Question

Pythagorean theorem and trig!

Answer 1

15. $\bf{\underline{Given:-}}$

• $\sf{Hypotenuse = \sqrt{10}}$
• $\sf{Length\:of\:one\:leg = \sqrt{3}}$

$\bf{\underline{To\:Find:-}}$

Length of other leg.

$\bf{\underline{Assumption:-}}$

Let one of the given legs be base.

Let the other leg (i.e., Perpendicular) be x.

$\bf{\underline{Solution:-}}$

According to Pythagoras Theorem,

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

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

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

$\sf{\implies Perpendicular = \sqrt{(\sqrt{10})^2 - (\sqrt{3})^2}}$

$\sf{\implies Perpendicular = \sqrt{10-3}}$

$\sf{\implies Perpendicular = \sqrt{7}}$

$\sf{\therefore}$ The length of other leg is $\sf{\sqrt{7}}$

$\sf{\underline{Therefore,\: Option\: [1]\: is\: the\: correct \:answer.}}$

_________________________________________

16. $\bf{\underline{Given:-}}$

• Hypotenuse = 25
• Base = 24

$\bf{\underline{To\:Find:-}}$

Area of the right triangle.

$\bf{\underline{Solution:-}}$

According to Pythagoras Theorem,

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

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

$\sf{\implies Perpendicular = \sqrt{(25)^2-(24)^2}}$

By using the identity $\sf{\underline{(a)^2-(b)^2 = (a+b)(a-b)}}$

$\sf{\implies Perpendicular = \sqrt{(25+24)(25-24)}}$

$\sf{\implies Perpendicular = \sqrt{49\times1}}$

$\sf{\implies Perpendicular = \sqrt{49}}$

$\sf{\implies Perpendicular = 7}$

Now,

$\sf{Semi-perimeter\:of\:the\:triangle = \dfrac{a+b+c}{2}}$

$\sf{Here \:a = 25\:b=24\:c=7}$

Therefore,

$\sf{s = \dfrac{25+24+7}{2}}$

$\sf{\implies s = \dfrac{56}{2}}$

$\sf{\implies s = 28}$

According to Heron's Formula,

$\sf{Area = \sqrt{s(s-a)(s-b)(s-c)}}$

= $\sf{Area = \sqrt{28(28-25)(28-24)(28-7)}}$

= $\sf{Area = \sqrt{28\times3\times4\times21}}$

= $\sf{Area = \sqrt{4\times7\times3\times4\times3\times7}}$

= $\sf{Area = 3\times4\times7}$

= $\sf{Area = 84\:sq.units}$

$\sf{\underline{Therefore, \:Option\: [2] \:is \:the\: correct \:answer.}}$

$\bf{\underline{Additional\:Information:-}}$

• We can also find the area of the triangle by applying the Formula $\sf{Area = \dfrac{1}{2}\times base\times height\:\:sq.units}$

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

Answer 2

$15. \huge\bf{\underline{Given:-}}$

$\boxed{Hypotenuse = \sqrt{10}}$

$\sf{Length\:of\:one\:leg =\sqrt{3}}$

$\bf\huge{\underline{To\:Find:-}}$

$\bf{\underline{Assumption:-}}$

$\huge\boxed{\underline{Solution:-}}$

$According \: to \: Pythagoras \: Theorem,$

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

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

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

$\boxed{\implies Perpendicular = \sqrt{(\sqrt{10})^2 - (\sqrt{3})^2}}$

$\boxed{\implies Perpendicular = \sqrt{10-3}}$

$\boxed{\implies Perpendicular = \sqrt{7}}$

$\sf{\therefore}The \: length \: of \: other \: leg \: is \sf{\sqrt{7}}$

$\huge\boxed{\underline{Therefore,\: Option\: [1]\: is\: the\: correct \:answer.}}$