To find the shaded area in triangle ABC, first determine that since all angles are 60°, the triangle is equilateral with equal sides. Using the cosine rule in triangle CPQ with angle 60°, solve for the side lengths: CQ = 10 and CP = 20. Then apply the area formula for any triangle: Area = 1/2 × a × b × sin(θ), where a and b are side lengths and θ is the included angle. Substituting the values: Area = 1/2 × 10 × 20 × sin(60°) = 50√3.
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Can you find the shaded area?Ajouté :
Can you find this shaded area in triangle ABC?
It is A B C This angle is 60° This angle is 60° So, this angle it will be also 60° That means AB BC and AC they will be equal Suppose A and CQ is BC - BQ and BC is A BQ is 15 So, CQ is A - 15 and CP is CA AP and CA is A AP is 5 CP is A - 5 and this angle is 60° and >> Now, in triangle CPQ, cos 60° it will be base CQ over hypotenuse is CP.
And cos 60° is 1 over 2.
And CQ is A minus 15 over CP is A minus 5.
And if we cross multiply, then A minus 5 it will be two times A minus 15.
And A minus 5 is 2A minus 30.
And minus 5 plus 30 it will be 2A minus A.
So, 25 it is A.
So, A is 25.
And A minus 15 is 25 minus 15 that will be 10.
And A minus 5 is 25 minus 5 that will be 20.
And now in any triangle ABC if AB is A AC is B and this angle is theta then area of ABC is 1/2 times A times B times sin theta So area of CPQ it will be 1/2 times CQ times CP times sin 60° And it is 1/2 times CQ is 10 CP is 20 and sin 60° is root 3 over 2 And 2 * 5 is 10 2 * 10 is 20 So it is 5 times 10 times root 3 that will be 50 root 3 root three.
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