Mahabharat ko Apne vichar mein likhiye
Answers
महाभारत
पांडव सदैव धर्म के मार्ग में चलते थे
पर दुर्योधन कभी भी धर्म के मार्ग नहीं अपनाया
Explanation:
Given that
DE=4cm
=5.5cm
DF=5cm
Step of construction
(i) Draw a line segment EF=5.5cm
(ii) With F as centre and radius 5cm draw an arc
(iii) With E as centre & radius 4cm draw another arc cutting the previous arc at D
(iv) Join FD and ED then triangle DEF is the required triangle.
Steps for construction :
Step 1. Draw a line segment PQ=3.5 cm.
Step 2. With P as centre and radius =4.5 cm, draw an arc.
Step 3. With Q as centre and radius =6.5 cm, draw an arc to cut the arc in the previous step at point S
Step 4. Join PS and QS
Step 5. With S as centre and radius =5 cm, draw an arc.
Step 3. With Q as centre and radius =4 cm, draw an arc to cut the arc in the previous step at point R
Step 4. Join SR and QR
Hence, required quadrilateral PQRS is constructed.
Given that
DE=4cm
=5.5cm
DF=5cm
Step of construction
(i) Draw a line segment EF=5.5cm
(ii) With F as centre and radius 5cm draw an arc
(iii) With E as centre & radius 4cm draw another arc cutting the previous arc at D
(iv) Join FD and ED then triangle DEF is the required triangle.
Steps for construction :
Step 1. Draw a line segment PQ=3.5 cm.
Step 2. With P as centre and radius =4.5 cm, draw an arc.
Step 3. With Q as centre and radius =6.5 cm, draw an arc to cut the arc in the previous step at point S
Step 4. Join PS and QS
Step 5. With S as centre and radius =5 cm, draw an arc.
Step 3. With Q as centre and radius =4 cm, draw an arc to cut the arc in the previous step at point R
Step 4. Join SR and QR
Hence, required quadrilateral PQRS is constructed.
Given that
DE=4cm
=5.5cm
DF=5cm
Step of construction
(i) Draw a line segment EF=5.5cm
(ii) With F as centre and radius 5cm draw an arc
(iii) With E as centre & radius 4cm draw another arc cutting the previous arc at D
(iv) Join FD and ED then triangle DEF is the required triangle.
Steps for construction :
Step 1. Draw a line segment PQ=3.5 cm.
Step 2. With P as centre and radius =4.5 cm, draw an arc.
Step 3. With Q as centre and radius =6.5 cm, draw an arc to cut the arc in the previous step at point S
Step 4. Join PS and QS
Step 5. With S as centre and radius =5 cm, draw an arc.
Step 3. With Q as centre and radius =4 cm, draw an arc to cut the arc in the previous step at point R
Step 4. Join SR and QR
Hence, required quadrilateral PQRS is constructed.
Given that
DE=4cm
=5.5cm
DF=5cm
Step of construction
(i) Draw a line segment EF=5.5cm
(ii) With F as centre and radius 5cm draw an arc
(iii) With E as centre & radius 4cm draw another arc cutting the previous arc at D
(iv) Join FD and ED then triangle DEF is the required triangle.
Steps for construction :
Step 1. Draw a line segment PQ=3.5 cm.
Step 2. With P as centre and radius =4.5 cm, draw an arc.
Step 3. With Q as centre and radius =6.5 cm, draw an arc to cut the arc in the previous step at point S
Step 4. Join PS and QS
Step 5. With S as centre and radius =5 cm, draw an arc.
Step 3. With Q as centre and radius =4 cm, draw an arc to cut the arc in the previous step at point R
Step 4. Join SR and QR
Hence, required quadrilateral PQRS is constructed.
Given that
DE=4cm
=5.5cm
DF=5cm
Step of construction
(i) Draw a line segment EF=5.5cm
(ii) With F as centre and radius 5cm draw an arc
(iii) With E as centre & radius 4cm draw another arc cutting the previous arc at D
(iv) Join FD and ED then triangle DEF is the required triangle.
Input to the given function f is denoted by t; input to its Laplace transform F is denoted by s. By default, the domain of the function f=f(t) is the set of all non- negative real numbers. The domain of its Laplace transform depends on f and can vary from a function to a function. L(fInput to the given function f is denoted by t; input to its Laplace transform F is denoted by s. By default, the domain of the function f=f(t) is the set of all non- negative real numbers. The domain of its Laplace transform depends on f and can vary from a function to a function. L(f