SSC (English Medium) १० वीं कक्षा - Maharashtra State Board Question Bank Solutions for Geometry Mathematics 2

Find distance between point A(7, 5) and B(2, 5).

Find distance of point A(6, 8) from origin.

Find distance between points O(0, 0) and B(–5, 12).

Find distance between point Q(3, –7) and point R(3, 3)

Solution: Suppose Q(x₁, y₁) and point R(x₂, y₂)

x₁ = 3, y₁ = –7 and x₂ = 3, y₂ = 3

Using distance formula,

d(Q, R) = `sqrt(square)`

∴ d(Q, R) = `sqrt(square - 100)`

∴ d(Q, R) =  `sqrt(square)`

∴ d(Q, R) = `square`

Find distance between point A(–1, 1) and point B(5, –7):

Solution: Suppose A(x₁, y₁) and B(x₂, y₂)

x₁ = –1, y₁ = 1 and x₂ = 5, y₂ = –7

Using distance formula,

d(A, B) = `sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2`

∴ d(A, B) = `sqrt(square +[(-7) + square]^2`

∴ d(A, B) = `sqrt(square)`

∴ d(A, B) = `square`

If the distance between point L(x, 7) and point M(1, 15) is 10, then find the value of x.

Find distance CD where C(–3a, a), D(a, –2a).

Show that the point (11, –2) is equidistant from (4, –3) and (6, 3).

Show that P(–2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle.

Show that the point (0, 9) is equidistant from the points (–4, 1) and (4, 1).

Show that the points (0, –1), (8, 3), (6, 7) and (–2, 3) are vertices of a rectangle.

Show that the points (2, 0), (–2, 0) and (0, 2) are vertices of a triangle. State the type of triangle with reason.

Show that A(1, 2), (1, 6), C(1 + 2`sqrt(3)`, 4) are vertices of an equilateral triangle.

Seg OA is the radius of a circle with centre O. The coordinates of point A is (0, 2) then decide whether the point B(1, 2) is on the circle?

Using distance formula decide whether the points (4, 3), (5, 1) and (1, 9) are collinear or not.

If a and b are natural numbers and a > b If (a² + b²), (a² – b²) and 2ab are the sides of the triangle, then prove that the triangle is right-angled. Find out two Pythagorean triplets by taking suitable values of a and b.

In ΔABC, ∠ABC = 90° and ∠ACB = θ. Then write the ratios of sin θ and tan θ from the figure.

Prove that, The areas of two triangles with the same height are in proportion to their corresponding bases. To prove this theorem start as follows:

1. Draw two triangles, give the names of all points, and show heights.
2. Write 'Given' and 'To prove' from the figure drawn.

Prove that sec θ + tan θ = `cos θ/(1 - sin θ)`.

Proof: L.H.S. = sec θ + tan θ

= `1/square + square/square`

= `square/square`  ......`(∵ sec θ = 1/square, tan θ = square/square)`

= `((1 + sin θ) square)/(cos θ  square)`  ......[Multiplying `square` with the numerator and denominator]

= `(1^2 - square)/(cos θ  square)`

= `square/(cos θ  square)`

= `cos θ/(1 - sin θ)` = R.H.S.

∴ L.H.S. = R.H.S.

∴ sec θ + tan θ = `cos θ/(1 - sin θ)`

Find distance between points P(– 5, – 7) and Q(0, 3).

By distance formula,

PQ = `sqrt(square + (y_2 - y_1)^2`

= `sqrt(square + square)`

= `sqrt(square + square)`

= `sqrt(square + square)`

= `sqrt(125)`

= `5sqrt(5)`