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

As shown in figure, LK = `6sqrt(2)` then

1. MK = ?
2. ML = ?
3. MN = ?

In ΔABC, ∠ABC = 90°, ∠BAC = ∠BCA = 45°. If AC = `9sqrt(2)`, then find the value of AB.

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`

In ΔABC, AB = 9 cm, BC = 40 cm, AC = 41 cm. State whether ΔABC is a right-angled triangle or not. Write reason.

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 the given figure, triangle PQR is right-angled at Q. S is the mid-point of side QR. Prove that QR² = 4(PS² – PQ²).

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.

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.

In the adjoining figure, a tangent is drawn to a circle of radius 4 cm and centre C, at the point S. Find the length of the tangent ST, if CT = 10 cm.

In a right angled triangle, right-angled at B, lengths of sides AB and AC are 5 cm and 13 cm, respectively. What will be the length of side BC?