Question

find the missing side of a right triangle if
area 120cm ,base 40cm
​

Answer 1

Area = 120 sq cm

Base = 40 cm

Area of right angled triangle =

$\frac{1}{2} \times base \times height \\ \\ 120 = \frac{1}{2} \times 40 \times height \\ \\ 120 = 20 \times height \\ \\ \frac{120}{20} = height \\ \\ height = 60 \: cm$

Now according to Pythagoras theorum:-

${ (hypotenuse)}^{2} = {(base)}^{2} + {(height)}^{2} \\ \\ {(hypotenuse)}^{2} = {40}^{2} + {60}^{2} \\ \\ {(hypotenuse)}^{2} = 1600 + 3600 \\ \\ {(hypotenuse)}^{2} = 5200 \\ \\ hypotenuse = \sqrt{5200} \\ \\ hypotenuse = 20 \sqrt{13} \: cm$

So the sides of triangle are:-

20 cm, 60 cm and 20√13 cm.

Hope it helps.

Answer 2

$\huge{\underline{\underline{\red{♡Solution→}}}}$

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$\bold{\huge{\underline{\underline{\rm{ Given :}}}}}$

Area (a) = 120

Area (a) = 120 Base (b) = 40

$\bold{\huge{\underline{\underline{\rm{ </strong><strong>To\</strong><strong>:</strong><strong>find</strong><strong> :}}}}}$

Height = ?

Height = ? Hypotenuse = ?

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$\</strong><strong>p</strong><strong>u</strong><strong>r</strong><strong>p</strong><strong>l</strong><strong>e</strong><strong>{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$a = \frac{1}{2} \times base \times height \\ 120 = \frac{1}{2} \times 40 \times h \\ 120 = 20 \times h \\ h = \frac{120}{20}$

$</strong><strong>\</strong><strong>b</strong><strong>o</strong><strong>x</strong><strong>e</strong><strong>d</strong><strong>{</strong><strong>h</strong><strong> = 6</strong><strong>}</strong><strong>$

Using the Pythagoras theorem →

$( {hypotenuse)}^{2} = ( {base)}^{2} + ( {height)}^{2}$

$( {hyp.)}^{2} = ( {40)}^{2} + ( {6)}^{2} \\ = 1600 + 36 \\ = 1636$

$hyp. = \sqrt{1636}$

$</strong><strong>\</strong><strong>b</strong><strong>o</strong><strong>x</strong><strong>e</strong><strong>d</strong><strong>{</strong><strong>hyp</strong><strong>. = </strong><strong>40.4474968</strong><strong>}$

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