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

Prove that `sqrt((1 + sin A)/(1 - sin A))` = sec A + tan A.

Δ AMT ∼ ΔAHE. In  Δ AMT, MA = 6.3 cm, ∠MAT = 120°, AT = 4.9 cm, `(MA)/(HA) = 7/5`. construct  Δ AHE. 

If sin θ = `1/2`, then find the value of θ. 

Draw seg AB = 6.8 cm and draw perpendicular bisector of it. 

Find the co-ordinates of the centroid of the Δ PQR, whose vertices are P(3, –5), Q(4, 3) and R(11, –4) 

If cosθ = `5/13`, then find sinθ. 

Verify that the points A(–2, 2), B(2, 2) and C(2, 7) are the vertices of a right-angled triangle. 

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°.

If sec θ = `25/7`, find the value of tan θ.

Solution:

1 + tan² θ = sec² θ

∴ 1 + tan² θ = `(25/7)^square`

∴ tan² θ = `625/49 - square`

= `(625 - 49)/49`

= `square/49`

∴ tan θ = `square/7` ........(by taking square roots)

If sin θ + sin² θ = 1 show that: cos² θ + cos⁴ θ = 1

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

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

ΔPQR ~ ΔABC, `(PR)/(AC) = 5/7`, then

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