CBSE Class 8 Mathematics Understanding Quadrilaterals Assignments Set 10

Question. Two angles of a triangle are \( 60^\circ \), \( 80^\circ \). Find the third angle.
Answer:
The sum of all three angles in any triangle is always \( 180^\circ \).
We are given two angles: \( 60^\circ \) and \( 80^\circ \).
Let us call the third angle \( x \).
We can write this as:
\( 60^\circ + 80^\circ + x = 180^\circ \)
\( \implies 140^\circ + x = 180^\circ \)
\( \implies x = 180^\circ - 140^\circ \)
\( \implies x = 40^\circ \)
So, the third angle is \( 40^\circ \).
In simple words: All three angles in a triangle add up to \( 180^\circ \). Subtract the two angles you know from \( 180^\circ \) to find the last one.
Exam Tip: Always write down the Angle Sum Property of a triangle before starting your calculations to score full marks.

Question. Three angles of a quadrilateral are \( 40^\circ \), \( 70^\circ \) and \( 105^\circ \). Find the fourth angle.
Answer:
The sum of all four interior angles in any quadrilateral is \( 360^\circ \).
Let us call the fourth angle \( y \).
Adding all the angles gives:
\( 40^\circ + 70^\circ + 105^\circ + y = 360^\circ \)
\( \implies 215^\circ + y = 360^\circ \)
\( \implies y = 360^\circ - 215^\circ \)
\( \implies y = 145^\circ \)
So, the fourth angle is \( 145^\circ \).
In simple words: A four-sided shape has angles that add up to \( 360^\circ \). Add the three angles you know and subtract that total from \( 360^\circ \).

Exam Tip: Clearly state that the sum of angles of a quadrilateral is \( 360^\circ \) as your reason.

Question. The angles of a quadrilateral are in the ratio \( 1:2:3:4 \). Find the smallest and the largest angle.
Answer:
Let the four angles be \( x \), \( 2x \), \( 3x \), and \( 4x \).
The sum of all angles in a quadrilateral is \( 360^\circ \).
We can write this as:
\( x + 2x + 3x + 4x = 360^\circ \)
\( \implies 10x = 360^\circ \)
\( \implies x = \frac{360^\circ}{10} \)
\( \implies x = 36^\circ \)
Now, we calculate the required angles:
The smallest angle is \( x = 36^\circ \).
The largest angle is \( 4x = 4 \times 36^\circ = 144^\circ \).
In simple words: Write the angles with \( x \) based on the ratio. Add them to equal \( 360^\circ \), find \( x \), and use it to find the smallest and largest angles.

Exam Tip: Do not stop after finding \( x \). Remember to calculate both the smallest and largest angles to get full credit.

Question. The measure of two adjacent angles of a quadrilateral are 120 and \( 40^\circ \). If the other two angles are equal, Find their measures.
Answer:
The sum of all four interior angles in any quadrilateral is \( 360^\circ \).
The given angles are \( 120^\circ \) and \( 40^\circ \).
Let each of the other two equal angles be \( z \).
We can write the equation:
\( 120^\circ + 40^\circ + z + z = 360^\circ \)
\( \implies 160^\circ + 2z = 360^\circ \)
\( \implies 2z = 360^\circ - 160^\circ \)
\( \implies 2z = 200^\circ \)
\( \implies z = \frac{200^\circ}{2} \)
\( \implies z = 100^\circ \)
So, each of the other two angles measures \( 100^\circ \).
In simple words: Name the two equal angles as the same letter. Add them with the known angles to get \( 360^\circ \), then solve for that letter.

Exam Tip: Clearly define your variable at the start. For example, write "Let each of the equal angles be \( z \)."

Question. A quadrilateral has all its four angles equal. Find the measure of each angle.
Answer:
Let each angle of the quadrilateral be \( a \).
The sum of all four angles in any quadrilateral is \( 360^\circ \).
We can write:
\( a + a + a + a = 360^\circ \)
\( \implies 4a = 360^\circ \)
\( \implies a = \frac{360^\circ}{4} \)
\( \implies a = 90^\circ \)
So, each angle of the quadrilateral measures \( 90^\circ \).
In simple words: Since all four angles are equal, just divide the total of \( 360^\circ \) by 4 to find each angle.

Exam Tip: A quadrilateral with four equal angles of \( 90^\circ \) is a rectangle or a square. Stating this adds extra value to your answer.

Question. The sum of two angles of a quadrilateral is \( 210^\circ \). If the other two angles are in the ratio 1:2. Find the angles.
Answer:
Let the other two angles be \( y \) and \( 2y \).
The sum of all four angles in a quadrilateral is \( 360^\circ \).
The sum of the first two angles is \( 210^\circ \).
We can write:
\( 210^\circ + y + 2y = 360^\circ \)
\( \implies 210^\circ + 3y = 360^\circ \)
\( \implies 3y = 360^\circ - 210^\circ \)
\( \implies 3y = 150^\circ \)
\( \implies y = \frac{150^\circ}{3} \)
\( \implies y = 50^\circ \)
Now we find the two angles:
The first angle is \( y = 50^\circ \).
The second angle is \( 2y = 2 \times 50^\circ = 100^\circ \).
So, the other two angles are \( 50^\circ \) and \( 100^\circ \).
In simple words: Subtract the sum of the first two angles from \( 360^\circ \). Then split the rest using the 1:2 ratio to find the last two angles.

Exam Tip: Check your final values at the end: \( 210^\circ + 50^\circ + 100^\circ = 360^\circ \). This ensures your answer is correct.

