Question

Find the missing lengths: LK=15 and KH=9, find KI and HI.

Answer 1

Answer:

KI=11.25 and HI=6.75

Step-by-step explanation:

Consider the below figure attached with this question.

According to Pythagoras Theorem:

$base^2+perpendicular^2=hypotenuse^2$

Use Pythagoras in triangle HKL

$LH^2+KH^2=LK^2$

$(LH)^2+9^2=15^2$

$(LH)^2+81=225$

$(LH)^2=144$

Taking square root on both sides.

$LH=12$

Let length of HI be x.

LI = 12+x

Use Pythagoras theorem in ΔKLI,

$(KI)^2+15^2=(12+x)^2$

$(KI)^2+225=x^2+24x+144$

$(KI)^2=x^2+24x+144-225$

$(KI)^2=x^2+24x-81...(1)$

Use Pythagoras theorem in ΔHKI,

$(KI)^2=x^2+9^2$

$(KI)^2=x^2+81...(2)$

From (1) and (2) we get

$x^2+81=x^2+24x-81$

$24x=162$

$x=\dfrac{162}{24}=6.75$

Hence, the measure of HI is 6.75 units.

Substitute x=6.75 in equation (2).

$(KI)^2=(6.75)^2+81$

$(KI)^2=126.5625$

Taking square root on both sides.

$KI=\sqrt{126.5625}$

$KI=11.25$

Hence, the measure of KI is 11.25 units.