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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.

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boffeemadrid

Answer:

[tex]A=225+64+120+120=529cm^2[/tex]

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 [tex]\frac{30}{\sqrt{2}}cm[/tex]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 [tex]\frac{16}{\sqrt{2}}cm[/tex]long in measure.

Now, areas of triangles are:

For triangle AOB, the area is:

[tex]A=\frac{1}{2}(\frac{30}{\sqrt{2}})(\frac{30}{\sqrt{2}})=225cm^2[/tex]

For triangle COD, the area is:

[tex]A=\frac{1}{2}(\frac{16}{\sqrt{2}})(\frac{16}{\sqrt{2}})=64cm^2[/tex]

For triangle BOC, the area is:

[tex]A=\frac{1}{2}(\frac{30}{\sqrt{2}})(\frac{16}{\sqrt{2}})=120cm^2[/tex]

For triangle AOD, the area is:

[tex]A=\frac{1}{2}(\frac{30}{\sqrt{2}})(\frac{16}{\sqrt{2}})=120cm^2[/tex]

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

[tex]A=225+64+120+120=529cm^2[/tex]

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aksnkj

Answer:

The area of the trapezium is 528.3 square centimeter.

Step-by-step explanation:

Consider an isosceles trapezium ABCD as shown in the figure. AB is shorter side, which is equal to 16 cm. CD is larger side, which is equal to 30 cm.

O is the intersection point of the diagonals. The diagonals intersect at right angle as given in the question.

Now, the trapezium is an isosceles type. So, the diagonals with intersect such that AO=BO and CO=DO.

Now, In triangle AOB, Angle O is [tex]90^\circ[/tex] and the sides AO and BO are equal. AB is equal to 16 cm. So, by Pythagorus Theorem,

[tex]AB^2=AO^2+BO^2\\16^2=2AO^2\\AO^2=128\\AO=11.3\rm cm[/tex]

So, AO=BO=11.3 cm.

Similarly, In triangle COD, Angle O is [tex]90^\circ[/tex] and the sides CO and DO are equal. CD is equal to 30 cm. So, by Pythagorus Theorem,

[tex]CD^2=CO^2+DO^2\\30^2=2CO^2\\CO^2=450\\CO=21.21\rm cm[/tex]

So, CO=DO=21.21 cm.

Now, the area of the triangle AOB will be,

[tex]ar(\Delta AOB)=\dfrac{1}{2}AO\times BO\\ar(\Delta AOB)=\dfrac{1}{2}11.3\times 11.3\\\\ar(\Delta AOB)=63.8 \rm cm^2[/tex]

Now, the area of the triangle BOC will be,

[tex]ar(\Delta BOC)=\dfrac{1}{2}CO\times BO\\ar(\Delta BOC)=\dfrac{1}{2}21.21\times 11.3\\\\ar(\Delta BOC)=119.8 \rm cm^2[/tex]

Now, the area of the triangle COD will be,

[tex]ar(\Delta COD)=\dfrac{1}{2}CO\times DO\\ar(\Delta COD)=\dfrac{1}{2}21.21\times 21.21\\\\ar(\Delta COD)=224.9 \rm cm^2[/tex]

Now, the area of the triangle AOD will be,

[tex]ar(\Delta AOD)=\dfrac{1}{2}AO\times DO\\ar(\Delta AOD)=\dfrac{1}{2}21.21\times 11.3\\\\ar(\Delta AOD)=119.8 \rm cm^2[/tex]

Thus, the area of the trapezium ABCD will be,

[tex]ar(ABCD)=ar (\Delta AOB)+ar (\Delta BOD)+ar (\Delta COD)+ar (\Delta AOD)\\ar(ABCD)=63.8+119.8+224.9+119.8\\ar(ABCD)=528.3 \rm cm^2[/tex]

Therefore, the area of the trapezium is 528.3 square centimeter.

For more details, refer the link:

https://brainly.com/question/19276993?referrer=searchResults

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