SSC (English Medium) 10th Standard - Maharashtra State Board Question Bank Solutions

In the above figure, line l || line m and line n is a transversal. Using the given information find the value of x. 

In the above figure, line AB || line CD || line EF, line l, and line m are its transversals. If  AC = 6, CE = 9. BD = 8, then complete the following activity to find DF. 

Activity :

`"AC"/"" = ""/"DF"`   (Property of three parallel lines and their transversal) 

∴ `6/9 = ""/"DF"`

∴ `"DF"  = "___"`

From given figure, In ∆ABC, AD ⊥ BC, then prove that AB² + CD² = BD² + AC² by completing activity.

Activity: From given figure, In ∆ACD, By pythagoras theorem

AC² = AD² + `square`

∴ AD² = AC² – CD²    ......(I)

Also, In ∆ABD, by pythagoras theorem,

AB² = `square` + BD²

∴ AD² = AB² – BD²    ......(II)

∴ `square` − BD² = AC² − `square`

∴ AB² + CD² = AC²+ BD²

A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the base of wall. Complete the given activity.

Activity: As shown in figure suppose

PR is the length of ladder = 10 m

At P – window, At Q – base of wall, At R – foot of ladder

∴ PQ = 8 m

∴ QR = ?

In ∆PQR, m∠PQR = 90°

By Pythagoras Theorem,

∴ PQ² + `square` = PR²    .....(I)

Here, PR = 10, PQ = `square`

From equation (I)

8² + QR² = 10²

QR² = 10² – 8²

QR² = 100 – 64

QR² = `square`

QR = 6

∴ The distance of foot of the ladder from the base of wall is 6 m.

Complete the following activity to find the length of hypotenuse of right angled triangle, if sides of right angle are 9 cm and 12 cm.

Activity: In ∆PQR, m∠PQR = 90°

By Pythagoras Theorem,

PQ² + `square` = PR²    ......(I)

∴ PR² = 9² + 12² 

∴ PR² = `square` + 144

∴ PR² = `square`

∴ PR = 15

∴ Length of hypotenuse of triangle PQR is `square` cm.

From given figure, in ∆PQR, if ∠QPR = 90°, PM ⊥ QR, PM = 10, QM = 8, then for finding the value of QR, complete the following activity.

Activity: In ∆PQR, if ∠QPR = 90°, PM ⊥ QR,  ......[Given]

In ∆PMQ, by Pythagoras Theorem,

∴ PM² + `square` = PQ²     ......(I)

∴ PQ² = 10² + 8² 

∴ PQ² = `square` + 64

∴ PQ² = `square`

∴ PQ = `sqrt(164)`

Here, ∆QPR ~ ∆QMP ~ ∆PMR

∴ ∆QMP ~ ∆PMR

∴ `"PM"/"RM" = "QM"/"PM"`

∴ PM² = RM × QM

∴ 10² = RM × 8

RM = `100/8 = square`

And,

QR = QM + MR

QR = `square` + `25/2 = 41/2`

Find the diagonal of a rectangle whose length is 16 cm and area is 192 sq.cm. Complete the following activity.

Activity: As shown in figure LMNT is a reactangle.
∴ Area of rectangle = length × breadth

∴ Area of rectangle = `square` × breadth

∴ 192 = `square` × breadth

∴ Breadth = 12 cm

Also,

∠TLM = 90°    ......[Each angle of reactangle is right angle]

In ∆TLM,

By Pythagoras theorem

∴ TM² = TL² + `square`

∴ TM² = 12² + `square`

∴ TM² = 144 + `square`

∴ TM² = 400

∴ TM = 20

A congruent side of an isosceles right angled triangle is 7 cm, Find its perimeter

In the figure, if the chord PQ and chord RS intersect at point T, prove that: m∠STQ = `1/2` [m(arc PR) + m(arc SQ)] for any measure of ∠STQ by filling out the boxes

Proof: m∠STQ = m∠SPQ + `square`  .....[Theorem of the external angle of a triangle]

 = `1/2` m(arc SQ) + `square`   .....[Inscribed angle theorem]

= `1/2 [square + square]`

In figure, chord EF || chord GH. Prove that, chord EG ≅ chord FH. Fill in the blanks and write the proof.

Proof: Draw seg GF.

∠EFG = ∠FGH     ......`square`    .....(I)

∠EFG = `square`   ......[inscribed angle theorem] (II)

∠FGH = `square`   ......[inscribed angle theorem] (III)

∴ m(arc EG) = `square`  ......[By (I), (II), and (III)]

chord EG ≅ chord FH   ........[corresponding chords of congruent arcs]

The angle inscribed in the semicircle is a right angle. Prove the result by completing the following activity.

Given: ∠ABC is inscribed angle in a semicircle with center M

To prove: ∠ABC is a right angle.

Proof: Segment AC is a diameter of the circle.

∴ m(arc AXC) = `square`

Arc AXC is intercepted by the inscribed angle ∠ABC

∠ABC = `square`      ......[Inscribed angle theorem]

= `1/2 xx square`

∴ m∠ABC = `square`

∴ ∠ABC is a right angle.

Prove that angles inscribed in the same arc are congruent.

Given: In a circle with center C, ∠PQR and ∠PSR are inscribed in same arc PQR. Arc PTR is intercepted by the angles.

To prove: ∠PQR ≅ ∠PSR.

Proof:

m∠PQR = `1/2 xx ["m"("arc PTR")]`         ......(i) `square`

m∠`square` = `1/2 xx ["m"("arc PTR")]`   ......(ii) `square`

m∠`square` = m∠PSR      .....[By (i) and (ii)]

∴ ∠PQR ≅ ∠PSR

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)`