Question

if in a right-angle triangle measure of hypotenuse =8cm measure of perpendicular =6cm then the measure of base is

this is fill in the blank plz answer this right

Answer 1

Solution

Hypotenuse(H) = 8cm

Perpendicular(P) = 6cm

Base(B) = ?

By Pathagoras theorem

$\bigstar \: \sf\large{\underline{\boxed{\sf{(H)²=(P)²+(B)²}}} \: \bigstar}$

Substituting known values in the formula

$\sf\implies{(8)²=(6)²+(B)²}$

$\sf\implies{(B)²+(6)²=(8)²}$

$\sf\implies{(B)²=(8)²-(6)²}$

$\sf\implies{(B)²=64-36}$

$\sf\implies{(B)²=28}$

$\sf\implies{B= \sqrt{28}}$

$\sf\implies{B=2 \sqrt{7 \:}cm}$

Therefore, base of the right-angle ∆ is 2√7cm or 5.29cm(approx).

Answer 2

✰ Given :-

• Perpendicular = 6 cm
• Hypotenuse = 8 cm

✰ To Find :-

• The measure of the base

✰ Concept :-

In this question, we will use the concept of "Pythagoras Theorem" which states that square of Hypotenuse is equal to the sum of square of base and square of perpendicular, i.e.

$\boxed{ \sf \: ✰ \: {hypotenuse}^{2} = {perpendicular}^{2} + {base}^{2} }$

Now, to get the required answer we can simply put the given values and solve it. So, let's do it! :D

✰ Solution :-

$\sf \leadsto \: {8}^{2} = {6}^{2} + {b}^{2} \: \: \: \: \: \\ \sf \leadsto \: {b} \: = \sqrt{64 - 36} \: \\ \sf \leadsto \: {b} = \sqrt{28} \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \leadsto \: b = 2 \sqrt{7} \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \therefore \: base \: = 2 \sqrt{7} \: or \: 5.29 \: cm \: approx$

Thanks :)