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

Answer:

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

Step-by-step 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

$\dfrac{AB}{AE}$ = $\dfrac{AC}{AD}$ = $\dfrac{BC}{ED}$

Or,  $\dfrac{AB}{AE}$ = $\dfrac{BC}{ED}$

So, $\dfrac{7.2 + x}{7.2}$ = $\dfrac{10.2}{5.2}$

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 = $\frac{36}{5.2}$

I.e x = 6.9

Now in Δ BED

BE² + ED² = BD²

Or, 6.9² + 5.2² = BD²

Or, BD² = 74.65

∴  BD = $\sqrt{74.65}$

I.e BD = 8.64

Or, BD = 8.6

Similarly for y

$\dfrac{AC}{AD}$ = $\dfrac{BC}{ED}$

Or,  $\dfrac{6.9+y}{6.9}$ = $\dfrac{10.2}{5.2}$

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 = $\frac{34.5}{5.2}$

I.e y = 6.6

Now in Δ CED

CD² + ED² = CE²

Or, 6.6² + 5.2² = CE²

Or, CE² = 70.6

∴  CE = $\sqrt{70.6}$

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

Answer 2

Answer:

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