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

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Geometry Mathematics 2
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ΔAMT ~ ΔAHE. In ΔAMT, AM = 6.3 cm, ∠MAT = 120°, AT = 4.9 cm, `"AM"/"HA" = 7/5`, then construct ΔAMT and ΔAHE

[4] Geometric Constructions
Chapter: [4] Geometric Constructions
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ΔRHP ~ ΔNED, In ΔNED, NE = 7 cm. ∠D = 30°, ∠N = 20°, `"HP"/"ED" = 4/5`, then construct ΔRHP and ∆NED

[4] Geometric Constructions
Chapter: [4] Geometric Constructions
Concept: undefined >> undefined

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ΔABC ~ ΔPBR, BC = 8 cm, AC = 10 cm , ∠B = 90°, `"BC"/"BR" = 5/4` then construct ∆ABC and ΔPBR

[4] Geometric Constructions
Chapter: [4] Geometric Constructions
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The distance between point P(2, 2) and Q(5, x) is 5 cm, then the value of x ______

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
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The distance between points P(–1, 1) and Q(5, –7) is ______

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
Concept: undefined >> undefined

If the length of the segment joining point L(x, 7) and point M(1, 15) is 10 cm, then the value of x is ______

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
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Find distance between point A(– 3, 4) and origin O

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
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Find distance between point A(7, 5) and B(2, 5)

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
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Find distance of point A(6, 8) from origin

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
Concept: undefined >> undefined

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

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
Concept: undefined >> undefined

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`

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
Concept: undefined >> undefined

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`

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
Concept: undefined >> undefined

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

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
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Find distance CD where C(– 3a, a), D(a, – 2a)

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
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Show that the point (11, – 2) is equidistant from (4, – 3) and (6, 3)

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
Concept: undefined >> undefined

If the point P (6, 7) divides the segment joining A(8, 9) and B(1, 2) in some ratio, find that ratio

Solution:

Point P divides segment AB in the ratio m: n.

A(8, 9) = (x1, y1), B(1, 2 ) = (x2, y2) and P(6, 7) = (x, y)

Using Section formula of internal division,

∴ 7 = `("m"(square) - "n"(9))/("m" + "n")`

∴ 7m + 7n = `square` + 9n

∴ 7m – `square` = 9n – `square`

∴ `square` = 2n

∴ `"m"/"n" = square`

[4] Geometric Constructions
Chapter: [4] Geometric Constructions
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Show that P(– 2, 2), Q(2, 2) and R(2, 7) are vertices of a right angled triangle

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
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Show that the point (0, 9) is equidistant from the points (– 4, 1) and (4, 1)

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
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Show that the points (0, –1), (8, 3), (6, 7) and (– 2, 3) are vertices of a rectangle.

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
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Show that the points (2, 0), (– 2, 0) and (0, 2) are vertices of a triangle. State the type of triangle with reason

[5] Co-ordinate Geometry
Chapter: [5] Co-ordinate Geometry
Concept: undefined >> undefined
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