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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$56$

Vedclass · Answer

(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 $\Delta PQR$.

Similar Questions

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = x+1$,$MB = x$,$AN = 3x-3$ and $NC = 4x-10$,then $x = \dots$

$\Delta ABC \sim \Delta PQR$ for the correspondence $ABC \leftrightarrow PQR$. If $AB = 8$,$PQ = 14$ and the perimeter of $\Delta ABC$ is $20$,find the perimeter of $\Delta PQR$.

If in the figure, $O$ is the point of intersection of two chords $AB$ and $CD$ such that $OB = OD$, then triangles $OAC$ and $ODB$ are

$O$ is the point of intersection of the diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$. Through $O$,a line segment $PQ$ is drawn parallel to $AB$ meeting $AD$ in $P$ and $BC$ in $Q$. Prove that $PO = QO$.

In $\Delta ABC$,$D$ is the midpoint of $\overline{BC}$,$AB = 7$,$AC = 5$ and $AD = 5$. Then $BC = \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module