SSC (Marathi Semi-English) 10th Standard Board Exam [इयत्ता १० वी] - Maharashtra State Board Question Bank Solutions

ΔRHP ~ ΔNED, In ΔNED, NE = 7 cm, ∠D = 30°, ∠N = 20° and `"HP"/"ED" = 4/5`. Then construct ΔRHP and ΔNED

ΔPQR ~ ΔABC. In ΔPQR, PQ = 3.6cm, QR = 4 cm, PR = 4.2 cm. Ratio of the corresponding sides of triangle is 3 : 4, then construct ΔPQR and ΔABC

Construct an equilateral ∆ABC with side 5 cm. ∆ABC ~ ∆LMN, ratio the corresponding sides of triangle is 6 : 7, then construct ΔLMN and ΔABC

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

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

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

The distance between point P(2, 2) and Q(5, x) is 5 cm, then the value of x ______

The distance between points P(–1, 1) and Q(5, –7) is ______

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 ______

Find distance between point A(– 3, 4) and origin O

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)

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`

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