In Delta PQR,m angle P = m angle Q + m angle | vedclass

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(B) Given that in $\Delta PQR$,$m \angle P = m \angle Q + m \angle R$.We know that the sum of all angles in a triangle is $180^\circ$,so $m \angle P +

Vedclass · Accepted Answer

$56$

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(B) Given that in $\Delta PQR$,$m \angle P = m \angle Q + m \angle R$.We know that the sum of all angles in a triangle is $180^\circ$,so $m \angle P + m \angle Q + m \angle R = 180^\circ$.Substituting $m \angle Q + m \angle R = m \angle P$,we get $m \angle P + m \angle P = 180^\circ$,which implies $2(m \angle P) = 180^\circ$,so $m \angle P = 90^\circ$.Thus,$\Delta PQR$ is a right-angled triangle with the hypotenuse $QR = 25$ and one side $PQ = 7$.Using the Pythagorean theorem,$PQ^2 + PR^2 = QR^2$.$7^2 + PR^2 = 25^2$.$49 + PR^2 = 625$.$PR^2 = 625 - 49 = 576$.$PR = \sqrt{576} = 24$.The perimeter of $\Delta PQR = PQ + QR + PR = 7 + 25 + 24 = 56$.

In $\Delta PQR$,$m \angle P = m \angle Q + m \angle R$,$PQ = 7$ and $QR = 25$. Find the perimeter of $\Delta PQR$.

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