Mathematics, 19.07.2019 14:00 barbar12345

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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Answer from: Imamdiallo18

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

Find the area of an isosceles trapezoid, if the lengths of its bases are 16 cm, and 30 cm, and the d

Answer from: Quest

13=7.8514=3.5feet15=10.99feet16=2& 317=418=about 9ft19=7.8520=4in

step-by-step explanation:

Ineed y’all to me with 13-20 so i can get this done early

Answer from: Quest

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

$ab^2+ac^2=bc^2$

$ab^2+ac^2=x^2$ )

triangle abd and cba are similar,

$\frac{ab}{bc}=\frac{bd}{ab}$

$ab^2=bc\cdot bd$

$25=bc\cdot bd$ (2)

triangle cad and cba are similar,

$\frac{ac}{bc}=\frac{cd}{ac}$

$ac^2=bc\cdot dc$

$7\times 7=bc\cdot dc$

$49=bc\cdot dc$ )

using (1), (2) and (3), we get

$25+49=x^2$

$74=x^2$

hence proved.

therefore option 2 is correct.

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