NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals (Ex 8.2)
Exercise 8.2 of Class 9 Maths Chapter 8 Quadrilaterals is built on one simple idea called the mid-point theorem. If you join the middle points of any two sides of a triangle, the line you get is parallel to the third side, and it is exactly half as long. That one rule, along with its converse, solves every question in this exercise. The converse says that a line drawn through the middle point of one side, parallel to another side, will cut the third side exactly in half.
This exercise has 7 questions, and almost all of them ask you to prove something. A common type is this: take any quadrilateral, join the middle points of its four sides in order, and show what shape you get. Every question here is solved step by step, with the reason written next to each step, which is how CBSE gives marks. Students can revise on their own, teachers can use the same steps in class, and parents can check homework at home. A free printable PDF of the full exercise is also available.
NCERT Solutions Class 9 Maths Chapter 8 Quadrilaterals: Exercise 8.2
Q1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC is a diagonal. Show that: (i) SR ∥ AC and SR = ½ AC (ii) PQ = SR (iii) PQRS is a parallelogram.
Answer:
Given: ABCD is a quadrilateral. P, Q, R and S are the mid-points of AB, BC, CD and DA. AC is a diagonal.
To prove: (i) SR ∥ AC and SR = ½ AC (ii) PQ = SR (iii) PQRS is a parallelogram.
Proof:
(i) In â–³ADC, S and R are the mid-points of DA and DC.
By the mid-point theorem, the line joining the mid-points of two sides of a triangle is parallel to the third side and is half of it.
So, SR ∥ AC and SR = ½ AC … (i)
(ii) In â–³ABC, P and Q are the mid-points of AB and BC.
By the mid-point theorem,
PQ ∥ AC and PQ = ½ AC … (ii)
From (i) and (ii), both SR and PQ are equal to half of AC.
So, PQ = SR
(iii) From (i), SR ∥ AC. From (ii), PQ ∥ AC.
Two lines parallel to the same line are parallel to each other.
So, PQ ∥ SR. Also, PQ = SR.
In quadrilateral PQRS, one pair of opposite sides is equal and parallel.
Idea used: Mid-point theorem.
So, PQRS is a parallelogram. Hence proved.
Q2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.
Answer:
Given: ABCD is a rhombus. P, Q, R and S are the mid-points of AB, BC, CD and DA.
To prove: PQRS is a rectangle.
Construction: Join AC and BD.
Proof:
In â–³ABC, P and Q are the mid-points of AB and BC.
By the mid-point theorem, PQ = ½ AC and PQ ∥ AC … (i)
In â–³ADC, S and R are the mid-points of AD and DC.
By the mid-point theorem, SR = ½ AC and SR ∥ AC … (ii)
From (i) and (ii), PQ = SR and PQ ∥ SR.
One pair of opposite sides is equal and parallel, so PQRS is a parallelogram.
Now, in â–³ABD, P and S are the mid-points of AB and AD.
By the mid-point theorem, PS = ½ BD and PS ∥ BD … (iii)
Let the diagonals AC and BD meet at O. Let PS cut AC at M and PQ cut BD at N.
PN is a part of PQ and MO is a part of AC. Since PQ ∥ AC, we get PN ∥ MO.
MP is a part of PS and ON is a part of BD. Since PS ∥ BD, we get MP ∥ ON.
Both pairs of opposite sides are parallel, so PMON is a parallelogram.
In a rhombus, the diagonals cut each other at right angles. So, ∠MON = 90°.
Opposite angles of a parallelogram are equal. So, ∠P = ∠MON = 90°.
PQRS is a parallelogram in which one angle is 90°.
Idea used: Mid-point theorem + diagonals of a rhombus are perpendicular.
So, PQRS is a rectangle. Hence proved.
Q3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.
Answer:
Given: ABCD is a rectangle. P, Q, R and S are the mid-points of AB, BC, CD and DA.
To prove: PQRS is a rhombus.
Construction: Join AC and BD.
Proof:
In â–³ABC, P and Q are the mid-points of AB and BC.
By the mid-point theorem, PQ = ½ AC and PQ ∥ AC … (i)
In â–³ADC, S and R are the mid-points of AD and DC.
By the mid-point theorem, SR = ½ AC and SR ∥ AC … (ii)
From (i) and (ii), PQ = SR and PQ ∥ SR.
One pair of opposite sides is equal and parallel, so PQRS is a parallelogram.
