1.

The altitude of a right angled triangle is 7 cm less than its base. If the hypotenuse is 35 cm, find the other two sides of the triangle..​

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in a right angled triangle

hypotenuse = 35cm

base = bcm

ALTITUDE = b-7cm

we know that...by Pythagoras theorem sum of squares of base and perpendicular is EQUAL to square of hypotenuse..

=> H²=p²+b²

\implies \: {35}^{2} = {b}^{2} + {(b - 7)}^{2} \\ \implies \: 1225 = {b }^{2} + {b}^{2} + 49 - 14b \\ \implies \: 2 {b}^{2} - 14b = 1225 - 49 \\ \implies \: 2 {b}^{2} - 14b = 1176 \\ \implies \: 2 {b}^{2} - 14b - 1176 = 0 \\ \bold{now \: divide \: the \: equation \: by \: 2} \\ \\ \implies \: {b}^{2} - 7b - 588 = 0

it's in the form of a quadratic equation ax²+bx+c = 0

so now solving the equation by quadratic formula..

d = {b}^{2} - 4ac \\ \implies \: d = {( - 7)}^{2} - 4 \times ( - 588) \\ \implies \: d = 49 + 2352 \\ \implies \: d = 2401

so discriminant = 2401

now..we know that..

\bold{x = \frac{ - b \: ± \: \sqrt{d} }{2a} }

so,

\bold{b = \frac{- ( - 7)± \sqrt{2401}}{2} } \\ \implies \: b = \frac{7±49}{2} \\ \implies \: b = \frac{7 + 49}{2} \: and \: b = \frac{7 - 49}{2} \\ \implies \: b = \frac{56}{2} \: and \: b = \frac{42}{2} \\ \implies \: b = 28 \: and \: b = 21

so possible measurements of base is 28cm and 21cm

→ altitude = b-7

→ altitude = 28-7 or 21-7

→ altitude = 21 or 14cm

now we have to varify

H² = p²+b²

{(35)}^{2} = {(28)}^{2} + {(21)}^{2} \\ \implies \: 1225 = 784 \: + \: 441 \\ \implies \: 1225 = 1225 \\ → lhs = rhs \\ so \: base \: = 28cm \: and \: altitude \: = 21cm

now we have to varify Pythagoras theorem with value of base is 21cm

{(35)}^{2} = {(21)}^{2} + {(14)}^{2} \\ \implies \: 1225 = 441 + 196 \\ \implies \: 1225 = 637 \\ →lhs ≠ \: rhs \\ so \: value \: of \: base \: is \: 21 \: is \: not \: possible \\ for \: a \: right \: angled \: triangle

hence,

sides of triangle are :-

base = 28cm

altitude = 21cm

hypotenuse = 35cm


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