$\Sigma M_D = 0$

$6R_A = 5(12) + 3(20)$

$R_A = 20 \, \text{ kN}$

$\Sigma M_A = 0$

$6R_D = 1(12) + 3(20)$

$R_D = 12 \, \text{ kN}$

**At joint A**

$\Sigma F_V = 0$
$\frac{\sqrt{21}}{5}F_{AG} = R_A$

$\frac{\sqrt{21}}{5}F_{AG} = 20$

$F_{AG} = 21.82 \, \text{ kN compression}$

$\Sigma F_H = 0$
$F_{AB} = \frac{2}{5}F_{AG}$

$F_{AB} = \frac{2}{5}(21.82)$

$F_{AB} = 8.73 \, \text{ kN tension}$

**At joint G**

$\Sigma F_V = 0$
$\frac{\sqrt{21}}{5}F_{BG} + 12 = \frac{\sqrt{21}}{5}F_{AG}$

$\frac{\sqrt{21}}{5}F_{BG} + 1 2 = \frac{\sqrt{21}}{5}(21.82)$

$F_{BG} = 8.73 \, \text{ kN tension}$

$\Sigma F_H = 0$
$F_{FG} = \frac{2}{5}F_{AG} + \frac{2}{5}F_{BG}$

$F_{FG} = \frac{2}{5}(21.82) + \frac{2}{5}(8.73)$

$F_{FG} = 12.22 \, \text{ kN compression}$

**At joint B**

$\Sigma F_V = 0$
$\frac{\sqrt{21}}{5}F_{BF} = \frac{\sqrt{21}}{5}F_{BG}$

$F_{BF} = F_{BG}$

$F_{BF} = 8.73 \, \text{ kN compression}$

$\Sigma F_H = 0$
$F_{BC} = F_{AB} + \frac{2}{5}F_{BG} + \frac{2}{5}F_{BF}$

$F_{BC} = 8.73 + \frac{2}{5}(8.73) + \frac{2}{5}(8.73)$

$F_{BC} = 15.71 \, \text{ kN tension}$

**At joint F**

$\Sigma F_V = 0$
$\frac{\sqrt{21}}{5}F_{CF} + \frac{\sqrt{21}}{5}F_{BF} = 20$

$\frac{\sqrt{21}}{5}F_{CF} + \frac{\sqrt{21}}{5}(8.73) = 20$

$F_{CF} = 13.09 \, \text{ kN compression}$

$\Sigma F_H = 0$
$F_{EF} + \frac{2}{5}F_{CF} = \frac{2}{5}F_{BF} + F_{FG}$

$F_{EF} + \frac{2}{5}(13.09) = \frac{2}{5}(8.73) + 12.22$

$F_{EF} = 10.48 \, \text{ kN compression}$

**At joint C**

$\Sigma F_V = 0$
$\frac{\sqrt{21}}{5}F_{CE} = \frac{\sqrt{21}}{5}F_{CF}$

$F_{CE} = F_{CF}$

$F_{CE} = 13.09 \, \text{ kN tension}$

$\Sigma F_H = 0$
$F_{CD} + \frac{2}{5}F_{CE} + \frac{2}{5}F_{CF} = F_{BC}$

$F_{CD} + \frac{2}{5}(13.09) + \frac{2}{5}(13.09) = 15.71$

$F_{CD} = 5.24 \, \text{ kN tension}$

**At joint E**

$\Sigma F_V = 0$
$\frac{\sqrt{21}}{5}F_{DE} = \frac{\sqrt{21}}{5}F_{CE}$

$F_{DE} = F_{CE}$

$F_{DE} = 13.09 \, \text{ kN compression}$

$\Sigma F_H = 0$
$F_{EF} = \frac{2}{5}F_{CE} + \frac{2}{5}F_{DE}$

$10.48 = \frac{2}{5}(13.09) + \frac{2}{5}(13.09)$

$10.5 = 10.5$ *check*

**At joint D**

$\Sigma F_V = 0$
$R_D = \frac{\sqrt{21}}{5}F_{DE}$

$12 = \frac{\sqrt{21}}{5}(13.09)$

$12 = 12$ *check*

$\Sigma F_{H} = 0$
$F_{CD} = \frac{2}{5}F_{DE}$

$5.24 = \frac{2}{5}(13.09)$

$5.24 = 5.24$ *check*

**Summary**

*F*_{AB} = 8.73 kN tension

*F*_{AG} = 21.82 kN compression

*F*_{BC} = 15.71 kN tension

*F*_{BF} = 8.73 kN compression

*F*_{BG} = 8.73 kN tension

*F*_{CD} = 5.24 kN tension

*F*_{CE} = 13.09 kN tension

*F*_{CF} = 13.09 kN compression

*F*_{DE} = 13.09 kN compression

*F*_{EF} = 10.48 kN compression

*F*_{FG} = 12.22 kN compression
How come that the hypotenuse in the solution became 5m where it is 2.5m?

What is shown in the solution is slope. 1/2.5 = 2/5. If you are not used in solving with slope, go with angles.

Got it. Thank you

Great

how to find slope triangle please help

Use your scientific calculator that is capable of displaying fraction. If you input 1/2.5 the calculator will output 2/5.