The K-Pop band has a fast-growing fanbase in Kerala, with members of BTS Kerala ARMY holding a flash mob to mark the band’s sixth anniversaryપ્રશ્ન 1.
કડવા લીમડાની છાયા કેવી હોય છે?
(a) તૂરી
(b) શીતળ
(c) ગરમ
(d) આકરી
ઉત્તર :
(b) શીતળReleased: 2019
Album: Winter Bear
Artist: V
Given :
AB is parallel to DC ,AB = 11 cm, DC = 7 cm, AD = BC = 6 cm
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment AB = 11 cm.
Step 3 : Mark E on AB such that $$\overline{AE}$$ = 7 cm ( since DC = 7 cm)
Step 4 : With E and B as centres and (AD = EC = 6 cm) radius 6 cm draw two arcs. Let them cut at C.
Step 5 : Join BC and $$\overline{EC}$$ .
Step 6 : With C and A as centres draw two arcs of radii 7 cm and 6 cm respectively and let them cut at D.
Step 7 : Join AD and $$\overline{CD}$$ . ABCD is the required isosceles trapezium.
Step 8 : From D draw DF⊥AB and measure the length of DF. DF = h = 5.6 cm. AB = a = 11 cm and CD = b = 7 cm.
Calculation of area:
In the isosceles trapezium ABCD, a = 11 cm, b = 7 cm and h = 5.6 cm.
Area of the isosceles trapezium ABCD = 21h(a+b)
=21(5.6)(11+7)
=21×5.6×18
= 50.4 cm2
A particle of mass 5×10−5kg is placed at lowest point of smooth parabola x2=40y (x and y in m) (in rad/s). If its is displaced slightly such that it is constrained to move along parabola, angular frequency of oscillation will be approximately (g=10m/s2.
Answers & Comments
Answer:
mark me as brainlist
Explanation:
The K-Pop band has a fast-growing fanbase in Kerala, with members of BTS Kerala ARMY holding a flash mob to mark the band’s sixth anniversaryપ્રશ્ન 1.
કડવા લીમડાની છાયા કેવી હોય છે?
(a) તૂરી
(b) શીતળ
(c) ગરમ
(d) આકરી
ઉત્તર :
(b) શીતળReleased: 2019
Album: Winter Bear
Artist: V
Given :
AB is parallel to DC ,AB = 11 cm, DC = 7 cm, AD = BC = 6 cm
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment AB = 11 cm.
Step 3 : Mark E on AB such that $$\overline{AE}$$ = 7 cm ( since DC = 7 cm)
Step 4 : With E and B as centres and (AD = EC = 6 cm) radius 6 cm draw two arcs. Let them cut at C.
Step 5 : Join BC and $$\overline{EC}$$ .
Step 6 : With C and A as centres draw two arcs of radii 7 cm and 6 cm respectively and let them cut at D.
Step 7 : Join AD and $$\overline{CD}$$ . ABCD is the required isosceles trapezium.
Step 8 : From D draw DF⊥AB and measure the length of DF. DF = h = 5.6 cm. AB = a = 11 cm and CD = b = 7 cm.
Calculation of area:
In the isosceles trapezium ABCD, a = 11 cm, b = 7 cm and h = 5.6 cm.
Area of the isosceles trapezium ABCD = 21h(a+b)
=21(5.6)(11+7)
=21×5.6×18
= 50.4 cm2
A particle of mass 5×10−5kg is placed at lowest point of smooth parabola x2=40y (x and y in m) (in rad/s). If its is displaced slightly such that it is constrained to move along parabola, angular frequency of oscillation will be approximately (g=10m/s2.
Verified answer
Answer:
સીરામણ
Explanation:
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