Question

In the given triangle, ∠AED ∼ ∠ ABC, AD = 6.9, AE = 7.2, DE = 5.2, and BC = 10.2. Find the measure of BD and CE. Round your answer to the nearest tenth.

Answer 1

By using geometric calculations,the measure of BD and CE are 6.9 and 7.4 respectively.

Answer 2

The measure of side BD is 8.6 and The measure of side CE is 8.4

Explanation:

Given as :

The Triangle is ABC with side AB , BC , CA

And The points E and D is on the side AB and AC

So, AED is a Triangle

And Δ AED
`\sim` Δ ABC

The measure of side AD = 6.9

The measure of side AE = 7.2

The measure of side ED = 5.2

The measure of side BC = 10.2

Let The The measure of side EB = x

And The measure of side DC = y

So, From similarity property

`(AB)/(AE)` =
`(AC)/(AD)` =
`(BC)/(ED)`

Or,
`(AB)/(AE)` =
`(BC)/(ED)`

So,
`(7.2 + x)/(7.2)` =
`(10.2)/(5.2)`

Or, 5.2 × ( 7.2 + x ) = 10.2 × 7.2

Or, 37.44 + 5.2 x = 73.44

Or, 73.44 - 37.44 = 5.2 x

∴ x =
`(36)/(5.2)`

I.e x = 6.9

Now in Δ BED

BE² + ED² = BD²

Or, 6.9² + 5.2² = BD²

Or, BD² = 74.65

∴ BD =
`√(74.65)`

I.e BD = 8.64

Or, BD = 8.6

Similarly for y

`(AC)/(AD)` =
`(BC)/(ED)`

Or,
`(6.9+y)/(6.9)` =
`(10.2)/(5.2)`

Or, 5.2 × ( 6.9 + y ) = 10.2 × 6.9

Or, 35.88 + 5.2 y = 70.38

or, 5.2 y = 70.8 - 35.88

Or, 5.2 y = 34.5

∴ y =
`(34.5)/(5.2)`

I.e y = 6.6

Now in Δ CED

CD² + ED² = CE²

Or, 6.6² + 5.2² = CE²

Or, CE² = 70.6

∴ CE =
`√(70.6)`

I.e CE = 8.40

Or, CE = 8.4

Hence The measure of side BD is 8.6 and The measure of side CE is 8.4 Answer