Question

The base of a right triangle is 16 cm and hypotenuse is 34 cm, then area of the triangle is

Answer 1

Answer:

hi there here is your answer

Step-by-step explanation:

area=1/2*b*h

you need to use Pythagoras

34^2=16^2+b^2

therefore b=30

1/2*16*30=240 cm^2

Answer 2

GivEn:

• Base of Right angle triangle = 16 cm

• Hypotenuse of Right angle triangle = 34 cm

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To find:

• Area of Triangle.

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SoluTion:

$\bf Here \begin{cases} & \text{Base = \bf{16 cm}} \\ & \text{Hypotenuse = \bf{34 cm}} \end{cases}$

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Using Pythagoras theorem,

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$\;\;\sf\star\;(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2$

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$:\implies\sf (34)^2 = (16)^2 + (P)^2$

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$:\implies\sf (P)^2 = (34)^2 - (16)^2$

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$:\implies\sf (P)^2 = 1156 - 256$

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$:\implies\sf (P)^2 = 900$⠀

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$\;\;\;\sf \underline{ Taking\;sqrt.\;both\;side}$

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$:\implies\sf \sqrt{(P)^2} = \sqrt{900}$

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$:\implies\sf \red{P = 30\;cm}$

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$\therefore$ Height of Right angle triangle is 30 cm.

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⋆ DIAGRAM:

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$\setlength{\unitlength}{1.6cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{C}}\put(10.6,1){\large\sf{B}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(8.4,0.7){\sf{\large{16 cm}}}\put(7.4,2){\sf{\large{30 cm}}}\put(9.3,2){\sf{\large{34 cm}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\qbezier(9.8,1)(9.7,1.25)(10,1.4)\put(9.4,1.2){\sf\large{}}\end{picture}$

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Now, As we know that,

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★ Area of Right angle triangle is,

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$\star\;{\boxed{\sf{\purple{Area = \dfrac{1}{2} \times base \times height}}}}$

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$\;\;\;\small\sf \underline{Now,\;put\;the\;given\;values,}$

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$:\implies\sf Area = \dfrac{1}{ \cancel{2}} \times \cancel{16} \times 30$

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$:\implies\sf Area = 8 \times 30$

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$:\implies{\underline{\boxed{\sf{\pink{240\;cm^2}}}}}\;\bigstar$

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$\therefore$ Hence, Area of Right angle triangle is 240 cm².

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★ Additional Information:

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• Perimeter of Triangle = sum of its all sides( a + b + c )

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• Area of ∆ using Heron's formula = $\sf \sqrt{s(s - a)(s - b)(s - b)}$