Class 9 Maths: Quadrilaterals – Extra Questions with Answers
Q. 1 If diagonal AC in quadrilateral ABCD bisects diagonal BD, then quadrilateral ABCD is a ______.
Check Solution
Ans: D
Solution: If AC bisects BD, then the diagonals bisect each other. This is a property of parallelograms.
Answer: parallelogram
Q. 2 If all the angles of a quadrilateral are 90 degrees, what type of quadrilateral is it?
Check Solution
Ans: C
Solution: A quadrilateral with all angles equal to 90 degrees is a rectangle. If all sides are also equal, it’s a square.
Answer: Rectangle
Q. 3 In parallelogram PQRS, diagonals PR and QS intersect at point O. If PR = 10 cm, QS = 24 cm, and PO = 5 cm, what is the length of SO?
Check Solution
Ans: B
Solution:
Since the diagonals of a parallelogram bisect each other, PO = OR and QO = OS. Given PO = 5 cm, and PR = 10 cm, then OR = 5 cm. Also, PR = PO + OR = 10 cm. Since the length of QS = 24 cm, and O is the midpoint, then SO = QS/2.
SO = QS/2 = 24/2 = 12 cm.
Answer: 12 cm
Q. 4 In parallelogram PQRS, PQ = 12 cm. The altitudes corresponding to sides PQ and PS are respectively 9 cm and 6 cm. If PS is $x$ cm, then the value of $3x$ is:
Check Solution
Ans: B
Solution:
Area of parallelogram = base × height.
Using PQ as base, Area = PQ × 9 = 12 × 9 = 108 sq cm.
Using PS as base, Area = PS × 6 = x × 6.
Equating the two expressions for area, 6x = 108, so x = 18.
3x = 3 × 18 = 54.
Answer: 54
Q. 5 A quadrilateral has interior angles A, B, C, and D. If angle A is 40 degrees and the ratio of the angles A:B:C:D is 2:4:5:x, what is the value of x?
Check Solution
Ans: D
Solution:
The sum of interior angles in a quadrilateral is 360 degrees.
Given A = 40 degrees, and A:B:C:D = 2:4:5:x.
Since A corresponds to the ratio 2, we can find the multiplier: 40/2 = 20.
Therefore, B = 4 * 20 = 80 degrees, and C = 5 * 20 = 100 degrees.
So, D = x * 20.
The sum of angles is 360: A + B + C + D = 360
40 + 80 + 100 + 20x = 360
220 + 20x = 360
20x = 140
x = 7
Answer: 7
Q. 6 The line segments joining the midpoints of the adjacent sides of a square form what shape?
Check Solution
Ans: D
Solution: Connecting the midpoints of adjacent sides creates four right triangles. The hypotenuses of these triangles form the sides of the new shape.
Answer: Square
Q. 7 The base of an isosceles triangle is 12 cm and its area is 36 cm$^2$. What is the perimeter of the triangle?
Check Solution
Ans: D
Solution:
Let the height of the triangle be h.
Area = (1/2) * base * height
36 = (1/2) * 12 * h
h = 6 cm
Since the triangle is isosceles, the height bisects the base, forming two right triangles with legs 6 cm and 6 cm.
The hypotenuse (equal side) = sqrt(6^2 + 6^2) = sqrt(72) = 6 * sqrt(2) cm
Perimeter = 12 + 2 * 6*sqrt(2) = 12 + 12*sqrt(2) cm
Answer: 12 + 12√2 cm
Q. 8 The consecutive angles of a parallelogram are in the ratio 3:7. What are the measures of the four angles?
Check Solution
Ans: C
Solution:
Let the angles be 3x and 7x. Consecutive angles of a parallelogram are supplementary.
So, 3x + 7x = 180
10x = 180
x = 18
The angles are 3(18) = 54 degrees, and 7(18) = 126 degrees.
The opposite angles are equal, so the angles are 54, 126, 54, and 126 degrees.
Answer: 54, 126, 54, 126
Q. 9 The difference between the lengths of the parallel sides of a trapezium is 6 cm, the perpendicular distance between these sides is 12 cm and the area of the trapezium is 108 cm$^2$. Find the length of each of the parallel sides.
Check Solution
Ans: A
Solution:
Let the parallel sides be $a$ and $b$, where $a – b = 6$. The area of the trapezium is given by $\frac{1}{2}(a+b)h$, where $h=12$.
So, $108 = \frac{1}{2}(a+b) \times 12$.
Then, $108 = 6(a+b)$.
Dividing by 6, $a+b = 18$.
We have $a-b = 6$ and $a+b = 18$.
Adding the equations, $2a = 24$, so $a = 12$.
Then $b = 18-12 = 6$.
Answer: 12 cm and 6 cm
Q. 10 If AD is the median of $\triangle ABC$ and P is a point on AC such that $ar(\triangle ADP) : ar(\triangle ABD) = 2:5$, then $ar(\triangle PDC) : ar(\triangle ABC)$ is?
Check Solution
Ans: B
Solution:
Since AD is a median, $ar(\triangle ABD) = ar(\triangle ADC)$.
Given $ar(\triangle ADP) : ar(\triangle ABD) = 2:5$, we have $ar(\triangle ADP) = \frac{2}{5} ar(\triangle ABD)$.
Then, $ar(\triangle PDC) = ar(\triangle ADC) – ar(\triangle ADP) = ar(\triangle ABD) – \frac{2}{5} ar(\triangle ABD) = \frac{3}{5} ar(\triangle ABD)$.
Since $ar(\triangle ABD) = \frac{1}{2} ar(\triangle ABC)$, we have $ar(\triangle PDC) = \frac{3}{5} \cdot \frac{1}{2} ar(\triangle ABC) = \frac{3}{10} ar(\triangle ABC)$.
Therefore, $ar(\triangle PDC) : ar(\triangle ABC) = \frac{3}{10} : 1$.
Answer: 3/10
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