Question. In the adjoining figure, ABCD is a quadrilateral. The bisectors of \( \angle A \) and \( \angle B \) meet at P. If \( \angle C = 80^\circ \) and \( \angle D = 50^\circ \). Find \( \angle APB \).
Answer:
In the quadrilateral ABCD, the sum of all interior angles is \( 360^\circ \).
\( \angle A + \angle B + \angle C + \angle D = 360^\circ \)
We are given \( \angle C = 80^\circ \) and \( \angle D = 50^\circ \).
Substituting these values:
\( \angle A + \angle B + 80^\circ + 50^\circ = 360^\circ \)
\( \implies \angle A + \angle B + 130^\circ = 360^\circ \)
\( \implies \angle A + \angle B = 360^\circ - 130^\circ \)
\( \implies \angle A + \angle B = 230^\circ \)
Since AP and BP are angle bisectors:
\( \angle PAB = \frac{1}{2} \angle A \)
\( \angle PBA = \frac{1}{2} \angle B \)
In triangle APB, the sum of angles is \( 180^\circ \):
\( \angle APB + \angle PAB + \angle PBA = 180^\circ \)
\( \implies \angle APB + \frac{1}{2} \angle A + \frac{1}{2} \angle B = 180^\circ \)
\( \implies \angle APB + \frac{1}{2} (\angle A + \angle B) = 180^\circ \)
Substituting \( \angle A + \angle B = 230^\circ \):
\( \angle APB + \frac{1}{2} (230^\circ) = 180^\circ \)
\( \implies \angle APB + 115^\circ = 180^\circ \)
\( \implies \angle APB = 180^\circ - 115^\circ \)
\( \implies \angle APB = 65^\circ \).
In simple words: Find the sum of the bottom two angles. Halve that sum because of the angle bisectors, then subtract from \( 180^\circ \) to find the angle at P.

Exam Tip: Remember this shortcut formula: \( \angle APB = \frac{1}{2} (\angle C + \angle D) \). It will help you quickly check your work.

Question. In the adjoining figure, \( \angle A = 60^\circ \), \( OE \perp AC \) and \( OD \perp AB \). Find \( \angle DOE \).
Answer:
From the given diagram, ADOE forms a quadrilateral.
We are given:
\( \angle A = 60^\circ \)
Since \( OE \perp AC \), we have \( \angle AEO = 90^\circ \).
Since \( OD \perp AB \), we have \( \angle ADO = 90^\circ \).
The sum of all four interior angles in quadrilateral ADOE is \( 360^\circ \).
So, we can write:
\( \angle A + \angle ADO + \angle DOE + \angle AEO = 360^\circ \)
\( \implies 60^\circ + 90^\circ + \angle DOE + 90^\circ = 360^\circ \)
\( \implies 240^\circ + \angle DOE = 360^\circ \)
\( \implies \angle DOE = 360^\circ - 240^\circ \)
\( \implies \angle DOE = 120^\circ \).
In simple words: The four-sided shape ADOE has two corners that are right angles (\( 90^\circ \)). Add all the known corners and subtract from \( 360^\circ \) to find the missing angle.

Exam Tip: Be sure to write down why the angles are \( 90^\circ \) by mentioning that the lines are perpendicular.

Question. In the adjoining figure find x.
Answer:
The sum of all exterior angles of any polygon is always \( 360^\circ \).
Let us list the given exterior angles of quadrilateral ABCD:
At vertex A: \( 120^\circ \)
At vertex D: \( 70^\circ \)
At vertex C: \( 60^\circ \)
At vertex B: \( x^\circ \)
Since the sum of these exterior angles is \( 360^\circ \), we get:
\( 120^\circ + 70^\circ + 60^\circ + x^\circ = 360^\circ \)
\( \implies 250^\circ + x = 360^\circ \)
\( \implies x = 360^\circ - 250^\circ \)
\( \implies x = 110^\circ \).
In simple words: All the outer angles around any flat shape add up to \( 360^\circ \). Add the three known outer angles and subtract from \( 360^\circ \).

Exam Tip: Be careful to identify whether the given angle is interior or exterior. All the angles shown here are exterior angles.

Question. In the adjoining figure find x.
Answer:
The given shape ABCDE is a 5-sided polygon, which is a pentagon.
The sum of all interior angles of an \( n \)-sided polygon is:
Sum \( = (n - 2) \times 180^\circ \)
For a pentagon (\( n = 5 \)):
Sum \( = (5 - 2) \times 180^\circ = 3 \times 180^\circ = 540^\circ \)
The interior angles of this pentagon are:
\( \angle D = 40^\circ \), \( \angle E = 100^\circ \), \( \angle C = 100^\circ \), \( \angle A = x^\circ \), and \( \angle B = x^\circ \).
Adding these angles together:
\( 40^\circ + 100^\circ + 100^\circ + x^\circ + x^\circ = 540^\circ \)
\( \implies 240^\circ + 2x = 540^\circ \)
\( \implies 2x = 540^\circ - 240^\circ \)
\( \implies 2x = 300^\circ \)
\( \implies x = \frac{300^\circ}{2} \)
\( \implies x = 150^\circ \).
In simple words: A five-sided shape has interior angles adding up to \( 540^\circ \). Add the three angles you know, subtract from \( 540^\circ \), and divide by 2 to find \( x \).

Exam Tip: First use the formula \( (n-2) \times 180^\circ \) to find the sum of angles for any polygon. This helps you get started correctly.