30 points available
for circle k and circle l, segment j m is a common internal tangent. if jk = 3, lm = 9 and jm = 20, find kl. show all work.

Step-by-step explanation:

Triangles JKN MLN are similar.

There is a right angle at J and M. <JNK and <MNL are vertically opposite. By AA the triangles are similar.

Find JN

JK = 3

LM = 9

JK/LM = JN/(20 - JN)

3/9 = JN/(20 - JN)

1/3 = JN/(20 - JN)

20 - JN = 3JN

20 = 4JN

JN = 5

NM = 20 - 5

NM = 15

Find KN

KN^2 = JN^2 + JK^2

JK = 3

JN = 5

KN = ?

KN^2 = JK^2 + JN^2

KN^2 = 3^2 + 5^2

KN^2 = 34

KN = sqrt(34)

Find LN

LN = ?

MN = 15

LM = 9

LN = sqrt(15^2 + 9^2)

LN = sqrt(306)

LN = 3*sqrt(34)

Find LK

LK = sqrt(34) + 3sqrt(34)

LK = 4 sqrt(34)

answer: the area of the parallelogram below is 42 square meters.

- officialprodixy ☕

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.