Question

2. In Fig., ABCD is a trapezium whose parallel sides ABand CD are respectively 9 cm and 5 cm. If E and F arerespectively the mid-points of BC and AD respectivelyand EF BA, find the length of EF.​

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{\boxed { \huge \mathcal\red{ solution}}}}$

See in the Figure....

For the Trapezium ABEF ,

base(AB)=9 cm

Another parallel side(EF)= X cm (let)

height (FH)=(h/2)cm

therefore area of the Trapezium ABEF

$\implies \frac{1}{2} \times (FH) \times (AB + EF) \\ \implies \: \frac{1}{2} \times \frac{h}{2} \times (9 + x) \\ \implies \frac{h}{4} \times (9 + x) \: cm {}^{2}$

For the Trapezium CDFE ,

base(EF)=X cm

another parallel side(CD)=5 cm

height (DI)=(h/2)cm

therefore area of the Trapezium CDFE,

$\implies \frac{1}{2} \times (DI) \times (FE + CD) \\ \implies \: \frac{1}{2} \times \frac{h}{2} \times (9 + x) \\ \implies \frac{h}{4} \times (x + 5) \: cm {}^{2}$

Now for the Trapezium ABCD

base(AB)=9 cm

another parallel side(CD)=5 cm

height (CJ)=h (let)

therefore area of the Trapezium ABCD

$\implies \frac{1}{2} \times (CJ) \times (AB + CD) \\ \implies \: \frac{1}{2} \times h \times (9 + 5) \\ \implies \frac{h}{2} \times (14) \: cm {}^{2}$

Now area of Trapezium ABCD =area of the Trapezium ABEF + area of the Trapezium CDFE

$\implies \frac{h}{2} \times (14) = \frac{h}{4} \times (9 + x) \: + \frac{h}{4} \times (x + 5) \: \\ \implies \frac{h}{2} \times (14)=\frac{h}{4}(x+9+x+5)\\ \implies 2x+14=14\times2\\ \implies\boxed{\red {x=7\:cm}}$

$\therefore$ length of EF=7 cm

$\underline{ \huge\mathfrak{hope \: this \: helps \: you}}$

$\mathcal{ \#\mathcal{answer with quality }\: \: \& \: \: \#BAL }$