Application of Vedic Maths For Class 9 and 10 in Mensuration (क्षेत्रफल तथा आयतन में वैदिक गणित का अनुप्रयोग कक्षा 9 तथा 10)
वैदिक गणित का कक्षा नवमी तथा दशमी के लिए क्षेत्रमिति में अनुप्रयोग (Application of Vedic-Ganit For Class X & IX in Menstruation )
सूत्र – एकाधिकेन पूर्वेण –
पहले से एक अधिक के द्वारा
(One more than the existing one)
उपसूत्र – अन्त्योर्दशकेअपि –
अंतिम अंको का योग दस।
(Sum of last digits is ten.)
गोवर्धन पीठ, पूरी के 143 वें शंकराचार्य जगत गुरु स्वामी भारती कृष्ण तीर्थ जी महाराज द्वारा रचित वैदिक गणित के उपरोक्त सूत्र (formula) के प्रयोग के द्वारा विशेष(special) परिस्थितियों (conditions) में गुणा (multiply) किया जा सकता है जब किसी दो संख्याओं के इकाई-अंकों (ones-place) का योग (Sum) दस (ten) हो तथा दहाई (tens) या शेष (rest) अंक (digits) समान ( equal) हो।
जैसे – 25 × 25, 38 × 32, 46 × 44, 53 × 57, 69 ×61, 105 × 105
उपरोक्त उदाहरण (example) में इकाई अंकों का योग दस है – 5 + 5, 8 + 2, 6 + 4, 3 + 7, 9 + 1 इत्यादि तथा दहाई या शेष अंक समान है – 2 – 2, 3 – 3, 4 – 4, 5 – 5, 6 – 6, 10 – 10
बीजगणितीय विश्लेषण ( Algebraic analysis)
(X5)² = ( 10 X + 5)²
= ( 10 X + 5) ( 10 X + 5)
= 10 X ( 10 X + 5) + 5 ( 10 X + 5)
= 100 X² + 50 X + 50 X + 25
= 100 X² + 100X + 25
= 100 X ( X + 1) + 25
= X ( X + 1) सैकड़ा (hundreds) + 25
उदाहरण (Example)
(1) 294 × 296 = 29 × (29 + 1) / 24 = 870 24
(2) 405 × 405 = 40 × ( 40 + 1) / 25 = 1640 25
(3) 499 × 491 = 49 × (49 + 1) / 09 = 2450 09
सूत्र – (2)
11, 111 से गुणन (Multiplication with 11, 111)
उदाहरण (Example)
121 × 11
= 0 1 2 1 0
= 0+1 / 1+2 / 2+1 / 1+0
= 1 3 3 1
उदाहरण (Example)
121 × 111
= 0 0 1 2 1 0 0
= 0+0+1 / 0+1+2 / 1+2+1 / 2+1+0 / 1+0+0
= 1 3 4 3 1
सूत्र – (3)
विलोकनम्
निरीक्षण तथा परीक्षण के द्वारा
By Observation
पृष्ठीय क्षेत्रफल (Surface area)
Question :- (1)
एक खिलौना जिसकी त्रिज्या 3. 5 सेमी. है जोकि समान त्रिज्या वाले अर्धगोले के उपर स्थित है जिसकी कुल ऊँचाई 15. 5 सेमी. है का संपूर्ण पृष्ठीय क्षेत्रफल ज्ञात किजिये।
( A toy is in the form of a cone of radius 3. 5 cm. mounted on a hemisphere of the same radius. Total height of the toy is 15. 5 cm. Find the total surface area of the toy.)
ठोस का आयतन (Volume of Solid)
Solutions :-
Radius = 3.5 cm.
Total height = 15.5 cm.
Height of cone = (15.5 – 3.5) = 12 cm
(Slant height)² = (Radius)² + (Height)²
= (12)² + (3.5)²
= 144 + 12.25
= 156. 25 = (12.5)²
Slant height = 12.5 cm.
Surface area of toy = C S A of Cone + C S A of Hemisphere
= pi r l + 2 pi r²
= pi r ( l + 2r)
= 22/7 × 3.5 ( 12.5 + 2× 3.5)
= 11 × 19.5 = 214.5 cm²
आयतन (Volume)
Question :- (2)
एक ठोस खिलौने का आयतन ज्ञात किजिये जोकि शंकु के उपर अर्धगोला के रुप में निर्मित है, जिसकी त्रिज्या 10. 5 सेमी. तथा कुल ऊँचाई 42 सेमी. है।
(Find the volume of a solid toy which is in the form of cone with surmounted by a hemisphere with radius 10. 5 cm. and it’s total height 42 cm.)
हल (Solution) :-
त्रिज्या (r) = 10. 5 cm.
ऊँचाई (H) = 42 cm.
शंकु की ऊँचाई (h) = 31.5 cm
Vol. of solid toy = Vol. of cone + Vol. of hemisphere
= 1/3 pi r² h + 2/3 pi r³
= 1/3 pi r² ( h + r)
= 1/3 × 22/7 × (10.5)² ( 10.5 + 31.5)
= 1/3 × 22/7 × 110.25 × 42
= 4851 cm³
{ = 22 × 110.25 × 2
= 11 × 441.00
= 4851 cm³}
Exercise
(1) A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
(2) A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs. 500 per m².
(3) From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².
(4) A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as show in Fig. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
(5) A vessel is in the form of an inverted cone. its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm are dropped into the vessel, one-forth of the water flows out. Find the number of lead shots dropped in the vessel.
(6) A 20 m deep well with diameter 7 m is dug and the earth from digging is evenly spread out to form a platform 22 m by 14 m. Find the height of the platform.
(7) A well of diamter 3 m is dug 14 m deep. The earth taken out of it has been spread evenly all around it in shape of a circular ring of width 4 m to form an embankment. Find the height of the embankment.
(8) How many silver coins, 1.75 cm in diameter and of thickness 2 mm, must be melted to form a cuboid of dimensions 5.5 cn × 10 cm × 3.5 cm?
Application of Vedic Maths For Class X IX in Menstruation (क्षेत्रफल तथा आयतन में वैदिक गणित का अनुप्रयोग कक्षा IX तथा X)
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