In AABC, 2B = 90°, AB = (2x + 1) cm and BC = (x + 1) cm. If the area

1.	In AABC, 2B = 90°, AB = (2x + 1) cm and BC = (x + 1) cm. If the area of the AABC is60 cm?, find its perimeter.
Answer» Correct Question: In ∆ ABC, B = 90°, AB= (2x + 1) cm, and BC = (x + 1) cm. If the area of the ∆ ABC is 60 cm², find it's PERIMETER? ⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ $\setlength{\unitlength}{1.9cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{B}}\put(10.6,1){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(8.8,0.7){\sf{\large{(x + 1)}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\put(7.1,1.8){\sf (2x + 1)}\end{picture}$ ⠀ $\frak{Here} \begin{cases} & \sf{Base = \bf{x + 1}} \\ & \sf{Perpendicular = \bf{2x + 1}} \end{cases}\\ \\$ We KNOW that, ⠀ ${\boxed{\sf{\purple{Area_{\; \triangle} = \dfrac{1}{2} \times Base \times Height}}}}\\ \\$ Now, Putting values, ⠀⠀ $:\implies\sf \dfrac{1}{2} \times (x + 1) \times (2x + 1) = 60\\ \\$ $:\implies\sf \dfrac{1}{2} (2x^2 + 3x + 1) = 60\\ \\$ $:\implies\sf 2x^2 + 3x + 1 = 120\\ \\$ $:\implies\sf 2x^2 + 3x - 119 = 0\\ \\$ So, ⠀⠀ $:\implies\sf x = \dfrac{- 3 \pm \sqrt{3^2 - 4(2)(-119)}}{2(2)}\\ \\$ $:\implies\sf x = \dfrac{- 3 \pm \sqrt{9 + 952}}{4}\\ \\$ $:\implies\sf x = \dfrac{- 3 \pm 31}{4}\\ \\$ $:\implies\sf x = 7, \dfrac{-34}{4}\\ \\$ Here, value of x can't be negative. ⠀⠀ So, ⠀⠀ $:\implies{\boxed{\frak{\pink{x = 7}}}}\;\bigstar\\ \\$ Therefore, ⠀⠀ AB, (2x + 1) = 2 × 7 = 1 = 15 cm BC, (x + 1) = 7 + 1 = 8 cm ⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ ☯ Now, finding AC which is Hypotenuse of ∆ ABC, ⠀⠀ ★ Using Pythagoras Theorem, ⠀⠀ ${\boxed{\sf{\purple{H^2 = B^2 + P^2}}}}\\ \\$ $:\implies\sf H^2 = 8^2 + 15^2\\ \\$ $:\implies\sf H^2 = 64 + 225\\ \\$ $:\implies\sf H^2 = 289\\ \\$ $:\implies\sf \sqrt{H^2} = \sqrt{289}\\ \\$ $:\implies\bf H = 17\;cm\\ \\$ ⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ ☯ Now, Finding Perimeter of ∆ ABC, ⠀⠀ $:\implies\sf Perimeter = AB + BC + CD\\ \\$ $:\implies\sf Perimeter = 15 + 8 + 17\\ \\$ $:\implies{\boxed{\frak{\pink{Perimeter = 40\;cm}}}}\;\bigstar\\ \\$ $\therefore\;{\underline{\sf{Hence,\;The\; perimeter\;of\; \triangle\;ABC\;is\; \bf{40\;cm}.}}}$

In AABC, 2B = 90°, AB = (2x + 1) cm and BC = (x + 1) cm. If the area of the AABC is60 cm?, find its perimeter.

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Correct Question:

In ∆ ABC, B = 90°, AB= (2x + 1) cm, and BC = (x + 1) cm. If the area of the ∆ ABC is 60 cm², find it's PERIMETER?

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$\setlength{\unitlength}{1.9cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{B}}\put(10.6,1){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(8.8,0.7){\sf{\large{(x + 1)}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\put(7.1,1.8){\sf (2x + 1)}\end{picture}$