Write the measures of the sides of Delta PQR | vedclass

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

Question

(QR, PR, PQ) In $\Delta PQR$,the sum of all interior angles is $180^{\circ}$.Therefore,$\angle P + \angle Q + \angle R = 180^{\circ}$.Substituting the

Vedclass · Accepted Answer

Vedclass · Answer

(QR, PR, PQ) In $\Delta PQR$,the sum of all interior angles is $180^{\circ}$.
Therefore,$\angle P + \angle Q + \angle R = 180^{\circ}$.
Substituting the given values: $40^{\circ} + \angle Q + 80^{\circ} = 180^{\circ}$.
$120^{\circ} + \angle Q = 180^{\circ}$,which gives $\angle Q = 60^{\circ}$.
We know that the side opposite to the smallest angle is the shortest,and the side opposite to the largest angle is the longest.
The angles in ascending order are $\angle P (40^{\circ}) < \angle Q (60^{\circ}) < \angle R (80^{\circ})$.
Therefore,the sides opposite to these angles in ascending order are $QR < PR < PQ$.

Write the measures of the sides of $\Delta PQR$ in ascending order,given that $\angle P = 40^{\circ}$ and $\angle R = 80^{\circ}$.

Similar Questions

In $\Delta ABC$,$AD$,$BE$,and $CF$ are its medians. Prove that $AB + BC + CA > AD + BE + CF$.

State whether the following statement is true or false:
$(1)$ In $\Delta PQR$,$\angle P = \angle Q = 95^{\circ}$ is possible.

In $\Delta ABC$,$\angle A + \angle B = 80^{\circ}$ and $\angle B + \angle C = 150^{\circ}$. Find the measure of each angle of $\Delta ABC$.

In the figure,$AD$ is the bisector of $\angle BAC$. Prove that $AB > BD$.

Write True or False and justify your answer:
$ABC$ and $BDE$ are two equilateral triangles such that $D$ is the mid-point of $BC$. Then $ar(\triangle BDE) = \frac{1}{4} ar(\triangle ABC).$

Vedclass Test Series

Exam Paper Generator

Online Exam Module