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

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 = $\frac{1}{2} \times base \times height$

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

$\implies 300 = EF\times6$

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

$\implies \bold{EF= 50 \: cm}$

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 = $\frac{1}{2} \times base \times height$

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

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

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

Thus, $\boxed{\bold{Area (\triangle GEF) = 50 \:cm}}$.

(iii) Finding Area (⎕ DEGF ):

From the given figure we can write as,

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

substituting Area (Δ DEF) = 300 cm² & Area (Δ GEF) = 500 cm²

⇒ Area (⎕ DEGF) = 300 cm² + 500 cm²

⇒ Area (⎕ DEGF) = 800 cm²

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

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Answer 2

Step-by-step explanation:

Given:

seg DH ⊥ side EF

seg GK ⊥ side EF

DH = 12 cm

GK = 20 cm

A (Δ DEF) = 300 cm

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

2

1

×base×height

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

2

1

×EF×12

\implies 300 = EF\times6⟹300=EF×6

\implies EF= \frac{300}{6}⟹EF=

6

300

\implies \bold{EF= 50 \: cm}⟹EF=50cm

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

EF=50cm

.

(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

2

1

×base×height

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

2

1

×50×20

\implies Area (\triangle GEF) = 50\times 10⟹Area(△GEF)=50×10

\implies Area (\triangle GEF) = 500 \: cm^2⟹Area(△GEF)=500cm

2

Thus, \boxed{\bold{Area (\triangle GEF) = 50 \:cm}}

Area(△GEF)=50cm

.

(iii) Finding Area (⎕ DEGF ):

From the given figure we can write as,

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

substituting Area (Δ DEF) = 300 cm² & Area (Δ GEF) = 500 cm²

⇒ Area (⎕ DEGF) = 300 cm² + 500 cm²

⇒ Area (⎕ DEGF) = 800 cm²

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

Area(quadDEGF)=800cm 2