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somvir jha

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  1. Asked: December 5, 2020In: Class 10 Maths

    Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio.

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    somvir jha
    Added an answer on February 14, 2023 at 10:07 am

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  2. Asked: November 21, 2020In: Class 10 Maths

    PQR is a triangle right angled at P and M is a point on QR such that PM perpendicular QR. Show that PM² = QM . MR.

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    somvir jha
    Added an answer on February 14, 2023 at 10:07 am

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  3. Asked: December 27, 2020In: Class 10 Maths

    ABC is an isosceles triangle right angled at C. Prove that AB² = 2AC².

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    somvir jha
    Added an answer on February 14, 2023 at 10:07 am

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  4. Asked: December 14, 2020In: Class 10 Maths

    ABC is an isosceles triangle with AC = BC. If AB² = 2AC² , prove that ABC is a right triangle.

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    somvir jha
    Added an answer on February 14, 2023 at 10:02 am

    Given that : AB² = 2AC² ⇒ AB² = AC² + AC² ⇒ AB² = AC² + AB² [Because AC = BC] These sides satisfy the pythagoras theorem. Hence, the triangle ABC is a right angled triangle.

    Given that : AB² = 2AC²
    ⇒ AB² = AC² + AC²
    ⇒ AB² = AC² + AB² [Because AC = BC]
    These sides satisfy the pythagoras theorem.
    Hence, the triangle ABC is a right angled triangle.

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  5. Asked: December 26, 2020In: Class 10 Maths

    ABC is an equilateral triangle of side 2a. Find each of its altitudes.

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    somvir jha
    Added an answer on February 14, 2023 at 10:02 am

    Let ABC, be any equilateral triangle with each sides of lenght 2a. Perpendicular AD is drawn from A to BC. We know that the altitude in equilateral triangle, bisects the opposite sides. Therefore, ∴ BD = DC = a In triangleADB, by Pythagoras theorem AB² = AD² + BD² ⇒ (2a)² = AD² + a² [Because AB = 2aRead more

    Let ABC, be any equilateral triangle with each sides of lenght 2a. Perpendicular AD is drawn from A to BC.
    We know that the altitude in equilateral triangle, bisects the opposite sides.
    Therefore, ∴ BD = DC = a
    In triangleADB, by Pythagoras theorem
    AB² = AD² + BD²
    ⇒ (2a)² = AD² + a² [Because AB = 2a]
    ⇒ 4a² = AD² + a²
    ⇒ AD² = 3a²
    ⇒ AD = √3a
    Hence, the lenght of each altitude is √3a.

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