Dear student,2a As ABC is right angled triangle, so we can use pythagoras theorem As, right angled is B so the opposite side is hypotenuse thus, AC is hypotenuse Therefore,AC2=AB2+BC2x+152=x+82+x+12We know that, (a+b)2=a2+b2+2ab⇒x2+30x+225=x2+16x+64+x2+2x+1⇒x2-12x-160=0⇒x2-20x+8x-160=0⇒x(x-20)+8(x-20)=0⇒(x+8)(x-20)=0x=-8,20Length can't be negative Thus x=20Sides are:x+15=35;x+1=21;x+8=28For other question, start a new threadRegards

Pls solve 2a and 2c

D e a r s t u d e n t, 2 a . A s A B C i s r i g h t a n g l e d t r i a n g l e, s o w e c a n u s e p y t h a g o r a s t h e o r e m . A s, r i g h t a n g l e d i s B s o t h e o p p o s i t e s i d e i s h y p o t e n u s e . t h u s, A C i s h y p o t e n u s e . T h e r e f o r e, A C^{2} = A B^{2} + B C^{2} {(x + 15)}^{2} = {(x + 8)}^{2} + {(x + 1)}^{2} W e k n o w t h a t, {(a + b)}^{2} = a^{2} + b^{2} + 2 a b \Rightarrow x^{2} + 30 x + 225 = x^{2} + 16 x + 64 + x^{2} + 2 x + 1 \Rightarrow x^{2} - 12 x - 160 = 0 \Rightarrow x^{2} - 20 x + 8 x - 160 = 0 \Rightarrow x (x - 20) + 8 (x - 20) = 0 \Rightarrow (x + 8) (x - 20) = 0 x = - 8, 20 L e n g t h c a n' t b e n e g a t i v e . T h u s x = 20 S i d e s a r e : x + 15 = 35; x + 1 = 21; x + 8 = 28 F o r o t h e r q u e s t i o n, s t a r t a n e w t h r e a d R e g a r d s

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