Q. 43.7( 3 Votes )

# The vertices of a triangle are the points with coordinates (3, 5), (9, 13), (10, 6). Prove that the triangle is isosceles. Calculate is area.

Answer :

The three vertices are (3, 5), (9, 13), (10, 6)

Length of first side = √((9 - 3)^{2} + (13 - 5)^{2})

⇒ Length of first side = √(6^{2} + 8^{2})

⇒ Length of first side = 10 units

Length of second side = √((10 - 9)^{2} + (6 - 13)^{2})

⇒ Length of second side = √(1^{2} + 7^{2})

⇒ Length of second side = √50 units

Length of third side = √((10 - 3)^{2} + (6 - 5)^{2})

⇒ Length of third side = √(7^{2} + 1^{2})

⇒ Length of third side = √50 units

Since the length of second and third sides are equal so the triangle is isosceles

Length of height

Length of height = √75 units

Area = 0.5 × base × height

⇒ Area = 12.5 × 1.732

⇒ Area = 12.5 × 1.732 = 10.768 sq. units

Rate this question :

A 2.6 metres long rod leans against a wall, its foot 1 metre from the wall. When the foot is moved a little away from the wall, its upper end slides the same length down. How much farther is the foot moved?

Kerala Board Mathematics Part-1

An isosceles triangle has to be made like this

The height should be 2 metres less than the base. What should be the length of its sides?

Kerala Board Mathematics Part-1Draw an isosceles right triangle of hypotenuse 4 centimetres.

Kerala Board Mathematics Part-1Draw an equilateral triangle of height 3 centimetres.

Kerala Board Mathematics Part-1The vertices of a triangle are the points with coordinates (3, 5), (9, 13), (10, 6). Prove that the triangle is isosceles. Calculate is area.

Kerala Board Mathematics Part-2Prove that in any parallelogram, the sum of the square of all sides is equal to the sum of the squares of the diagonals.

Kerala Board Mathematics Part-2