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

Draw seg AB of length 9.7 cm. Take a point P on it such that A-P-B, AP = 3.5 cm. Construct a line MN ⊥ sag AB through point P.

In the following figure, ray PT is the bisector of ∠QPR Find the value of x and perimeter of ∠QPR.

Draw the circumcircle of ΔPMT in which PM = 5.6 cm, ∠P = 60°, ∠M = 70°.

Choose the correct alternative:

______ number of tangents can be drawn to a circle from the point on the circle.

Choose the correct alternative:

In the figure ΔABC ~ ΔADE then the ratio of their corresponding sides is ______

Choose the correct alternative:

ΔPQR ~ ΔABC, `"PR"/"AC" = 5/7`, then

Choose the correct alternative:

∆ABC ∼ ∆AQR. `"AB"/"AQ" = 7/5`, then which of the following option is true?

Draw seg AB of length 9 cm and divide it in the ratio 3 : 2

∆ABC ~ ∆PBQ. In ∆ABC, AB = 3 cm, ∠B = 90°, BC = 4 cm. Ratio of the corresponding sides of two triangles is 7 : 4. Then construct ∆ABC and ∆PBQ

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

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`

In ΔABC, ∠ABC = 90°, ∠BAC = ∠BCA = 45°. If AC = `9sqrt(2)`, then find the value of AB.

Given: In the figure, point A is in the exterior of the circle with centre P. AB is the tangent segment and secant through A intersects the circle in C and D.

To prove: AB² = AC × AD

Construction: Draw segments BC and BD.

Write the proof by completing the activity.

Proof: In ΔABC and ΔADB,

∠BAC ≅ ∠DAB  .....becuase ______

∠______ ≅ ∠______  ......[Theorem of tangent secant]

∴ ΔABC ∼ ΔADB  .......By ______ test

∴ `square/square = square/square`   .....[C.S.S.T.]

∴  AB² = AC × AD

Proved.

From the information given in the figure, determine whether MP is the bisector of ∠KMN.

If ΔABC ∼ ΔDEF such that ∠A = 92° and ∠B = 40°, then ∠F = ?