Geometry Mathematics 2 Model set 3 by shaalaa.com 2024-2025 SSC (English Medium) 10th Standard Board Exam Question Paper Solution

What is the distance of the point (– 5, 4) from the origin?

3 units

`sqrt(14)` units

`sqrt(31)` units

`sqrt(41)` units

If two tangents TL and TM are drawn to a circle with centre C such that ∠LTM = 70°, then find ∠MCT.

30°

60°

45°

55°

What will be the value of sin 45° + `1/sqrt(2)`?

`1 + sqrt(2)`

`2sqrt(2)`

`1/sqrt(2)`

`sqrt(2)`

In the right-angled triangle ABC, Hypotenuse AC = 10 and side AB = 5, then what is the measure of ∠A?

30°

60°

90°

45°

What is the name of the point of intersection of coordinate axes?

Find the value of sin 45° + cos 45° + tan 45°.

Chord AB and chord CD of a circle with centre 0 are congruent. If m(arc AB) = 120°, then find the m(arc CD).

`square`ABCD is cyclic. If ∠B = 110°, then find measure of ∠D.

Prove that, The areas of two triangles with the same height are in proportion to their corresponding bases. To prove this theorem start as follows:

1. Draw two triangles, give the names of all points, and show heights.
2. Write 'Given' and 'To prove' from the figure drawn.

If the points P(1, 2), Q(0, 0) and R(x, y) are collinear, then find the relation between x and y.

Given points are P(1, 2), Q(0, 0) and R(x, y).

The given points are collinear, so the area of the triangle formed by them is `square`.

∴ `1/2 |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = square`

`1/2 |1(square) + 0(square) + x(square)| = square`

`square + square + square` = 0

`square + square` = 0

`square = square`

Hence, the relation between x and y is `square`.

Two circles with centres O and O' touch each other at point L. Prove that, a tangent through L bisects the common tangent AB of the two circles at point M.

Given: AB is a common tangent of the two circles that touch each other at point L. ML is a tangent through point L.

To prove: M is a mid-point of the tangent AB or MA = MB.

Proof: From the figure,

M is an external point that draws two tangents, MA and ML to the circle with the centre O.

So, `square` = `square`  ......(i)

Similarly, M draws two tangents ML and MB to the circle with the centre O'.

So, `square` = `square`  ......(ii)

From the equations (i) and (ii),

`square` = `square`

Hence, the tangent at the point L, bisects the common tangent, AB of the two circles at point M.

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.

Find the value of y, if the points A(3, 4), B(6, y) and C(7, 8) are collinear.

In a ΔABC, ∠CAB is an obtuse angle. P is the circumcentre of ∆ABC. Prove that ∠CAB – ∠PBC = 90°.

If length of the circular arc is 10 cm and the radius 3·5 cm, find the area of the sector of the circle.

Draw a circle of radius 4 cm. Draw a point 8 cm away from its centre and construct a pair of tangents.

Point P is the centre of the circle and AB is a diameter. Find the coordinates of points B if coordinates of point A and P are (2, – 3) and (– 2, 0) respectively.

Given: A`square` and P`square`. Let B (x, y)

The centre of the circle is the midpoint of the diameter.

∴ Mid point formula,

`square = (square + x)/square`

⇒ `square = square` + x

⇒ x = `square - square`

⇒ x = – 6

and `square = (square + y)/2`

⇒ `square` + y = 0

⇒ y = 3

Hence coordinates of B is (– 6, 3).

Prove that sec θ + tan θ = `cos θ/(1 - sin θ)`.

Proof: L.H.S. = sec θ + tan θ

= `1/square + square/square`

= `square/square`  ......`(∵ sec θ = 1/square, tan θ = square/square)`

= `((1 + sin θ) square)/(cos θ  square)`  ......[Multiplying `square` with the numerator and denominator]

= `(1^2 - square)/(cos θ  square)`

= `square/(cos θ  square)`

= `cos θ/(1 - sin θ)` = R.H.S.

∴ L.H.S. = R.H.S.

∴ sec θ + tan θ = `cos θ/(1 - sin θ)`

A monkey is climbing a rope of length 15 m in a circus. The rope is tied to a vertical pole from its top. Find the height of the pole, if the angle, the rope makes with the ground level is equal to 60°.

In the given figure, ΔPQR is a right-angled triangle with ∠PQR = 90°. QS is perpendicular to PR. Prove that pq = rx.

Construct an equilateral triangle of side 7 cm. Now, construct another triangle similar to the first triangle such that each of its sides are `5/7` times of the corresponding sides of the first triangle.

If one looks from a tower 10 m high at the top of a flag staff, the depression angle of 30° is made. Also, looking at the bottom of the staff from the tower, the angle of the depression made is of 60°. Find the height of the flag staff.

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)`

Use area theorem of similar triangles to prove congruency of two similar triangles with equal areas.

The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of the frustum complete the following activity.( \[\pi = \frac{22}{7}\]) 

A person starts his trip from home. He moves 24 km in south direction and then starts moving towards east. He travels 7 km in that direction and finally reaches his destination. How far is the destination from his home?

If m and n are real numbers and m > n, if m² + n², m² – n² and 2 mn are the sides of the triangle, then prove that the triangle is right-angled. (Use the converse of the Pythagoras theorem). Find out two Pythagorian triplets using convenient values of m and n.