure, that is ADXYQ,

so, total area = (84+120+24) cm².

= 224 cm².

NOW, THAT'S THE CORRECT ANSWER.