Question

Find the hypotenuse of a right triangle of sides 3cm and 4 cm.

Answer 1

Answer:

25

Step-by-step explanation:

hypotenous =base sq +side sq

9+16=25

Answer 2

Answer:

✯ Given :

➻ The base of a triangle is 3 cm and its height is 4 cm.

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✯ To Find :

➻ Hypotenuse of the triangle.

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✯ Solution :

Formula Used :

$\large{\orange{\bigstar}}\: \: \: \: {\underline{\boxed{\red{\sf{ Hypotenuse² = Height² + Base²}}}}}$

$\qquad{━━━━━━━━━━━━━━━━━━━━━━━━━━}$

Hypotenuse of Triangle :

$\longmapsto{\qquad{\sf{ Hypotenuse² = Height² + Base²}}}$

$\longmapsto{\qquad{\sf{ Hypotenuse² = 4² + 3²}}}$

$\longmapsto{\qquad{\sf{ Hypotenuse² = 16 + 9}}}$

$\longmapsto{\qquad{\sf{ Hypotenuse² = 25}}}$

$\longmapsto{\qquad{\sf{ Hypotenuse = \sqrt{25}}}}$

$\large{\blue{:\longmapsto{\underline{\boxed{\purple{\tt{ Hypotenuse = 5 \: cm}}}}}}}$

$\qquad{━━━━━━━━━━━━━━━━━━━━━━━━━━}$

Therefore :

❝ Hypotenuse of the triangle is 5 cm. ❞

${\green{\underline{▬▬▬▬▬}}{\pink{\underline{▬▬▬▬▬}}{\orange{\underline{▬▬▬▬▬}}{\purple{\underline{▬▬▬▬▬▬}}}}}}$

✯ Know more :

Formulas :

$\twoheadrightarrow{\pink{\sf{ Area{\small_{(Rectangle) }} = {\green{\sf{Length \times Breadth}}}}}}$

$\twoheadrightarrow{\pink{\sf{ Area{\small_{(Square) }}} = {\green{\sf{Side²}}}}}$

$\twoheadrightarrow{\pink{\sf{ Area{\small_{(Triangle) }}} = {\green{\sf{\dfrac{1}{2} \times b \times h}}}}}$

$\twoheadrightarrow{\pink{\sf{ Perimeter{\small_{(Rectangle) }}} = {\green{\sf{2(Length + Breadth)}}}}}$

$\twoheadrightarrow{\pink{\sf{ Perimeter{\small_{(Square) }}} = {\green{\sf{ 4 \times Side}}}}}$

$\twoheadrightarrow{\pink{\sf{ Perimeter{\small_{(Triangle) }}} = {\green{\sf{a + b + c}}}}}$

$\twoheadrightarrow{\pink{\sf{ Semi - Perimeter {\small_{(Triangle) }}} = {\green{\sf{\dfrac{a + b + c}{2}}}}}}$

$\twoheadrightarrow{\pink{\sf{ Area{\small_{(Rectangle) }}} = {\green{\sf{Length \times Breadth}}}}}$

$\twoheadrightarrow{\pink{\sf{ Perimeter{\small_{(Rhombus) }}} = {\green{\sf{ 4 \times Side}}}}}$

$\twoheadrightarrow{\pink{\sf{ Perimeter{\small_{(Square) }}} = {\green{\sf{ 4 \times Side}}}}}$

$\twoheadrightarrow{\pink{\sf{ TSA{\small_{(Cylinder) }}} = {\green{\sf{ 2πr(r + h) }}}}}$

$\twoheadrightarrow{\pink{\sf{ CSA{\small_{(Cylinder) }}} = {\green{\sf{ 2πrh }}}}}$

$\twoheadrightarrow{\pink{\sf{ Volume{\small_{(Cylinder) }}} = {\green{\sf{ πr²h}}}}}$

$\twoheadrightarrow{\pink{\sf{ Volume{\small_{(Cuboid) }}} = {\green{\sf{ Length \times Width \times Height}}}}}$

$\twoheadrightarrow{\pink{\sf{ Volume{\small_{(Cube) }}} = {\green{\sf{ Side³}}}}}$

${\green{\underline{▬▬▬▬▬}}{\pink{\underline{▬▬▬▬▬}}{\orange{\underline{▬▬▬▬▬}}{\purple{\underline{▬▬▬▬▬▬}}}}}}$