In Delta ABC,m angle A = m angle B + m angle | vedclass

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(B) Given that in $\Delta ABC$,$m \angle A = m \angle B + m \angle C$.We know that the sum of angles in a triangle is $180^{\circ}$,so $m \angle A + m

Vedclass · Accepted Answer

$56$

Vedclass · Answer

(B) Given that in $\Delta ABC$,$m \angle A = m \angle B + m \angle C$.We know that the sum of angles in a triangle is $180^{\circ}$,so $m \angle A + m \angle B + m \angle C = 180^{\circ}$.Substituting $m \angle B + m \angle C$ with $m \angle A$,we get $m \angle A + m \angle A = 180^{\circ}$,which implies $2 m \angle A = 180^{\circ}$,so $m \angle A = 90^{\circ}$.Since $\Delta ABC$ is a right-angled triangle with the right angle at $A$,$BC$ is the hypotenuse.By the Pythagorean theorem,$BC^2 = AB^2 + AC^2$.Given $AB = 7$ and $BC = 25$,we have $25^2 = 7^2 + AC^2$.$625 = 49 + AC^2$,so $AC^2 = 625 - 49 = 576$.Thus,$AC = \sqrt{576} = 24$.The perimeter of $\Delta ABC = AB + BC + AC = 7 + 25 + 24 = 56$.

In $\Delta ABC$,$m \angle A = m \angle B + m \angle C$. If $AB = 7$ and $BC = 25$,the perimeter of $\Delta ABC = ........$

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