Question

Q 1) The Perimeter of triangle is 72cm and its sides are in the ratio 3:4:5 . Find it's area and the Length of the altitude Corresponding to the longest side .

Q2) Find the base of an isosceles triangle whose area is 192cm² and the Length of one of the equal sides is 20cm.
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Answer 1

☞︎︎︎ Solution :

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↬ Q1)

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↱ Given :

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• Perimeter of triangle = 72cm

• Ratio of the sides = 3:4:5

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➝ Let the constant of proportionality be = k

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⇒ 3k + 4k + 5k = 72cm

⇒ 12k = 72cm

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$\large \: ⇒ \rm \: k = \dfrac{72}{12}$

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$\boxed{ \large \therefore \underline{\: \rm \: k = 6cm}}$

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↦ The sides are -

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• 3 × 6 = 18cm

• 4 × 6 = 24cm

• 5 × 6 = 30cm

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➢ Herons Formula :

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$\large \mapsto \boxed{\: \mathtt{ \sqrt{s(s - a)(s - b)(s - c)} }}$

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• Finding Area :

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$\large = \rm \mathtt{ \sqrt{36(36 - 18)(36 - 24)(36 - 30)} }$

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$\large ⇒ \mathtt{ \sqrt{36(18)(12)(6)} }$

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$\large ⇒ \mathtt{ \sqrt{(648)(72)} }$

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$\large ⇒ \mathtt{ \sqrt{(46656)} }$

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⇢ 216cm²

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➣ Finding altitude :

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• Height = h

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➦ We know ,

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$\: \: \: \: \: \: \: \: \:$ Area of triangle =

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$\large ➬ \: \boxed{ \: \: \mathtt{ \dfrac{1}{2} \times base \times height}}$

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• Base = 30cm

• Area = 216cm²

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$\large⇒ \mathtt{ \dfrac{1}{2} \times 30 \times h = 216 {cm}^{2} }$

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$\large⇒ \mathtt{ \dfrac{1}{ \cancel2} \times \cancel{30} \: 15 \times h = 216 {cm}^{2} }$

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• On Transposing The Terms :

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$\large⇒ \: \mathtt{h \: = \dfrac{216}{15} }$

$\boxed{\boxed{ \large \therefore \: \underline{\tt{altitude = 14.4cm}}}}$

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↬ Q2 )

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↱ Given :

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• Area = 192cm²

• Side¹ = 20cm

• Side² = 20cm

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$\large⇒ \: \mathtt{h = \sqrt{b - \dfrac{ {a}^{2} }{4} } }$

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$\: \: \: \: \: \: \: \: \:$ Area of triangle =

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$\large ➬ \: \boxed{ \: \: \mathtt{ \dfrac{1}{2} \times base \times height}}$

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$\large⇒ \mathtt{ \dfrac{1}{2} \times a\times h = \dfrac{1}{2} \times a \times \sqrt{b - \dfrac{ {a}^{2} }{4} } }$

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$\large⇒ \mathtt{ \dfrac{1}{2} \times {a}^{2} \times \sqrt{ \dfrac{ {b}^{2} }{ {a}^{2} } - \dfrac{1}{4} } }$

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• Area Given By -

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$\large⇒ \mathtt{ \dfrac{1}{2} \times {x}^{2} \times \sqrt{ \dfrac{ {20}^{2} }{ {x}^{2} } - \dfrac{1}{4} } }$

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$\large⇒ \mathtt{ \dfrac{ {x}^{2} }{2} \times \sqrt{ \dfrac{ {1600} - {x}^{2} }{ {4x}^{2} } } }$

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$\large⇒ 192 { \rm \: cm }^{2} = \\ \\ \large \: \: \: \: \: \mathtt{ \dfrac{ {x}^{2} }{2} \times \sqrt{ \dfrac{ {1600} - {x}^{2} }{ {4x}^{2} } } }$

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$\large⇒ 192 { }^{2} = \large \ \mathtt{ \dfrac{ {x}^{2} }{2} \times { \dfrac{ {1600} - {x}^{2} }{ {4x}^{2} } } }$

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⇒ 192² × 16 = x²(1600 - x²)

⇒ x⁴ - 1600x² + 589824 = 0

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★ Splitting The Middle Term -

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• t = x²

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⇒ t² - 1600t + 589824 = 0

⇒ t² - 1024t - 576t + 589824 = 0

⇒ t(t - 1024) - 576(t - 1024) = 0

⇒ t = 1024 or 576

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$\large⇒ \: \mathtt{ {x}^{2} = 1024 \: \: \: or \: \: \: 576}$

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$\large⇒ \: \mathtt{ {x}^{} = \sqrt{ 1024 }\: \: \: or \: \: \: \sqrt{ 576}}$

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$\large{ \boxed{\therefore \underline{\: \mathtt{ {x}^{} = 32cm \: \: \: or \: \: \:24cm} }}}$

Answer 2

Answer:

hii

Step-by-step explanation:

see answer above

so nice