Question

The hypotenuse of right angled triangle with its base and perpendicular of
lengths 3x, 4x respectively is: *
25x
7x
5x
O 16x​

Answer 1

Answer:

5x

Step-by-step explanation:

As we all know,

(Base)^2+(Perpendicular)^2=(Hypotenuse)^2

(3x)^2+(4x)^2=H^2

9x^2+16x^2=H^2

25x^2=H^2

H=root25x^2

H=5x

Answer 2

Given :

• Length of the Base of Triangle is 3x
• Perpendicular Length of Triangle is 4x

To Find :

• Hypotenuse of Right Angled Triangle.

Solution :

$\setlength{\unitlength}{1.9cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{B}}\put(10.6,1){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(8.6,0.7){\sf{\large{3 x}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\put(7.4,1.8){\sf 4 x}\put(9.4,2.2){\sf ? }\end{picture}$

Using Pythagoras Theoram :

$\longmapsto\tt\bf{{(H)}^{2}={(B)}^{2}+{(P)}^{2}}$

$\longmapsto\tt{{H}^{2}={3x}^{2}+{4x}^{2}}$

$\longmapsto\tt{{H}^{2}={9x}^{2}+{16x}^{2}}$

$\longmapsto\tt{{H}^{2}=25{x}^{2}}$

$\longmapsto\tt{H=\sqrt{25{x}^{2}}}$

$\longmapsto\tt\bf{H=5x}$

So , The Hypotenuse of the Triangle is 5x...