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

Question

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

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Given, ΔABC is an equilateral triangle the length of whose side is equal to 10 cm, and ΔDBC is right-angled at D and BD=8cm.From figure:Area of shaded region = Area of ΔABC−  Area of ΔDBC.....(1)Area of ΔABC:Area =43​​(side)2=43​​(10)2=43.30So area of ΔABC is 43.30cm2Area of right ΔDBC:Area =21​×base×height...(2)From Pythagoras Theorem:Hypotenuse2 = Base2 + Height2BC2=DB2+Height2100−64=Height236=Height2or Height =6equation (2)⇒Area =21​×8×6=24So area of ΔDBC is 24cm2Equation (1) implies Area of shaded region =43.30−24=19.30Therefore, Area of shaded region =19.3cm2

Question Text	In the given figure, $Δ A BC$ is an equilateral triangle the length of whose side is equal to 10 cm, and $Δ D BC$ is right-angled at D and $B D = 8 c m .$ Find the area of the shaded region. [Take $3 = 1.732.$ ]
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In the given figure, $Δ A BC$ is an equilateral triangle the length of whose side is equal to 10 cm, and $Δ D BC$ is right-angled at D and $B D = 8 c m .$ Find the area of the shaded region. [Take $3 = 1.732.$ ]

Text solutionVerified

Practice questions from similar books

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

Text solutionVerified

Practice questions from similar books

In the given figure, $Δ A BC$ is an equilateral triangle the length of whose side is equal to 10 cm, and $Δ D BC$ is right-angled at D and $B D = 8 c m .$ Find the area of the shaded region. [Take $3 = 1.732.$ ]