Chapter 9.3, Problem 4PSA

a.

To determine

To calculate: Thelength of OK

Answer to Problem 4PSA

The length of OK is $2\sqrt{15}$

Explanation of Solution

Given:

The measures of the given lengths are as follows:

JK = $12$

KM = $5$

Concept used:

Here, we are using the concept hypotenuse leg theorem.

Calculation:

According to property associated with altitudes in right angled triangle. The altitude to hypotenuse divides the triangles into two similar triangles.

Hence, $\Delta OJK\sim \Delta OKM$

So, $\frac{OK}{JK}=\frac{KM}{OK}$

  $O{K}^2=KM\times JK$

  →$O{K}^2=12\times 5$

  →$OK=\sqrt{60}$

  →$OK=2\sqrt{15}$

Conclusion: Hence, the length of OK is $2\sqrt{15}$

b.

To determine

To calculate: Thelength of KM

Answer to Problem 4PSA

The length of KM is $5$

Explanation of Solution

Given:

The measures of the given lengths are as follows:

OK = $3\sqrt{5}$

JK = $9$

Concept used:

Here, we are using the concept hypotenuse leg theorem.

Calculation:

According to property associated with altitudes in right angled triangle. The altitude to hypotenuse divides the triangles into two similar triangles.

Hence, $\Delta OJK\sim \Delta OKM$

So, $\frac{OK}{JK}=\frac{KM}{OK}$

  $O{K}^2=KM\times JK$

  →$45=9\times KM$

  →$KM=5$

Conclusion: Hence, the length of KM is $5$

c.

To determine

To calculate: Thelength of JM

Answer to Problem 4PSA

The length of JM is $6$

Explanation of Solution

Given: The measures of the given lengths are as follows:

JO = $3\sqrt{2}$

JK = $3$

Concept used:

Here, we are using the concept hypotenuse leg theorem and Pythagoras theorem.

Calculation:

Here, ${\left(JO\right)}^2={\left(OK\right)}^2+{\left(JK\right)}^2$

  $18={\left(OK\right)}^2+9$

  $OK=3$

Now, according to property associated with altitudes in right angled triangle. The altitude to hypotenuse divides the triangles into two similar triangles.

Hence, $\Delta OJK\sim \Delta OKM$

So, $\frac{OK}{JK}=\frac{KM}{OK}$

  $O{K}^2=KM\times JK$

  →$9=3\times KM$

  →$KM=3$

And we know that $JK=3$

So, $JM=3+3=6$

Conclusion: Hence, the length of JM is $6$

d.

To determine

To calculate: Thelength of OM

Answer to Problem 4PSA

The length of OM is $\sqrt{55}$

Explanation of Solution

Given:

The measures of the given lengths are as follows:

KM = $5$

JK = $6$

Concept used:

Here, we are using the concept hypotenuse leg theorem and Pythagoras theorem.

Calculation:

According to property associated with altitudes in right angled triangle. The altitude to hypotenuse divides the triangles into two similar triangles.

Hence, $\Delta OJK\sim \Delta OKM$

So, $\frac{OK}{JK}=\frac{KM}{OK}$

  $O{K}^2=KM\times JK$

  →${\left(OK\right)}^2=5\times 6$

  →$OK=3$

Here, ${\left(JO\right)}^2={\left(OK\right)}^2+{\left(JK\right)}^2$

  ${\left(JO\right)}^2=30+36$

  ${\left(JO\right)}^2=60$

And, ${\left(OM\right)}^2={\left(OK\right)}^2+{\left(KM\right)}^2$

  ${\left(OM\right)}^2=30+25$

  $OM=\sqrt{55}$

Conclusion: Hence, the length of OM is $\sqrt{55}$