PQRS is a cyclic quadrilateral. If ∠P is 4 times ∠R, and ∠S is 3 times ∠Q, then the average of ∠Q and ∠R is:
40.5°
A cyclic quadrilateral is a four-sided figure (quadrilateral) inscribed in a circle. A key property of cyclic quadrilaterals is that their opposite angles are supplementary. This means that the sum of each pair of opposite angles is always $180^\circ$. For the cyclic quadrilateral PQRS, this property tells us:
The question provides us with additional relationships between the angles:
Our goal is to find the average of $\angle Q$ and $\angle R$. The average of two values is their sum divided by 2, so we need to calculate $(\angle Q + \angle R) / 2$. To do this, we first need to find the values of $\angle Q$ and $\angle R$.
We know that $\angle P + \angle R = 180^\circ$ and $\angle P = 4 \angle R$. We can substitute the second equation into the first equation to solve for $\angle R$:
$\angle P + \angle R = 180^\circ$
Substituting $4 \angle R$ for $\angle P$:
$4 \angle R + \angle R = 180^\circ$
Combine the terms involving $\angle R$:
$5 \angle R = 180^\circ$
Now, divide both sides by 5 to find the value of $\angle R$:
$\angle R = \frac{180^\circ}{5}$
$\angle R = 36^\circ$
With the value of $\angle R$, we can find $\angle P$ using $\angle P = 4 \angle R$:
$\angle P = 4 \times 36^\circ$
$\angle P = 144^\circ$
Let's quickly check if $\angle P + \angle R = 180^\circ$: $144^\circ + 36^\circ = 180^\circ$. This confirms our calculation for $\angle R$ and $\angle P$ is correct.
Similarly, we know that $\angle Q + \angle S = 180^\circ$ and $\angle S = 3 \angle Q$. We can substitute the second equation into the first equation to solve for $\angle Q$:
$\angle Q + \angle S = 180^\circ$
Substituting $3 \angle Q$ for $\angle S$:
$\angle Q + 3 \angle Q = 180^\circ$
Combine the terms involving $\angle Q$:
$4 \angle Q = 180^\circ$
Now, divide both sides by 4 to find the value of $\angle Q$:
$\angle Q = \frac{180^\circ}{4}$
$\angle Q = 45^\circ$
With the value of $\angle Q$, we can find $\angle S$ using $\angle S = 3 \angle Q$:
$\angle S = 3 \times 45^\circ$
$\angle S = 135^\circ$
Let's quickly check if $\angle Q + \angle S = 180^\circ$: $45^\circ + 135^\circ = 180^\circ$. This confirms our calculation for $\angle Q$ and $\angle S$ is correct.
We have found that $\angle Q = 45^\circ$ and $\angle R = 36^\circ$. Now we can calculate their average:
Average = $\frac{\angle Q + \angle R}{2}$
Average = $\frac{45^\circ + 36^\circ}{2}$
Average = $\frac{81^\circ}{2}$
Average = $40.5^\circ$
| Angle | Value |
|---|---|
| ∠P | $144^\circ$ |
| ∠Q | $45^\circ$ |
| ∠R | $36^\circ$ |
| ∠S | $135^\circ$ |
Check opposite angles sum: $\angle P + \angle R = 144^\circ + 36^\circ = 180^\circ$. $\angle Q + \angle S = 45^\circ + 135^\circ = 180^\circ$. The properties of a cyclic quadrilateral are satisfied.
The average of $\angle Q$ and $\angle R$ is $40.5^\circ$.
| Property | Description |
|---|---|
| Definition | A quadrilateral whose vertices all lie on a single circle. |
| Opposite Angles | Opposite angles are supplementary (sum to $180^\circ$). |
| Exterior Angle | An exterior angle is equal to the interior opposite angle. |
| Ptolemy's Theorem | For a cyclic quadrilateral ABCD, $AC \cdot BD = AB \cdot CD + BC \cdot DA$. |
Quadrilateral: A polygon with four edges (sides) and four vertices (corners). The sum of the interior angles of any quadrilateral is always $360^\circ$.
Supplementary Angles: Two angles are supplementary if their sum is $180^\circ$. This is a key property used in solving problems involving cyclic quadrilaterals.
Inscribed Polygon: A polygon is inscribed in a circle if all of its vertices lie on the circle. A cyclic quadrilateral is an inscribed quadrilateral.
Understanding the properties of cyclic quadrilaterals, especially the supplementary nature of opposite angles, is crucial for solving problems like this one involving angle relationships.
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