निम्नलिखित वाक्य का रचना के आधार पर भेद पहचानकर लिखिए :
निम्नलिखित वाक्य का रचना के आधार पर भेद पहचानकर लिखिए :
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Solution and Explanation
वाक्य भेद: यह वाक्य समीकरण वाक्य है, क्योंकि यह एक तथ्य या विचार की जानकारी प्रस्तुत करता है।
Top Questions on व्याकरण
मैं तुमसे हमेशा पाँच साल बड़ा रहूँगा। (संयुक्त वाक्य में बदलिए।)
अपनी बात चटपट कहो और अपनी राह लो। (रचना की दृष्टि से वाक्य-भेद लिखिए।)
जैसे ही हँसी साठवें की जगह आठवीं लिखा वैसे ही सब नंबर गायब! (रचना की दृष्टि से वाक्य-भेद लिखिए।)
मेरे दरजे में आओगे, तो दोपहर पसीना आ जाएगा। (सरल वाक्य में रूपांतरित कीजिए।)
सफल खिलाड़ी का कोई निशाना खाली नहीं जाता। (मिश्र वाक्य में बदलिए।)
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.