Maharashtra SSC Geometry HOTS Questions with Answers – Pythagoras Theorem | Class 10 Board Exam 2027

Asked in: March 2025 and March 2026 board papers (same question with option (5, 12, 13) in June 2026, and with dates in March 2024)

Out of the following which is a Pythagorean triplet ?

Asked in: July 2025 board paper (same question with side 5 cm in June 2026)

Find the diagonal of square whose side is 10 cm.

Asked in: March 2024 and March 2026 board papers

Find the side of a square whose diagonal is \(10\sqrt{2}\) cm.

Asked in: July 2024 board paper (same type as a 1-mark question with MQ = 9, QP = 4 in March 2025)

In the figure, \(\angle PQR = 90^\circ\), seg QN \(\perp\) seg PR, PN = 9, NR = 16, find QN.

[Figure: right-angled \(\Delta PQR\) with the right angle at Q; QN is the perpendicular from Q to the hypotenuse PR.]

Asked in: March 2024 board paper

Find the length of the hypotenuse of a right-angled triangle if remaining sides are 9 cm and 12 cm.

Asked in: June 2026 board paper

Find the length of diagonal of a rectangle whose length is 35 cm and breadth is 12 cm.

Asked in: March 2026 board paper (a 45°–45°–90° activity was also asked in March 2026)

In \(\Delta RST\), \(\angle S = 90^\circ\), \(\angle T = 30^\circ\), RT = 12 cm, then find RS and ST.

Asked in: March 2025 board paper

Complete the following activity to prove that the sum of squares of diagonals of a rhombus is equal to the sum of the squares of the sides.

Given: \(\square PQRS\) is a rhombus. Diagonals PR and SQ intersect each other at point T. To prove: \(PS^2 + SR^2 + QR^2 + PQ^2 = PR^2 + QS^2\).

Activity: Diagonals of a rhombus bisect each other. In \(\Delta PQS\), PT is the median and in \(\Delta QRS\), RT is the median. \(\therefore\) by Apollonius theorem, \(PQ^2 + PS^2 = \text{____} + 2QT^2\) ...... (I); \(QR^2 + SR^2 = \text{____} + 2QT^2\) ...... (II). Adding (I) and (II), \(PQ^2 + PS^2 + QR^2 + SR^2 = 2(PT^2 + \text{____}) + 4QT^2 = 2(PT^2 + \text{____}) + 4QT^2\) (RT = PT) \(= 4PT^2 + 4QT^2 = (\text{____})^2 + (2QT)^2\) \(\therefore PQ^2 + PS^2 + QR^2 + SR^2 = PR^2 + \text{____}\).

Asked in: March 2024 board paper (same Apollonius type in June 2026)

Seg PM is a median of \(\Delta PQR\), PM = 9 and \(PQ^2 + PR^2 = 290\), then find QR.

Asked in: June 2026 board paper

In \(\Delta PQR\), point S is the midpoint of side QR. If PQ = 11, PR = 17, PS = 13, find QR.

Asked in: July 2025 board paper

Prove that, ‘In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of remaining two sides’.

Asked in: July 2024 board paper

\(\Delta ABC\) is an equilateral triangle. Point D is on seg BC such that \(BD = \dfrac{1}{5} BC\). Then prove that \(\dfrac{AD^2}{AB^2} = \dfrac{21}{25}\).

Asked in: March 2025 board paper

\(\square ABCD\) is a rectangle. Taking AD as a diameter, a semicircle AXD is drawn which intersects the diagonal BD at X. If AB = 12 cm, AD = 9 cm, then find the values of BD and BX.

Out of the following, which is a Pythagorean triplet?

In a 30°–60°–90° triangle, the side opposite to the 30° angle is 7 cm. The hypotenuse is

In \(\Delta ABC\), D is the midpoint of BC. If \(AB^2 + AC^2 = 50\) and BC = 8, then AD = ?

Find the height of an equilateral triangle whose side is 6 cm.

The sides of a triangle are 8 cm, 15 cm and 17 cm. Is it a right-angled triangle? Give reason.

A ladder 13 m long rests against a vertical wall and its top reaches a window 12 m above the ground. How far is the foot of the ladder from the wall?

In \(\Delta ABC\), \(\angle ACB = 90^\circ\), seg CD \(\perp\) side AB, A – D – B. If AD = 4 and DB = 9, find CD and AC.

In \(\Delta LMN\), \(\angle M = 90^\circ\), \(\angle N = 45^\circ\) and \(LN = 6\sqrt{2}\) cm. Find LM and MN.

Prove that, ‘In a triangle, if the square of one side is equal to the sum of the squares of the remaining two sides, then the angle opposite to the first side is a right angle’.

In \(\Delta PQR\), \(\angle PQR\) is an obtuse angle. Seg PS \(\perp\) line QR, with R – Q – S. Prove that \(PR^2 = PQ^2 + QR^2 + 2\,QR \times QS\).

The sides of a parallelogram are 6 cm and 8 cm and one of its diagonals is 12 cm. Find the length of the other diagonal.

Prove Apollonius theorem: In \(\Delta ABC\), if M is the midpoint of side BC, then \(AB^2 + AC^2 = 2AM^2 + 2BM^2\).

Two vertical poles of heights 11 m and 6 m stand 12 m apart on level ground. (i) Find the distance between their tops. (ii) A wire is tied from the top of the taller pole to a peg on the ground exactly midway between the poles. Find the length of the wire.

\(\Delta ABC\) is an equilateral triangle. Point D is on side BC such that \(BD = \dfrac{1}{3} BC\). Prove that \(9AD^2 = 7AB^2\).