Important Questions Class 9 Maths Chapter 12 Quadrilaterals
Important Questions Class 9 Maths Chapter 12 Quadrilaterals
Quadrilaterals is Chapter 12 of the Class 9 Maths textbook Ganita Manjari. A quadrilateral is a closed figure with four sides. The chapter covers the angle sum property, the types of quadrilaterals (parallelogram, rectangle, rhombus, square, trapezium and kite), the properties of a parallelogram and its diagonals, the conditions for a quadrilateral to be a parallelogram, and the mid-point theorem.
Below are 12 important questions from Quadrilaterals with complete answers written by Extramarks experts. You will practise proving that a figure is a parallelogram, finding unknown angles and sides, using the properties of a rhombus and a rectangle, and applying the mid-point theorem in triangles. Draw the figure yourself while reading each proof.
Q1. In ΔQRS, median RO is produced to P such that RO = PO.
Prove that PQRS is a parallelogram.
Answer:
In ΔPOS and ΔROQ PO = OR (Given) ∠POS = ∠ROQ (Vertical opposite angle) SO = QO (O is mid-point of QS) So, by SAS congruence criteria ΔPOS ≅ ΔROQ. By CPCT, PS = QR and ∠PSO = ∠RQO. As ∠PSO = ∠RQO. So, SQ is transversal to PS and QR. So, PS || RQ and PS = RQ. Quadrilateral in which opposite sides are parallel and equal is a parallelogram. Hence, PQRS is a parallelogram.
Q2.
Find the possible measures of each angle of the parallelogram.
Answer: In a parallelogram the opposite angles are equal so ∠A = ∠C and ∠B = ∠D. Given angles are equal, so:
So, value of x can be 2 or –2 .
For x = 2,
For x = –2,
Hence, two possible set of angles are possible.
Q3. ABCD is a quadrilateral. The angle bisectors of B and C intersect at P. Show that the sum of the angles of A and D is equal to twice the ∠BPC.
Answer:
In a quadriateral we have,
Hence proved.
Q4. ABCD is a quadrilateral in which AD || BC and AB = DC, if
∠DCE = 70°, find the measure of the other three angles.
Answer:
Q5. In the given figure, ABCD is a parallelogram. Compute the values of x and y.
Answer:
Q6. In ΔPQR, median PA is drawn and B is the midpoint of median PA.
A line joining QB is produced to meet PR at C.
Answer:
Construction: Draw AD || BC, where D is point on PR. In ΔAPD,
AD || BC and B is midpoint of AP. So, by converse of midpoint theorem. C is midpoint of PD. PC = CD (1) In ΔRQC, AD || QC and A is midpoint of QR. So, by converse of midpoint theorem. D is midpoint of CR. CD = DR (2) Combining (1) and (2), PC = CD = DR (3) Equating, 2PC = 2CD 2PC = CD + DR (from(3)) 2PC = CR
Q7. If A and B are the midpoints of sides PQ and RS, respectively of a parallelogram PQRS. The line joining PR and AB intersect at O. Show that PR and AB bisect each other at O.
Answer:
As A and B are the mid points of sides PQ and RS, by using midpoint theorem we can say that AB is parallel to PS and QR. In ΔAOP and ΔBOR
PA = BR (As A and B are midpoint of PQ and RS respectively)
(Alternate interior angles)
∠APO = ∠BRO
So, PO = OR and AO = OB by CPCT. Hence, PR and AB bisect each other at O.
Q8.
. Determine the values of and , and then find the ratio of the length of side AB to the length of side BC.
Answer:
Q9. If ABCD is a rhombus, where AD = 29 cm, BO = 21 cm, then what is the area of the given rhombus.
Answer:
Q10.
Answer: We know that sum of angles of a quadrilateral is 360o. Given three angles are in the ratio 5 : 5 : 2 and fourth angle = 72°. ⸫ 5x + 5x + 2x + 72° = 360° ⇒ 12x = 360° – 72° = 288° ⇒ x = 24° ⸫ Three angles of the given quadrilateral are 5 × 24°, 5 × 24° and 2 × 24°, = 120°, 120°, 48°.
