In the given figure, ∆ABC is an equilateral triangle the length of w

1.	In the given figure, ∆ABC is an equilateral triangle the length of whose side is equal to 10 cm, and ∆DBC is right-angled at D and BD = 8 cm. Find the area of the shaded region. [Take √3 = 1.732.]
Answer» ∆ABC is an equilateral triangle the length of whose side is equal to 10 cm, and ∆DBC is right-angled at D and BD = 8 cm. From figure: Area of shaded region = Area of ABC – Area of DBC …..(1) Area of △ABC: Area = √3/4 (side)² = √3/4 (10)² = 43.30 So area of △ABC is 43.30 cm² Area of right △DBC: Area = 1/2 x base x height …(2) From Pythagorean Theorem: Hypotenuse² = Base² + Height² BC² = DB² + Height² 100 – 64 = Height² 36 = Height² or Height = 6 equation (2) => Area = 1/2 x 8 x 6 = 24 So area of △DBC is 24 cm² Equation (1) implies Area of shaded region = 43.30 – 24 = 19.30 Therefore, Area of shaded region = 19.3 cm²

In the given figure, ∆ABC is an equilateral triangle the length of whose side is equal to 10 cm, and ∆DBC is right-angled at D and BD = 8 cm. Find the area of the shaded region. [Take √3 = 1.732.]

Answer»

∆ABC is an equilateral triangle the length of whose side is equal to 10 cm, and ∆DBC is right-angled at D and BD = 8 cm.

From figure:

Area of shaded region = Area of ABC – Area of DBC …..(1)

Area of △ABC:

Area = √3/4 (side)² = √3/4 (10)² = 43.30

So area of △ABC is 43.30 cm²

Area of right △DBC:

Area = 1/2 x base x height …(2)

From Pythagorean Theorem:

Hypotenuse² = Base² + Height²

BC² = DB² + Height²

100 – 64 = Height²

36 = Height²

or Height = 6

equation (2) =>

Area = 1/2 x 8 x 6 = 24

So area of △DBC is 24 cm²

Equation (1) implies

Area of shaded region = 43.30 – 24 = 19.30

Therefore, Area of shaded region = 19.3 cm²