आत्मकथन :
आत्मकथन :
Solution and Explanation
आजच्या काळात, मला खेळांमध्ये थोडा आनंद मिळतो. मी क्रीडांगणावर वेळ घालवायला आवडतो कारण तेथे शारीरिक आणि मानसिक दोन्ही प्रकारे ताजेतवाने होतो. खेळाचे महत्त्व फार मोठे आहे. त्याच्यात शरीराचे आरोग्य, एकाग्रता आणि शिस्त असते, हे आपल्याला समजते.
तथापि, आजकाल टीव्हीचा प्रभाव वाढत चालला आहे. प्रत्येकाच्या घरी टीव्ही आहे आणि लोक त्यावर सतत बघत असतात. यामुळे ते खेळांपेक्षा टीव्हीवर अधिक लक्ष देतात, ज्यामुळे क्रीडांगणाकडे दुर्लक्ष होऊ लागले आहे. क्रीडांगणावर बऱ्याच वेळा अतिक्रमण झालेले आहे. लोक या ठिकाणी येऊन आपले सामान ठेवतात, आणि खेळ खेळण्यासाठी जागा कमी होते.
आजच्या स्थितीत, क्रीडांगण हे एकेकाळी मोठे होते, पण अतिक्रमणामुळे ते कमी झाले आहे. मला खूप खंत वाटते की, पूर्वीच्या तुलनेत खेळण्याची जागा लहान झाली आहे. हे पाहून मी खूप दु:खी होतो. मला आशा आहे की, क्रीडांगणाची जागा लवकरच वाढवली जाईल आणि लोक खेळांमध्ये भाग घेतील.
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- 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
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- 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 \(\frac{A(\triangle ABD)}{A(\triangle ABC)}\).
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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.