SSC (Marathi Semi-English) 10th Standard Board Exam [इयत्ता १० वी] - Maharashtra State Board Important Questions for Geometry Mathematics 2

In the above figure `square`ABCD is a rectangle. If AB = 5, AC = 13, then complete the following activity to find BC.

Activity: ΔABC is a `square` triangle.

∴ By Pythagoras theorem

AB² + BC² = AC²

∴ 25 + BC² = `square`

∴ BC² = `square`

∴ BC = `square`

If a and b are natural numbers and a > b If (a² + b²), (a² – b²) and 2ab are the sides of the triangle, then prove that the triangle is right-angled. Find out two Pythagorean triplets by taking suitable values of a and b.

Construct two concentric circles with centre O with radii 3 cm and 5 cm. Construct a tangent to a smaller circle from any point A on the larger circle. Measure and write the length of the tangent segment. Calculate the length of the tangent segment using Pythagoras' theorem.

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

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

From the information in the figure, complete the following activity to find the length of the hypotenuse AC.

AB = BC = `square`

∴ ∠BAC = `square`

Side opposite angle 45° = `square/square` × Hypotenuse

∴ `5sqrt(2) = 1/square` × AC

∴ AC = `5sqrt(2) xx square = square`

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)`

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.

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.

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

There is a ladder of length 32 m which rests on a pole. If the height of pole is 18 m, determine the distance between the foot of ladder and the pole.

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

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.

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?

In the adjoining figure, O is the centre of the circle. From point R, seg RM and seg RN are tangent segments touching the circle at M and N. If (OR) = 10 cm and radius of the circle = 5 cm, then

1. What is the length of each tangent segment?
2. What is the measure of ∠MRO?
3. What is the measure of ∠MRN?

Seg RM and seg RN are tangent segments of a circle with centre O. Prove that seg OR bisects ∠MRN as well as ∠MON with the help of activity.

In the given figure, ∆QRS is an equilateral triangle. Prove that,

1. arc RS ≅ arc QS ≅ arc QR
2. m(arc QRS) = 240°.
   

In the given figure, in a circle with centre O, length of chord AB is equal to the radius of the circle. Find measure of each of the following.
(1) ∠ AOB (2)∠ ACB
(3) arc AB (4) arc ACB.

In the given figure, O is the centre of the circle and B is a point of contact. seg OE ⊥ seg AD, AB = 12, AC = 8, find (1) AD (2) DC (3) DE.