माध्यमभाषया सरलार्थं लिखत। (2 तः 1)
यथैव सकला नद्यः प्रविशन्ति महोदधिम्।
तथा मानवताधर्मः सर्वे धर्माः समाग्रतः॥
माध्यमभाषया सरलार्थं लिखत। (2 तः 1)
यथैव सकला नद्यः प्रविशन्ति महोदधिम्।
तथा मानवताधर्मः सर्वे धर्माः समाग्रतः॥
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माध्यमभाषया सरलार्थं लिखत। (2 तः 1)
मनुजा वाचनेनैव बोधनं विषयान् बहून्।
दक्षा भवन्ति कार्येषु वाचनेन बहुश्रुताः॥
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श्लोक का अर्थ संस्कृत में लिखिए : ब्राह्मे मुहूर्ते बुध्येत, धर्मार्थौ चानुचिन्तयेत् । कायक्लेशाँश्च तन्मूलान् वेदतत्वार्थमेव च ।।
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श्लोक की हिन्दी में व्याख्या कीजिए : अभिवादनशीलस्य, नित्यं वृद्धोपसेविनः । चत्वारि तस्य वर्धन्ते आयुर्विद्या यशोबलम् ।।
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Questions Asked in Maharashtra Class X Board exam
- Appreciation of the poem:
Read the following poem and write an appreciation of it with the help of the points given below:
Stopping by Woods on a Snowy Evening
Whose woods these are I think I know.
His house is in the village, though;
He will not see me stopping here
To watch his woods fill up with snow
My little horse must think it queer
To stop without a farmhouse near
Between the woods and frozen lake
The darkest evening of the year.
He gives his harness bells a shake
To ask if there is some mistake.
The only other sound’s the sweep
Of easy wind and downy flake.
The woods are lovely, dark and deep,
But I have promises to keep,
And miles to go before I sleep,
And miles to go before I sleep.
Robert Frost
- Maharashtra Class X Board - 2025
- Reading Comprehension
- Pick out the infinitive from the following sentence :
He was asking to go for the concert.- Maharashtra Class X Board - 2025
- Grammar and Language Usage
- Out of the following which is a Pythagorean triplet ?
In the following figure \(\triangle\) ABC, B-D-C and BD = 7, BC = 20, then find \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\).
- Maharashtra Class X Board - 2025
- Properties of Triangles
The radius of a circle with centre 'P' is 10 cm. If chord AB of the circle subtends a right angle at P, find area of minor sector by using the following activity. (\(\pi = 3.14\))
Activity :
r = 10 cm, \(\theta\) = 90\(^\circ\), \(\pi\) = 3.14.
A(P-AXB) = \(\frac{\theta}{360} \times \boxed{\phantom{\pi r^2}}\) = \(\frac{\boxed{\phantom{90}}}{360} \times 3.14 \times 10^2\) = \(\frac{1}{4} \times \boxed{\phantom{314}}\) <br>
A(P-AXB) = \(\boxed{\phantom{78.5}}\) sq. cm.