Question

Identify the similar triangles and find x. Then find the measures of the indicated sides.

Answer 1

Triangle JMK is similar to triangle LMN.

`x = 3`.

`\rm \overline{KM}= 6`.

`\rm \overline{NM} = 3`.

Explanation:

The angle
`\rm N\hat{M}L` is a right angle for it is complementary with another right angle,
`\rm K\hat{M}J`.

The diagram also indicates that angle
`\rm \hat{J}` is equal to angle
`\rm \hat{L}`. As a result,
`\rm \triangle JMK \sim \triangle LMN` for two of their angles are equal.

Consequently,

`\displaystyle \rm \frac{(\overline{MN})}{(\overline{MK})} = \frac{(\overline{LN})}{(\overline{JK})}`.

Let
`x` be the length of segment
`\rm MN`.

`\displaystyle (x)/(3+x) = (5)/(10)`.

Cross multiply. In other words, multiply both sides by
`10(3 + x)`.

`10x = 5(3 + x)`.

`x = 3`.

`\rm \overline{KM} = 3 + \mathnormal{x} = 6`.

`\rm \overline{MN} = \mathnormal{x} = 3`.

Answer 2

The similar triangles are Δ KMJ and Δ NML

The value of x is 3

KM = 6 and NM = 3

Explanation:

* Lets revise the cases of similarity

1) AAA similarity : two triangles are similar if all three angles in the first

triangle equal the corresponding angle in the second triangle

- Example : In ΔABC and ΔDEF, m∠A = m∠D, m∠B = m∠E and

m∠C= m∠F then ΔABC ≈ ΔDEF by AAA

2) AA similarity : If two angles of one triangle are equal to the

corresponding angles of the other triangle, then the two triangles

are similar.

- Example : In ΔPQR and ΔDEF, m∠P = m∠D, m∠R = m∠F then

ΔPQR ≈ ΔDEF by AA

3) SSS similarity : If the corresponding sides of two triangles are

proportional, then the two triangles are similar.

- Example : In ΔXYZ and ΔLMN, if

then the two triangles are similar by SSS

4) SAS similarity : In two triangles, if two sets of corresponding sides

are proportional and the included angles are equal then the two

triangles are similar.

- Example : In triangle ABC and DEF, if m∠A = m∠D and

then the two triangles are similar by SAS

* Now lets solve the problem

- ∠KMJ is a aright angle and M is on JL

∴ m∠JML = 180° ⇒ straight angle

∵ m∠JMK + m∠LMN = m∠JML

∴ 90° + m∠NML = 180° ⇒ subtract 90° from both sides

∴ m∠NML = 90°

- In Δ KMJ and ΔNML

∵ m∠KMJ = m∠NML ⇒ proved

∵ m∠KJM = m∠NLM ⇒ given

- By using the second case above (AA similarity)

∴ Δ KMJ ≈ Δ NML

* The similar triangles are Δ KMJ and Δ NML

- From similarity

∴ Their sides are proportion

∴
`(KM)/(NM)=(MJ)/(ML)=(KJ)/(NL)`

∵ KJ = 10 and NL = 5

∵ KM = 3 + x and NM = x

- Substitute these values in the proportion relation

∵
`(KM)/(NM)=(KJ)/(NL)`

∴
`(3+x)/(x)=(10)/(5)`

- By using cross multiplication

∴ 5(3 + x) = 10(x) ⇒ simplify

∴ 5(3) + 5(x) = 10x

∴ 15 + 5x = 10x ⇒ subtract 5x from both sides

∴ 15 = 5x ⇒ divide both sides by 5

∴ 3 = x

* The value of x is 3

∵ KM = 3 + x

∵ x = 3

∴ KM = 3 + 3 = 6

∵ NM = x

∴ NM = 3

* KM = 6 and NM = 3

- Check the ratio

∵ KM/NM = 6/3 = 2

∵ KJ/NL = 10/5 = 2

∴ The sides are proportion