Question

If hypotenuse = 5 cm and perpendicular = 3 cm, what is the length of the base​

Answer 1

Answer:

$\huge{\underline{\underline{\sf{\green{Answer:-}}}}}$

Perpendicular sides imply that this is, in fact, a right triangle. Applying the well known Pythagorean formula that relates the legs of a right triangle to its hypotenuse…we get that 32+42=h2 where h is stands for the length of the hypotenuse. Solving our previous equation for h we get that 9+16=h2 and so h2=25 which leads us to h=5 . When solving for h it is also noteworthy to realize that h=−5 would also be a solution to our equation. Since triangles can't have negative side lengths we always go for the positive solution. The 3−4−5 right triangle is one of the most standard right triangles out there. Some other noteworthy primitive

5−12−13

5−12−13 8−15−17

5−12−13 8−15−17 7−24−25

5−12−13 8−15−17 7−24−25 20−21−2

Find all integer triples of the form (a,b,c) that satisfy an+bn=cn for which

Find all integer triples of the form (a,b,c) that satisfy an+bn=cn for which 0 < a<b<c<30 .

Answer 2

Given:-

• Hypotenuse = 5 cm
• Perpendicular = 3 cm

To find:-

Length of base

Solution:-

We know

According To Pythagoras Theorem

$\sf{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}$

$\sf{\implies (Base)^2 = (Hypotenuse)^2 - (Perpendicular)^2}$

$\sf{\implies Base = \sqrt{(Hypotenuse)^2 - (Perpendicular)^2}}$

= $\sf{Base = \sqrt{(5)^2 - (3)^2}}$

= $\sf{Base = \sqrt{25 - 9}}$

= $\sf{Base = \sqrt{16}}$

= $\sf{Base = 4 \:cm}$

$\sf{\therefore The\:length\:of\:the\:Base\:is\:4\:cm}$

Pythagoras Theorem states that the square of the Hypotenuse of a right-angled triangle is always equal to the sum of square of base and perpendicular.

• $\sf{(Hypotenuse)^2 = (Perpendicular)^2 + (Base)^2}$

• $\sf{(Perpendicular)^2 = (Hypotenuse)^2 - (Base)^2}$

• $\sf{(Base)^2 = (Hypotenuse)^2 - (Base)^2}$