In â–³ABD, P and S are the mid-points of AB and AD.
By the mid-point theorem, PS = ½ BD … (iii)
In a rectangle, the diagonals are equal. So, AC = BD.
Half of equal lengths are also equal. So, ½ AC = ½ BD.
From (i) and (iii), PQ = PS.
PQRS is a parallelogram in which two next-to-each-other (adjacent) sides are equal.
Idea used: Mid-point theorem + diagonals of a rectangle are equal.
So, PQRS is a rhombus. Hence proved.
Q4. ABCD is a trapezium in which AB ∥ DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F. Show that F is the mid-point of BC.
Answer:
Given: ABCD is a trapezium in which AB ∥ DC. BD is a diagonal. E is the mid-point of AD. A line through E, parallel to AB, cuts BC at F.
To prove: F is the mid-point of BC, that is, BF = CF.
Proof:
Let the line EF cut the diagonal BD at O.
In △ABD, E is the mid-point of AD and EO ∥ AB (because EF ∥ AB).
By the converse of the mid-point theorem, a line drawn through the mid-point of one side of a triangle, parallel to another side, cuts the third side into two equal parts.
So, O is the mid-point of BD, that is, DO = OB.
Now, AB ∥ DC and EF ∥ AB.
Lines parallel to the same line are parallel to each other. So, EF ∥ DC, that is, OF ∥ DC.
In △BCD, O is the mid-point of BD and OF ∥ DC.
By the converse of the mid-point theorem, F is the mid-point of BC.
Idea used: Converse of the mid-point theorem, used twice.
So, BF = CF and F is the mid-point of BC. Hence proved.
Q5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively. Show that the line segments AF and EC trisect the diagonal BD.
(Trisect means to cut into three equal parts.)
Answer:
Given: ABCD is a parallelogram. E and F are the mid-points of AB and CD. AF and EC cut the diagonal BD at P and Q.
To prove: DP = PQ = QB.
Proof:
ABCD is a parallelogram, so AB ∥ CD and AB = CD.
E is the mid-point of AB, so AE = ½ AB.
F is the mid-point of CD, so CF = ½ CD.
Since AB = CD, we get AE = CF. Also, AE ∥ CF, because AB ∥ CD.
In quadrilateral AECF, one pair of opposite sides is equal and parallel.
So, AECF is a parallelogram, and therefore AF ∥ EC.
In △DQC, F is the mid-point of DC and PF ∥ QC (because AF ∥ EC).
By the converse of the mid-point theorem, P is the mid-point of DQ.
So, DP = PQ … (i)
In △APB, E is the mid-point of AB and EQ ∥ AP (because EC ∥ AF).
By the converse of the mid-point theorem, Q is the mid-point of PB.
So, PQ = QB … (ii)
From (i) and (ii), DP = PQ = QB.
Idea used: Converse of the mid-point theorem.
So, AF and EC cut BD into three equal parts, that is, they trisect the diagonal BD. Hence proved.
Q6. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that: (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB.
(The sign ⊥ means "is perpendicular to", that is, the two lines meet at 90°.)
Answer:
Given: In △ABC, ∠C = 90°. M is the mid-point of AB. MD ∥ BC, and D lies on AC.
To prove: (i) D is the mid-point of AC (ii) MD ⊥ AC (iii) CM = MA = ½ AB.
Proof:
(i) In △ABC, M is the mid-point of AB and MD ∥ BC.
By the converse of the mid-point theorem, MD cuts the third side AC into two equal parts.
So, AD = DC, that is, D is the mid-point of AC.
(ii) MD ∥ BC, and AC is a transversal (a line that cuts both of them).
When a transversal cuts two parallel lines, the corresponding angles are equal.
So, ∠MDA = ∠BCA.
But ∠BCA = 90°, because the triangle is right angled at C.
So, ∠MDA = 90°.
Therefore, MD ⊥ AC.
(iii) In â–³MDC and â–³MDA:
DC = DA (proved in part (i))
∠MDC = ∠MDA (each is 90°, from part (ii))
MD = MD (common side)
So, △MDC ≅ △MDA by the S.A.S. rule (Side-Angle-Side).
By C.P.C.T. (Corresponding Parts of Congruent Triangles), CM = MA.
M is the mid-point of AB, so MA = ½ AB.
Idea used: Converse of the mid-point theorem + S.A.S. congruence.
So, CM = MA = ½ AB. Hence proved.