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).
Answers
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 =
Thus, .
(ii) Finding A (Δ GEF):
In Δ GEF, we have
EF = base = 50 cm
GK = height = 20 cm
∴ Area of Δ GEF =
Thus, .
(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, .
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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