Quantitative Aptitude
TRIANGLES MCQs
Total Questions : 83
| Page 9 of 9 pages
Answer: Option B. -> $$3\sqrt 5 $$ cm
AD2 = BD × DC
AD2 = 5 × 9
AD = $$\sqrt {45} $$
AD = $$3\sqrt 5 $$ cm
AD2 = BD × DC
AD2 = 5 × 9
AD = $$\sqrt {45} $$
AD = $$3\sqrt 5 $$ cm
Answer: Option B. -> $$48\sqrt 3 $$ cm2
According to question,
⇒ ∵ ∠BGC = 60° (given)
⇒ ∠GBC = ∠GCB = x°
⇒ x° + x° + 60° = 180°
⇒ x = 60°
⇒ So ΔBGC is an equilateral triangle with side 8 cm each
Then,
Area pf triangle ΔBGC
= $$\frac{{\sqrt 3 }}{4}$$ a2
= $$\frac{{\sqrt 3 }}{4}$$ 82
= 16$${\sqrt 3 }$$ cm2
⇒ Area of ΔABC
= Area (ΔBGC + ΔAGC + ΔAGB)
⇒ Area of ΔABC = 3 × 16$${\sqrt 3 }$$
⇒ Area of ΔABC = 48$${\sqrt 3 }$$ cm2 {∵ ΔBGC = ΔAGC = ΔAGB}
According to question,
⇒ ∵ ∠BGC = 60° (given)
⇒ ∠GBC = ∠GCB = x°
⇒ x° + x° + 60° = 180°
⇒ x = 60°
⇒ So ΔBGC is an equilateral triangle with side 8 cm each
Then,
Area pf triangle ΔBGC
= $$\frac{{\sqrt 3 }}{4}$$ a2
= $$\frac{{\sqrt 3 }}{4}$$ 82
= 16$${\sqrt 3 }$$ cm2
⇒ Area of ΔABC
= Area (ΔBGC + ΔAGC + ΔAGB)
⇒ Area of ΔABC = 3 × 16$${\sqrt 3 }$$
⇒ Area of ΔABC = 48$${\sqrt 3 }$$ cm2 {∵ ΔBGC = ΔAGC = ΔAGB}
Answer: Option B. -> 6 sq. units
According to question,
As we know that
Given: area of ΔABC = 24 square units
As we know that
D, E and F are the midpoint of AB, AC and BC
∴ Area of ΔADE = area of ΔDBF
= area of ΔDEF = area of ΔEFC
∴ Area of ΔDEF = $$\frac{1}{4}$$ area of ΔABC
Area of ΔDEF = $$\frac{1}{4}$$ × 24
Area of ΔDEF = 6 sq. units
According to question,
As we know that
Given: area of ΔABC = 24 square units
As we know that
D, E and F are the midpoint of AB, AC and BC
∴ Area of ΔADE = area of ΔDBF
= area of ΔDEF = area of ΔEFC
∴ Area of ΔDEF = $$\frac{1}{4}$$ area of ΔABC
Area of ΔDEF = $$\frac{1}{4}$$ × 24
Area of ΔDEF = 6 sq. units
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