Q11.
Answer: Mid-point theorem talks about the mid-points of sides of a triangle. It states: The line segment joining the mid points of two sides of a triangle is parallel to the third side and half of it.
Here, we see that when we join the mid points D and E of sides AB and AC, we get a line segment DE parallel to the third side BC and when we produce DE to point F, such that DF = BC Then DE = EF.
Q12.
Answer:
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Q.1 A trapezium whose parallel sides are 25 cm and 10 cm. The non-parallel sides are 14 cm and 13 cm. find the area of the trapezium.
Marks:4 Ans
ABCD is a trapezium. Draw CE ||AD so that AECD is a parallelogram Area of field = area of parallelogram AECD+ area of triangle CEB Consider triangle CE Band apply Heron’s formula Area of triangle = s(sˆ ’a)(s ˆ’b)(sˆ ’c)a=14 cm b=15cmc=13 cms=a+b+c2=21AreaoftriangleCEB=21(21ˆ’14)(21ˆ’15)(21ˆ’13)=21—7—6—8=7—3—7—3—2—2—2—2=7—3—2—2=84 cm2Area of triangle CEB=12—base—altitude84=12—15—altitude Q base=15‡’84=12—15—altitude‡ altitude=16815=565 Height of parallelogram=565Areaofparallelogram=base—altitude=10—565=112 cm2ˆ´Totalareatotrapezium=112+84=196 cm2
 Q.2 An umbrella is made by stitching 10 triangular pieces of cloth of two different colours, each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella
Marks:3 Ans
Firstly, find the area of 1 triangular piece.     Let, a = 20 cm, b = 50 cm, c = 50 cms=a+b+c2=20+50+502=1202=60cmArea of one piece=ss-as-bs-c=6060-2060-5060-50=60×40×10×10=2006cm2Area of 5 pieces of each colour = 5 × 2006= 10006cm2.
 Q.3 Find area and perimeter of triangle whose sides are 8 cm ,19 cm and 15 cm.
Marks:3 Ans
Perimeter of triangle=8+9+15=42 cm We will find the area of the triangle by Heron’s formula.
Semi-perimeter, s=Perimeter2=422=21cma=8,b=19,c=15Area of triangle=ssasbsc=2121821192115=21×13×2×6=691cm2
 Q.4 What is the area of a rhombus with sides as 80 cm and the length of one diagonal as 120 cm
Marks:2 Ans
The rhombus can be drawn as below:
Area of a triangle with sides as a, b and c is given by: Area of triangle=abc Here, s=a+b+c2  =80+80+1202           =140          Area of triangle ABC=1401408014080140120                                   =140606020                                   =12007 cm2Area of rhombus ABCD=2×Area of triangle ABC                                       =2×12007 cm2                                      =24007 cm2
Hence, area of the required rhombus is 24007 cm2.
 Q.5 What is the area of a triangle with two sides as 10 cm and 20 cm, and perimeter as 45 cm
Marks:2 Ans
Area of a triangle with sides as a, b and c is given by: Area of triangle=abc
Here, s=a+b+c2Since, for perimeter, 45=10+20+c45=30+c  c=15 cm Area of triangle=452452104522045215                         =45225252152                         =75154cm2
Hence, area of the required triangle is 75154cm2.
FAQs (Frequently Asked Questions)
In order to perform well and advance their problem-solving abilities, the students are advised to thoroughly read through Important Questions Class 9 Mathematics Chapter 12 prior to the first term exams. The main objective of these solutions is to assist you in comprehending the ideas and swiftly resolving any difficulties you may encounter while engaging in an independent study.
 The NCERT solutions for Class 9 Mathematics Chapter 12 are available for students to use as a self-study resource on the Extramarks website.  Studying from these study materials will give you the right direction and help you understand all of the topics covered in each chapter.Students will be certain to receive high scores in exams by completing the questions and exercises given in our NCERT solutions.