find the hcf of the terms of the expreission 2b^3c + 4b^4c^2 + 16b^2c^2 also, write the factors of the expression
Answers
Answer:
Hint: Take the triangle as ABC. Draw base BC as 5 cm and construct the equilateral triangle with side 5 cm. Take angle bisectors of B and C. Find the center of the circle, and with the center O, draw a circle inscribed in triangle ABC.
Complete step-by-step answer:
Let us name the equilateral triangle as ABC.
First draw the line segment of 5 cm. Let us take BC = 5 cm. Then with center as B and C, draw two arcs of 5 cm length. These arcs intersect at point A.
Thus we got an equilateral triangle ABC by joining the points AB and AC.
As it’s an equilateral triangle, we know all the sides are equal which is 5 cm. Similarly all the angles of triangle ABC is similar, i.e. the angle is 60∘.
60∘.
Let us draw angle bisectors from B and C of the triangle and they meet at point O, i.e. angle bisectors of ∠B
∠B
and ∠C
∠C
intersect each other at O. From point O, we can say that OL⊥BC
OL⊥BC
, i.e. OL is perpendicular to the line segment BC.
Now with center O and radius as OL, draw a circle which touches the sides of ΔABC.
ΔABC.
Thus we got the required figure, where an equilateral triangle of side 5 cm is drawn and a circle is inscribed in it.
Thus if we measure the radius of the inscribed circle, OL = 1.4 cm.
Hence radius, OL = 1.4 cm.
Note: We can also take angle bisectors and A and C or even B and C. We will get the same center as given in the figure. Draw OL perpendicular to BC in order to get the radius, then the construction of circle becomes easier and the circle should touch the sides of the ΔABC.
Answer:the HCF is 2b2C and factors are 2b2C and b+2b2c +8c
Step-by-step explanation:
2b3c+4b4c2+16b2c2
Now take a common
2b2c(b+2b2c+8c)
HCF=2b2C factors=2b2Candb+2b2c+8c