Question

6x + 36
10x
m
2x + 63
s
x + 50
6x - 145
a
d
c
x =
overline{jm} =

x =
overline{ts} =

x =
overline{bc} =

Explanation:

First Figure (Left - Hand Side):

Step1: Set up the equation

Since the two right - angled triangles \( \triangle MJ\) and \( \triangle ML\) are congruent (by Hypotenuse - Leg congruence, as \( \angle J=\angle L = 90^{\circ}\), \( MJ\) and \( ML\) are the legs, and \( M\) is the common vertex for the hypotenuse - like segments), we have \(6x + 36=10x\).

Step2: Solve for \(x\)

Subtract \(6x\) from both sides: \(36 = 10x-6x\), so \(4x=36\), and \(x = 9\).

Step3: Find \(JM\)

Substitute \(x = 9\) into \(6x + 36\). Then \(JM=6\times9 + 36=54 + 36=90\).

Second Figure (Middle):

Step1: Set up the equation

Since \(TR\) is the perpendicular bisector of \(GS\) (given the tick - marks indicating \(GR = RS\) and \( \angle TRG=\angle TRS = 90^{\circ}\)), by the Hypotenuse - Leg congruence of \( \triangle TGR\) and \( \triangle TSR\), we have \(2x+63 = 5x\).

Step2: Solve for \(x\)

Subtract \(2x\) from both sides: \(63=5x - 2x\), so \(3x = 63\), and \(x = 21\).

Step3: Find \(TS\)

Substitute \(x = 21\) into \(2x + 63\). Then \(TS=2\times21+63=42 + 63 = 105\).

Third Figure (Right - Hand Side):

Step1: Set up the equation

Since \(BD\) is the perpendicular bisector of \(AC\) (given the tick - marks indicating \(AD = DC\) and \( \angle BDA=\angle BDC = 90^{\circ}\)), by the Hypotenuse - Leg congruence of \( \triangle ABD\) and \( \triangle CBD\), we have \(x + 50=6x-145\).

Step2: Solve for \(x\)

Subtract \(x\) from both sides: \(50=6x - x-145\), then \(50 + 145=5x\), so \(195 = 5x\), and \(x = 39\).

Step3: Find \(BC\)

Substitute \(x = 39\) into \(6x-145\). Then \(BC=6\times39-145=234 - 145 = 89\).

Answer:

For the first figure: \(x = 9\), \(JM = 90\).
For the second figure: \(x = 21\), \(TS = 105\).
For the third figure: \(x = 39\), \(BC = 89\).