In the given figure (not drawn to scale), AD is parallel to BC. JDK, GHCI, EABF are straight and parallel lines.
(i) ∠GHD – ∠HDC (ii) ∠BCI + ∠HAB.
The factors of 8a3 + b3 - 6ab + 1 are
A (2a + b - 1)(4a2 + b2 + 1 - 3ab - 2a)
B (2ab - b + 1)(4a2 + b2 - 4ab + 1 - 2a + b)
C (2a + b + 1)(4a2 + b2 + 1 - 2ab - b - 2a)
D (2a - 1 + b)(4a2 + 1 - 4a - b - 2ab)
please also give the solution of this question
Following are the steps of construction of a ΔPQR, given that QR = 3 cm, ∠PQR = 45° and QP - PR = 2 cm. Arrange them and select the correct option.
(i) Make an angle XQR = 45° at point Q of base QR.
(ii) Join SR and draw the perpendicular bisector of SR say AB.
(iii) Draw the base QR of length 3 cm.
(iv) Let bisector AB intersect QX at P. Join PR.
(v) Cut the line segment QS = QP - PR = 2 cm from the ray QX.
A (iii) → (ii) → (i) → (v) → (iv)
B (iii) → (i) → (ii) → (v) → (iv)
C (iii) → (i) → (ii) → (iv) → (v)
D (iii) → (i) → (v) → (ii) → (iv)