Geometry Mathematics 2 Model set 4 by shaalaa.com 2024-2025 SSC (English Medium) 10th Standard Board Exam Question Paper Solution

If tan θ = `12/5`, then 5 sin θ – 12 cos θ = ?

`119/13`

0

1

`1/13`

Find the ∠ADE, if ∠BDF = 60° and ADB is the tangent to the circle with centre C.

30°

60°

45°

90°

If the perimeter of two similar triangles is in the ratio 2 : 3, what is the ratio of their sides?

4 : 9

2 : 3

`sqrt(2)` : `sqrt(3)`

3 : 2

Find the centroid of the ΔABC whose vertices are A(–2, 0), B(7, –3) and C(6, 2).

`(11/3, 1/3)`

`(11/3, (-1)/3)`

`((-11)/3, (-1)/3)`

`((-11)/3, 1/3)`

If sinθ = cosθ, then what will be the measure of angle θ?

In the given figure, if sin θ = `7/13`, which angle will be θ?

The sum of two angles of a triangle is 150°, and their difference is 30°. Find the angles.

In which quadrant, does the abscissa, and ordinate of a point have the same sign?

AB, BC and AC are three sides of a right-angled triangle having lengths 6 cm, 8 cm and 10 cm, respectively. To verify the Pythagoras theorem for this triangle, fill in the boxes:

ΔABC is a right-angled triangle and ∠ABC = 90°.

So, by the Pythagoras theorem,

`square` + `square` = `square`

Substituting 6 cm for AB and 8 cm for BC in L.H.S.

`square` + `square` = `square` + `square`

= `square` + `square`

= `square`

Substituting 10 cm for AC in R.H.S.

`square` = `square`

= `square`

Since, L.H.S. = R.H.S.

Hence, the Pythagoras theorem is verified.

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.

In the given figure, ΔLMN is similar to ΔPQR. To find the measure of ∠N, complete the following activity.

Given: ΔLMN ∼ ΔPQR

Since ΔLMN ∼ ΔPQR, therefore, corresponding angles are equal.

So, ∠L ≅ `square`

⇒ ∠L = `square`

We know, the sum of angles of a triangle = `square`

∴ ∠L + ∠M + ∠N = `square`

Substituting the values of ∠L and ∠M in equation (i),

`square` + `square` + ∠N = `square`

∠N + `square` = `square`

∠N = `square` – `square`

∠N = `square`

 Hence, the measure of ∠N is `square`.

A pole of height 30 m is observed from a point. The angle of depression of the point is 30°. Find the distance of the point from the base of the pole.

In what ratio does the Y-axis divide the line segment P(– 3, 1) and Q(6, 2)?

A circle of radius 3 cm with centre O and a point L outside the circle is drawn, such that OL = 7 cm. From the point L, construct a pair of tangents to the circle. Justify LM and LN are the two tangents.

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

In the adjoining figure, ΔADB ∼ ΔBDC. Prove that BD² = AD × DC.

Radius of a circle is 10 cm. Measure of an arc of the circle is 54°. Find the area of the sector associated with the arc. (π = 3.14)

Given: The radius of a circle (r) = `square`

Measure of an arc of the circle (θ) = `square`

Area of the sector = `θ/360^circ xx square`

= `square/360^circ xx square xx square xx square`

= `square xx square xx square`

= 47.10 cm²

Prove that: (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ

Proof: L.H.S. = (sec θ – cos θ) (cot θ + tan θ)

= `(1/square - cos θ) (square/square + square/square)` ......`[∵ sec θ = 1/square, cot θ = square/square and tan θ = square/square]`

= `((1 - square)/square) ((square + square)/(square  square))`

= `square/square xx 1/(square  square)`  ......`[(∵ square + square = 1),(∴ square = 1 - square)]`

 = `square/(square  square)`

= tan θ.sec θ

= R.H.S.

∴ L.H.S. = R.H.S.

∴ (sec θ – cos θ) (cot θ + tan θ) = tan θ.sec θ

In the figure, PQRS is a square with side 10 cm. The sectors drawn with P and R as centres form the shaded figure. Find the area of the shaded figure. (Use π = 3.14)

In the given figure, S is a point on side QR of ΔPQR such that ∠QPR = ∠PSR. Use this information to prove that PR² = QR × SR.

A 7 m broad pathway goes around a circular park with a circumference of 352 m. Find the area of road.

The top of a banquet hall has an angle of elevation of 45° from the foot of a transmission tower and the angle of elevation of the topmost point of the tower from the foot of the banquet hall is 60°. If the tower is 60 m high, find the height of the banquet hall in decimals.

If two consecutive angles of cyclic quadrilateral are congruent, then prove that one pair of opposite sides is congruent and other is parallel.

A cylinder and a cone have equal bases. The height of the cylinder is 2 cm and the area of its base is 64 cm². The cone is placed upon the cylinder volume of the solid figure so formed is 400 cm³. Find the total height of the figure.

In the given figure, triangle ABC is a right-angled at B. D is the mid-point of side BC. Prove that AC² = 4AD² – 3AB².

In an isosceles triangle PQR, the length of equal sides PQ and PR is 13 cm and base QR is 10 cm. Find the length of perpendicular bisector drawn from vertex P to side QR.

A milk container of height 16 cm is made of metal sheet in the form of frustum of a cone with radii of its lower and upper ends as 8 cm and 20 cm respectively. Find the cost of milk at the rate of ₹ 22 per litre which the container can hold.