The altitude of a right angled triangle is 7 cm less than its base. If

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 .plzzzz telllll
Answer» $\huge \bold \color{green}{ \underline { \underline \red{required \: answer :-}}}$ 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

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 .plzzzz telllll

Answer»

$\huge \bold \color{green}{ \underline { \underline \red{required \: answer :-}}}$

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

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