Question

3. In the figure, seg DH 1 side EF and seg GK 1 side
EF. If DH=12 cm, GK = 20 cm and
A(ADEF)= 300 cm?.
Find (i) EF (ii) A(AGEF) (iii) A(ODEGF).
D
K
F
H
G​

Answer 1

Correct Question:

In the figure, seg DH ⊥ side EF and seg GK ⊥ side EF. If DH = 12 cm, GK = 20 cm and

A ( Δ DEF ) = 300 cm².

Find ( i ) EF ( ii ) A ( Δ GEF )

( iii ) A ( ⎕ DEGF ).

Answer:

i. $\boxed{\red{\sf\:EF\:=\:50\:cm}}$

ii. $\boxed{\red{\sf\:A\:(\:\triangle\:GEF\:)\:=\:500\:cm^{2}}}$

iii. $\boxed{\red{\sf\:A\:(\:\square\:DEGF\:)\:=\:800\:cm^{2}}}$

Step-by-step-explanation:

i. EF

$\sf\:A\:(\:\triangle\:DE F\:)\:=\:\frac{1}{2}\:\times\:base\:\times\:height\\\\\implies\sf\:A\:(\:\triangle\:DE F\:)\:=\:\frac{1}{2}\:\times\:EF\:\times\:DH\\\\\implies\sf\:300\:=\:\frac{1}{2}\:\times\:EF\:\times\:12\\\\\implies\sf\:EF\:=\:\dfrac{300\:\times\:\cancel{2}}{\cancel{12}}\\\\\implies\sf\:EF\:=\:\cancel{\frac{300}{6}}\\\\\implies\boxed{\red{\sf\:EF\:=\:50\:cm}}$

$\rule{200}{1}$

ii. A ( Δ GEF )

$\sf\:A\:(\:\triangle\:GEF\:)\:=\:\frac{1}{2}\:\times\:base\:\times\:height\\\\\implies\sf\:A\:(\:\triangle\:GEF\:)\:=\:\frac{1}{2}\:\times\:EF\:\times\:GK\\\\\implies\sf\:A\:(\:\triangle\:GEF\:)\:=\:\frac{1}{\cancel2}\:\times\:50\:\times\:\cancel{20}\\\\\implies\sf\:A\:(\:\triangle\:GEF\:)\:=\:50\:\times\:10\\\\\implies\boxed{\red{\sf\:A\:(\:\triangle\:GEF\:)\:=\:500\:cm^{2}}}$

$\rule{200}{1}$

iii. A ( ⎕ DEGF )

$\sf\:A\:(\:\square\:DEGF\:)\:=\:A\:(\:\triangle\:DE F\:)\:+\:A\:(\:\triangle\:GEF\:)\:\:\:\:-\:-\:-\:[\:From\:figure\:]\\\\\implies\sf\:A\:(\:\square\:DEGF\:)\:=\:300\:+\:500\\\\\implies\boxed{\red{\sf\:A\:(\:\square\:DEGF\:)\:=\:800\:cm^{2}}}$