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 DEGF).
​

Answer 1

Given:

seg DH ⊥ side EF  

seg GK ⊥ side EF

DH = 12 cm

GK = 20 cm  

A (Δ DEF) = 300 cm ²

To find:

(i) EF  

(ii) A (ΔGEF)  

(iii) A (⎕ DEGF)

Solution:

(i) Finding EF:

In Δ DEF, we have

EF = base  

DH = height = 12 cm

Area (Δ DEF) = 300 cm²

∴ Area of Δ DEF =   $\frac{1}{2} \times base\times height$

$\implies 300 = \frac{1}{2} \times EF \times 12$

$\implies 300 = EF \times 6$

$\implies EF = \frac{300}{6}$

 $\implies\bold{EF = 50\:cm}$

Thus,   $\boxed{\bold{EF = 50 \:cm}}$.

(ii) Finding A (Δ GEF):

In Δ GEF, we have

EF = base = 50 cm

GK = height = 20 cm

Area of Δ GEF =   $\frac{1}{2} \times base\times height$

$\implies Area (\triangle \:GEF)= \frac{1}{2} \times 50 \times 20$

$\implies Area (\triangle \:GEF)= 50 \times 10$

$\implies\bold{ Area (\triangle \:GEF)= 500\:cm^2}$

Thus, $\boxed{\bold{Area (\triangle\:GEF = 500 \:cm^2)}}$ .

(iii) Finding Area (⎕ DEGF ):

From the attached figure we can write as,

Area (⎕ DEGF) = Area (Δ DEF) + Area (Δ GEF)

by substituting Area (Δ DEF) = 300 cm² & Area (Δ GEF) = 500 cm² , we get

⇒ Area (⎕ DEGF) = 300 cm² + 500 cm²

⇒ Area (⎕ DEGF) = 800 cm²

Thus, $\boxed{\bold{Area \:(quad\: DEGF) = 800\:cm^2}}$ .

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