State whether the following sentences are right or wrong
India and Brazil have the same seasons at the same time.
India and Brazil have the same seasons at the same time.
Show Hint
Solution and Explanation
India and Brazil have the same seasons at the same time.
Correct Answer: False
Step 1: Seasonal Differences
India is in the Northern Hemisphere, while Brazil is in the Southern Hemisphere. Therefore, the two countries experience opposite seasons. For example, when it is summer in India (March–June), it is winter in Brazil.
Step 2: Explanation of Seasons
The difference is caused by the tilt of the Earth's axis and its orbit around the Sun. When the Northern Hemisphere tilts toward the Sun, it experiences summer; the Southern Hemisphere experiences winter — and vice versa.
\[ \textbf{India and Brazil have opposite seasons due to Earth's axial tilt.} \]
Top Questions on Climate
- Give geographical reasons for the following
Most of the Himalayan rivers are perennial in nature. - Explain any three basic differences between rural and urban settlements in India.
- Subduction zone is also known as
- Which one of the following is known as 'doctor' wind?
- Roaring forties represent ____ .
Questions Asked in Maharashtra Class X Board exam
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: 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 \(\triangle\)PQS, PT is the median and in \(\triangle\)QRS, RT is the median.
\(\therefore\) by Apollonius theorem,
\[\begin{aligned} PQ^2 + PS^2 &= \boxed{\phantom{X}} + 2QT^2 \quad \dots \text{(I)} \\ QR^2 + SR^2 &= \boxed{\phantom{X}} + 2QT^2 \quad \dots \text{(II)} \\ \text{Adding (I) and (II),} \quad PQ^2 + PS^2 + QR^2 + SR^2 &= 2(PT^2 + \boxed{\phantom{X}}) + 4QT^2 \\ &= 2(PT^2 + \boxed{\phantom{X}}) + 4QT^2 \quad (\text{RT = PT}) \\ &= 4PT^2 + 4QT^2 \\ &= (\boxed{\phantom{X}})^2 + (2QT)^2 \\ \therefore \quad PQ^2 + PS^2 + QR^2 + SR^2 &= PR^2 + \boxed{\phantom{X}} \\ \end{aligned}\]
- Maharashtra Class X Board - 2025
- Quadrilaterals
- Prove that tangent segments drawn from an external point to a circle are congruent.
- Draw a circle with radius 4.1 cm. Construct tangents to the circle from a point at a distance 7.3 cm from the centre.
- Maharashtra Class X Board - 2025
- Constructions
- \(\triangle\)ABC \(\sim\) \(\triangle\)PQR. In \(\triangle\)ABC, AB = 3.6 cm, BC = 4 cm and AC = 4.2 cm. The corresponding sides of \(\triangle\)ABC and \(\triangle\)PQR are in the ratio 2 : 3, construct \(\triangle\)ABC and \(\triangle\)PQR.
- Maharashtra Class X Board - 2025
- Constructions
- \(\square\)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.