Question

Formula solve in all lesson and Class 9

Answer 1

Geometric Figure Area Perimeter

Rectangle A= l × w P = 2 × (l+w )

Triangle A = (1⁄2) × b × h P = a + b + c

Trapezoid A = (1⁄2) × h × (b1+ b2) P = a + b + c + d

Parallelogram A = b × h P = 2 (b + h)

Circle A = π r2 C = 2 π r

Algebraic Identities For Class 9

(a+b)2=a2+2ab+b2

(a−b)2=a2−2ab+b2

(a+b)(a–b)=a2–b2

(x+a)(x+b)=x2+(a+b)x+ab

(x+a)(x–b)=x2+(a–b)x–ab

(x–a)(x+b)=x2+(b–a)x–ab

(x–a)(x–b)=x2–(a+b)x+ab

(a+b)3=a3+b3+3ab(a+b)

(a–b)3=a3–b3–3ab(a–b)

(x+y+z)2=x2+y2+z2+2xy+2yz+2xz

(x+y–z)2=x2+y2+z2+2xy–2yz–2xz

(x–y+z)2=x2+y2+z2–2xy–2yz+2xz

(x–y–z)2=x2+y2+z2–2xy+2yz–2xz

x3+y3+z3–3xyz=(x+y+z)(x2+y2+z2–xy–yz−xz

x2+y2=12[(x+y)2+(x–y)2]

(x+a)(x+b)(x+c)=x3+(a+b+c)x2+(ab+bc+ca)x+abc

x3+y3=(x+y)(x2–xy+y2)

x3–y3=(x–y)(x2+xy+y2)

x2+y2+z2−xy–yz–zx=12[(x−y)2+(y−z)2+(z−x)2]<