Question

The altitude of a right triangle is 7cm
less than its base. If the hypoteneuse is
13cm. find the other two sides.
19​

Answer 1

Given :

• Let's base of triangle = x cm

• Altitude of triangle = x - 7 cm

• Hypotenuse of triangle = 13 cm

To Find :

• Base of triangle and altitude of triangle

Solution :

By using Pythagoras theorem

$\large\implies\boxed {\boxed {\sf {h}^{2} = {b}^{2} + {p}^{2} }}\\ \\ \sf \implies {(13)}^{2} = {(x)}^{2} + {(x - 7)}^{2} \\ \\ \sf \implies169 = {x}^{2} + {x}^{2} + 49 - 14x \\ \\ \sf \implies169 - 49 = 2 {x}^{2} - 14 \\ \\ \sf \implies120 = 2{x}^{2} - 14x \\ \\ \sf \implies 2{x}^{2} - 14x - 120 = 0 \\ \\ \sf \implies {x}^{2} - 7x - 60 = 0$

By splitting middle term

$\sf \implies{x}^{2} - 12x + 5x - 60 = 0 \\ \\ \sf \implies x(x - 12) + 5(x - 12) = 0 \\ \\ \sf \implies(x + 5)(x - 12) = 0 \\ \\ \sf \implies x = - 5 \\ \\ \sf \implies x = 12$

Here x have two values. But length can't be negative. So value of x is 12

Base oftriangle = 12 cm

Altitude of triangle = 12 - 7 = 5 cm

Answer 2

Answer:

⋆ DIAGRAM :

$\setlength{\unitlength}{1.5cm}\begin{picture}(6,2)\linethickness{0.5mm}\put(7.7,2.9){\large\sf{A}}\put(7.7,1){\large\sf{B}}\put(10.6,1){\large\sf{C}}\put(8,1){\line(1,0){2.5}}\put(8,1){\line(0,2){1.9}}\qbezier(10.5,1)(10,1.4)(8,2.9)\put(7.1,2){\sf{\large{(n - 7)}}}\put(9,0.7){\sf{\large{n}}}\put(9.4,1.9){\sf{\large{13 cm}}}\put(8.2,1){\line(0,1){0.2}}\put(8,1.2){\line(3,0){0.2}}\end{picture}$

$\rule{120}{1}$

Let the Base be n and Altitude be (n - 7) of Right Angled Triangle respectively.

$\underline{\bigstar\:\sf{By\: Pythagoras\: Theorem :}}$

$:\implies\sf (Hypotenuse)^2=(Altitude)^2+(Base)^2\\\\\\:\implies\sf (AC)^2=(AB)^2+(BC)^2\\\\\\:\implies\sf (13)^2=(n-7)^2+(n)^2\\\\\\:\implies\sf 169 = n^2 + 49 - 14n + n^2\\\\\\:\implies\sf 2n^2 - 14n - 120 = 0\\\\\\:\implies\sf 2(n^2 - 7n - 60) = 0\\\\\\:\implies\sf n^2 - 7n - 60 = 0\\\\\\:\implies\sf n^2 - (12 - 5)n - 60 = 0\\\\\\:\implies\sf n^2 - 12n + 5n - 60 = 0\\\\\\:\implies\sf n(n - 12) + 5(n - 12) = 0\\\\\\:\implies\sf (n - 12)(n + 5) = 0\\\\\\:\implies \green{\sf n = 12 }\quad \sf or \quad \red{ n = -\: 5}$

$\rule{100}{0.9}$

$\underline{\bigstar\:\sf{Sides\:of\: Triangle :}}$

$\bullet\:\:\textsf{Base = n = \textbf{12 cm}}\\\bullet\:\:\textsf{Altitude = (n - 5) = \textbf{5 cm}}$

⠀

$\therefore\:\underline{\textsf{Hence, Base \& Altitude are \textbf{12 cm \& 5 cm} respectively.}}$

$\rule{200}{2}$

⠀⠀⠀⠀Shortcut Trick

$\begin{tabular}{|c |c | c|}\cline{1-3}p/b & b/p & h \\\cline{1-3}3 & 4 & 5 \\5 & 12 &13\\8 & 15&17 \\7 & 24&25\\\cline{1-3}\end{tabular}$

This is Pythagorean Triplet, & If there's a Right Angled Triangle. Then it will must follow Pythagorean Triplet.

⠀

In this Question Hypotenuse is Given as 13, then it will must follow 5, 12, 13.

⠀

Now we will Check it.

• Hypotenuse = 13 cm

• Altitude = (Base – 7)

⠀

As Base is larger number ATQ, then we will take 12 for it & then.

⇢ Altitude = (Base – 7)

⇢ Altitude = (12 – 7)

⇢ Altitude = 5

⠀

So, we can see this is following the first Pythagorean Triplet.

$\bullet\:\:\textsf{Hypotenuse = \textbf{13 cm}}\\\bullet\:\:\textsf{Base = \textbf{12 cm}}\\\bullet\:\:\textsf{Altitude = \textbf{5 cm}}$