Question

Find the length of the midline KH in Trapezoid SQUAD.​

Answer 1

Question

• Find the length of the midline KH in Trapezoid SQUAD.

Answer

$\sf\red{the \: length \: of \: the \: middle \: kh \: is \: \frac{a \: + \: b}{2} units}$

• use the midpoint formula to find the coordinates of points K and H

$\begin{gathered}{\boxed{\begin{array}{cccc} \sf{For \: point \: K} & \\ & \sf \\ \sf{Q( 0, 0 )\: U( c, d )}\\ \tiny \times = \times \frac{ \times 1 \: + y \: 2}{2} \: \: y = \frac{y \: 1 + y \: 2}{2} \\ \tiny{ \times = \frac{0 \: + \: c}{2} \: y = \frac{0 \: + \: d}{2} } \\ \tiny{ \times = \frac{c}{2} \: \: \: y = \frac{d}{2} } \\ \tiny \red{the \: coordinates \: of \: point \: K \: are \: ( \frac{c}{2} . \frac{d}{2} )} \end{array}}}\end{gathered}$

$\begin{gathered}{\boxed{\begin{array}{cccc} \sf{For \: point \: H} \\ \\ \tiny{b),d] D(a,0)} \\ \tiny \times = \frac{ \times 1 \times 2}{2} \: \: y = \frac{y \: 1 + y \: 2}{2} & \\ \tiny{ \times = \frac{a + (c + b)}{2} \: \: y = \frac{0 + d}{2} } &\\ \tiny{ \times = \frac{a + b + c}{2} \: \: \: y = \frac{d}{2} } \\ \tiny\red{ The\: coordinates\: of\: point \:H\: are ( \frac{a + b + c}{2} , \frac{d}{2} ) }\end{array}}}\end{gathered}$

• b. use the distance for mula to find the length of KH

$\sf \: k( \frac{c}{2} . \frac{d}{2} ) \: \: \: h( \frac{a + b + c}{2} . \frac{d}{2} )$

$\sf \: d = \sqrt{(x2 - \times 1) {}^{2} + (y2 - y1) {}^{2} }$

$\sf \: kh = \sqrt{ \frac{[a \: + \: b \: + \: c}{2} - \frac{c}{2} {}^{2} + ( \frac{d}{2} - \frac{d}{2}) ^{2} }$

$\sf \: kh = \sqrt{ (\frac{a + b}{2}) {}^{2} + (0) {}^{2} }$

$\sf \: kh = \sqrt{( \frac{a + b}{2}) {}^{2} }$

$\sf \: kh = \frac{a + b}{2}$

Final Answer

$\sf\red{the \: length \: of \: the \: middle \: kh \: is \: \frac{a \: + \: b}{2} units}$