Question

An aerial camera is suspended from a blimp and positioned at D. The camera needs to cover 125 meters of ground distance. If the camera hangs 10 meters below the blimp and the blimp attachment is 20 meters in length, at what altitude from D to B should the camera be flown?

A blimp over triangle EDF with height of 10 meters and FE equals 20 meters and triangle ADC with height BD and AC equals 125 meters. Triangles share point D.

A. 31.25 m
B. 62.5 m
C. 150 m
D. 250 m

Answer 1

The correct option is B.

Explanation:

Given information: In ΔEDF, FE=20 m and height = 10 m. In ΔADC, AC=125 m.

From the given information, we conclude that AC║EF.

In ΔEDF and ΔADC,

`\angle E=\angle A` (Alternate interior angles)

`\angle EDF=\angle ADC` (Vertically opposite angle)

By AA rule of similarity,

`\triangle EDF\sim \triangle ADC`

The corresponding sides of two similar triangles are similar. So in ΔEDF and ΔADC,

`(base)/(height)=(FE)/(h)=(AC)/(DB)`

`(20)/(10)=(125)/(DB)`

`2=(125)/(DB)`

On cross multiplication, we get

`2DB=125`

Divide both sides by 2.

`(2DB)/(2)=(125)/(2)`

`DB=62.5`

Therefore the correct option is B.

Answer 2

B. 62.5 m

Explanation:

∠EDF and ∠ADC are vertical angles, and therefore equal.

EF and AC are parallel, so ∠DEF and ∠DAC are alternate interior angles, as well as ∠DFE and ∠DCA. Therefore, each pair is equal.

From this, we can say ΔDEF and ΔDAC are similar triangles. So we can write a proportion:

10 / 20 = DB / 125

DB = 62.5