Question

Find the area of an isosceles trapezoid, if the lengths of its bases are 16 cm, and 30 cm, and the diagonals are perpendicular to each other.

Answer 1

Answer:

$A=225+64+120+120=529cm^2$

Step-by-step explanation:

Let A, B, C and D are the vertices of the trapezoid, so that the trapezoid is ABCD.

Let AB be the longer base of 30 cm and CD be the shorter base of 16 cm.

Let the point O be the intersection of diagonals, Then the triangles  AOB,  BOC,  COD  and  AOD  are right angled triangles.

Now, from ΔAOD and ΔBOC, we have

∠AOD=∠BOC=90°

AD=BC (sides of trapezium)

thus, by RHS rule ΔAOD≅ΔBOC

similarly, ΔDOC≅ΔAOB.

Since the trapezoid ABCD has equal lateral sides, the triangle AOB is isosceles right angled triangle.

And since it base AB is 30 cm,  its sides (the legs)  AO and  BO  are $\frac{30}{\sqrt{2}}cm$long in measure.

Similarly, the triangle COD is isosceles right angled triangle. and since the base CD is 16 cm,  its sides (the legs) CO and  DO  are $\frac{16}{\sqrt{2}}cm$long in measure.

Now, areas of triangles are:

For triangle AOB, the area is:

$A=\frac{1}{2}(\frac{30}{\sqrt{2}})(\frac{30}{\sqrt{2}})=225cm^2$

For triangle COD, the area is:

$A=\frac{1}{2}(\frac{16}{\sqrt{2}})(\frac{16}{\sqrt{2}})=64cm^2$

For triangle BOC, the area is:

$A=\frac{1}{2}(\frac{30}{\sqrt{2}})(\frac{16}{\sqrt{2}})=120cm^2$

For triangle AOD, the area is:

$A=\frac{1}{2}(\frac{30}{\sqrt{2}})(\frac{16}{\sqrt{2}})=120cm^2$

since,  total area of the trapezoid is the sum of areas of triangles, therefore we have

$A=225+64+120+120=529cm^2$

Answer 2

We'll have this following figure drawn in the picture. Since we have an isosceles trapezoid, the diagonals are congruent. According to the conditions of the problem, these diagonals are perpendicular and then we can claim that we'll have right-isosceles triangles in this case. Indeed, ∠OBC = ∠FBD = 45°. Then, in the triangle BFD. BF=FD=23. It means the height is 23 cm. Obtaining this fact, we can easily calculate the area. $A= \frac{16+30}{2}23=529 cm^{2}$