Question

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

Answer 1

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

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$\frak{Here} \begin{cases} & \sf{Base = \bf{x + 1}} \\ & \sf{Perpendicular = \bf{2x + 1}} \end{cases}$

We know that,

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${\boxed{\sf{\purple{Area_{\; \triangle} = \dfrac{1}{2} \times Base \times Height}}}}$

Now, Putting values,

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$:\implies\sf \dfrac{1}{2} \times (x + 1) \times (2x + 1) = 60$

$:\implies\sf \dfrac{1}{2} (2x^2 + 3x + 1) = 60$

$:\implies\sf 2x^2 + 3x + 1 = 120$

$:\implies\sf 2x^2 + 3x - 119 = 0$

So,

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$:\implies\sf x = \dfrac{- 3 \pm \sqrt{3^2 - 4(2)(-119)}}{2(2)}$

$:\implies\sf x = \dfrac{- 3 \pm \sqrt{9 + 952}}{4}$

$:\implies\sf x = \dfrac{- 3 \pm 31}{4}$

$:\implies\sf x = 7, \dfrac{-34}{4}$

Here, value of x can't be negative.

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So,

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$:\implies{\boxed{\frak{\pink{x = 7}}}}\;\bigstar$

Therefore,

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• AB, (2x + 1) = 2 × 7 = 1 = 15 cm

• BC, (x + 1) = 7 + 1 = 8 cm

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☯ Now, finding AC which is Hypotenuse of ∆ ABC,

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★ Using Pythagoras Theorem,

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${\boxed{\sf{\purple{H^2 = B^2 + P^2}}}}$

$:\implies\sf H^2 = 8^2 + 15^2$

$:\implies\sf H^2 = 64 + 225$

$:\implies\sf H^2 = 289$

$:\implies\sf \sqrt{H^2} = \sqrt{289}$

$:\implies\bf H = 17\;cm$

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☯ Now, Finding Perimeter of ∆ ABC,

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$:\implies\sf Perimeter = AB + BC + CD$

$:\implies\sf Perimeter = 15 + 8 + 17$

$:\implies{\boxed{\frak{\pink{Perimeter = 40\;cm}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;The\; perimeter\;of\; \triangle\;ABC\;is\; \bf{40\;cm}.}}}$