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

In the figure, a circle with center C has m(arc AXB) = 100° then find central ∠ACB and measure m(arc AYB).

In the figure, seg AB is a diameter of a circle with centre O. The bisector of ∠ACB intersects the circle at point D. Prove that, seg AD ≅ seg BD. Complete the following proof by filling in the blanks.

Proof:

Draw seg OD.

∠ACB = `square`     ......[Angle inscribed in semicircle]

∠DCB = `square`    ......[CD is the bisector of ∠C]

m(arc DB) = `square`   ......[Inscribed angle theorem]

∠DOB = `square`   ......[Definition of measure of an arc](i)

seg OA ≅ seg OB     ...... `square` (ii)

∴ Line OD is `square` of seg AB    ......[From (i) and (ii)]

∴ seg AD ≅ seg BD

In the figure, chord LM ≅ chord LN, ∠L = 35°. 

Find
(i) m(arc MN)

(ii) m(arc LN)

In the figure, if O is the center of the circle and two chords of the circle EF and GH are parallel to each other. Show that ∠EOG ≅ ∠FOH

In the figure, ΔABC is an equilateral triangle. The angle bisector of ∠B will intersect the circumcircle ΔABC at point P. Then prove that: CQ = CA.

In the above figure, chord PQ and chord RS intersect each other at point T. If ∠STQ = 58° and ∠PSR = 24°, then complete the following activity to verify:

∠STQ = `1/2` [m(arc PR) + m(arc SQ)]

Activity: In ΔPTS,

∠SPQ = ∠STQ – `square`  ......[∵ Exterior angle theorem]

∴ ∠SPQ = 34°

∴ m(arc QS) = 2 × `square`° = 68°   ....... ∵ `square`

Similarly, m(arc PR) = 2∠PSR = `square`°

∴ `1/2` [m(arc QS) + m(arc PR)] = `1/2` × `square`° = 58°  ......(I)

But ∠STQ = 58°  .....(II) (given)

∴  `1/2` [m(arc PR) + m(arc QS)] = ∠______  ......[From (I) and (II)]

In the figure, the centre of the circle is O and ∠STP = 40°.

1. m (arc SP) = ? By which theorem?
2. m ∠SOP = ? Give reason.

ΔPQR, is a right angled triangle with ∠Q = 90°, QR = b, and A(ΔPQR) = a. If QN ⊥ PR, then prove that QN = `(2ab)/sqrt(b^4 + 4a^2)`

In the above figure, ∠L = 35°, find :

1. m(arc MN)
2. m(arc MLN)

Solution :

1. ∠L = `1/2` m(arc MN) ............(By inscribed angle theorem)
   ∴ `square = 1/2` m(arc MN)
   ∴ 2 × 35 = m(arc MN)
   ∴ m(arc MN) = `square`
2. m(arc MLN) = `square` – m(arc MN) ...........[Definition of measure of arc]
   = 360° – 70°
   ∴ m(arc MLN) = `square`

In the given figure, ∠MNP = 90°, seg NQ ⊥ seg MP, MQ = 9, QP = 4, find NQ.

In the given figure, ∠QPR = 90°, seg PM ⊥ seg QR and Q–M–R, PM = 10, QM = 8, find QR.

In the given figure, chord MN and chord RS intersect at point D.
(1) If RD = 15, DS = 4, MD = 8 find DN
(2) If RS = 18, MD = 9, DN = 8 find DS

In right-angled ΔABC, BD ⊥ AC. If AD= 4, DC= 9, then find BD.

In the given figure,
(1) m(arc CE) = 54°, m(arc BD) = 23°, find measure of ∠CAE.
(2) If AB = 4.2, BC = 5.4, AE = 12.0, find AD.
(3) If AB = 3.6, AC = 9.0, AD = 5.4, find AE.

As shown figure, ∠DFE = 90°, FG ⊥ ED, if GD = 8, FG = 12, then EG = ?

The value of 2tan45° – 2sin30° is ______.

In the figure, ΔPQR is right angled at Q, seg QS ⊥ seg PR. Find x, y.

If `sec alpha=2/sqrt3`  , then find the value of `(1-cosecalpha)/(1+cosecalpha)` where α is in IV quadrant.

Prove that sin⁶θ + cos⁶θ = 1 – 3 sin²θ. cos²θ.

If the sides of a triangle are 6 cm, 8 cm and 10 cm, respectively, then determine whether the triangle is a right angle triangle or not.