EXERCISE 10C
Question 1:
P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD. Show that
(i) PQ || AC and PQ $= \frac{1}{2}AC$
(ii) PQ || SR
(iii) PQRS is a parallelogram.
Answer 1:
Given: In quadrilateral ABCD, P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA.
To prove:
(i) PQ || AC and PQ$= \frac{1}{2}AC$
(ii) PQ || SR
(iii) PQRS is a parallelogram.
Proof:
(i)
In ΔABC,
Since, P and Q are the mid points of sides AB and BC, respectively. (Given)
⇒AC∥PQ and PQ$= \frac{1}{2}AC$ (Using mid-point theorem.) ...(1)
(ii) In ΔADC,
Since, S and R are the mid-points of AD and DC, respectively. (Given)
⇒SR∥AC and SR$= \frac{1}{2}AC$ (Using mid-point theorem.) ...(2)
From (1) and (2), we get
PQ || SR and equal .
(iii)
Construction : Join D to BIn ΔDCB
Since, R and Q are the mid points of sides DC and CB, respectively. (Given)
⇒RQ∥DB and RQ$= \frac{1}{2}$DB (Using mid-point theorem.) ....(3)
In ΔDAB
Since, S and P are the mid points of sides DA and AB, respectively. (Given)
⇒SP∥DB and SP$= \frac{1}{2}$DB (Using mid-point theorem.) ....(4)
From (3) and (4) , we get
PQ || SR and equal .
Question 2:
A square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.
Answer 2:
Given: In an isosceles right ∆ABC, CMPN is a square.
To prove: P bisects the hypotenuse AB i.e., AP = PB.
Proof:
In square CMPN,
∴ CM = MP = PN = CN (All sides are equal.)
Also, ∆ABC is an isosceles with AC = BC.
⇒ AN + NC = CM + MB
⇒ AN = MB (∵ CN = CM) ...(i)
Now,
In ∆ANP and ∆PMB,
AN = MB [From (i)]
∠ANP = ∠PMB = 90°
PN = PM (Sides of square CMPN)
∴ By SAS congruence criteria,
∆ANP ≅ ∆BMP
Hence, AP = PB (By CPCT)
Question 3:
In the adjoining figure, ABCD is a || gm in which E and F are the midpoints of AB and CD respectively. If GH is a line segment that cuts AD, EF and BC at G, P and H respectively, prove that GP = PH.
Answer 3:
In parallelogram ABCD, we have:
AD || BC and AB || DC
AB = AE + BE and DC = DF + FC
∴ AD || EF
Similarly, BEFC is also a parallelogram.
∴ EF || BC
∴ AD || EF || BC
Thus, AD, EF and BC are three parallel lines cut by the transversal line DC at D, F and C, respectively such that DF = FC.
These lines AD, EF and BC are also cut by the transversal AB at A, E and B, respectively such that AE = BE.
Similarly, they are also cut by GH.
∴ GP = PH (By intercept theorem)
Question 4:
M and N are points on opposite sides AD and BC of a parallelogram ABCD such that MN passes through the point of intersection O of its diagonals AC and BD. Show that MN is bisected at O.
Answer 4:
Given: A parallelogram ABCD
To prove: MN is bisected at O
Proof:
In OAM and OCN,
OA = OC (Diagonals of parallelogram bisect each other)
∠AOM = ∠CON (Vertically opposite angles)
∠MAO = ∠OCN (Alternate interior angles)
∴By ASA congruence criteria,
ΔOAM ≅ ΔOCN
⇒ OM = ON (CPCT)
Question 5:
In the adjoining figure, PQRS is a trapezium in which PQ || SR and M is the midpoint of PS. A line segment MN || PQ meets QR at N. Show that N is the midpoint of QR.
Answer 5:
Given: In trapezium PQRS, PQ || SR, M is the midpoint of PS and MN || PQ.
To prove: N is the midpoint of QR.
Construction: Join QS.
Proof:
In ∆SPQ,
Since, M is the mid-point of SP and MO || PQ.
Therefore, O is the mid-point of SQ. (By Mid-point theorem)
Similarly, in ∆SRQ,
Since, O is the mid-point of SQ and ON || SR (SR || PQ and MN || PQ)
Therefore, N is the mid-point of QR. (By Mid-point theorem)
Question 6:
In a parallelogram PQRS, PQ = 12 cm and PS = 9 cm. The bisector of ∠P meets SR in M. PM and QR both when produced meet at T. Find the length of RT.
Answer 6:
Given: In parallelogram PQRS, PQ = 12 cm and PS = 9 cm. The bisector of ∠SPQ meets SR at M.
Let ∠SPQ = 2x.
⇒ ∠SRQ = 2x and ∠TPQ = x.
Also, PQ ∥ SR
⇒ ∠TMR = ∠TPQ = x.
In △TMR, ∠SRQ is an exterior angle.
⇒ ∠SRQ = ∠TMR + ∠MTR
⇒ 2x = x + ∠MTR
⇒ ∠MTR = x
⇒ △TPQ is an isosceles triangle.
⇒ TQ = PQ = 12 cm
Now,
RT = TQ − QR
= TQ − PS
= 12 − 9
= 3 cm
Question 7:
In the adjoining figure, ABCD is a trapezium in which AB || DC and P, Q are the midpoints of AD and BC respectively. DQ and AB when produced meet at E. Also, AC and PQ intersect at R. Prove that (i) DQ = QE, (ii) PR || AB, (iii) AR = RC.
Answer 7:
Given: AB || DC, AP = PD and BQ = CQ
(i) In ∆QCD and ∆QBE, we have:
∠DQC = ∠BQE (Vertically opposite angles)
BQ = CQ (Q is the midpoints)
∴ ∆QCD ≅ ∆QBE
Hence, DQ = QE (CPCT)
(ii) Now, in ∆ADE, P and Q are the midpoints of AD and DE, respectively.
∴ PQ || AE
⇒ PQ || AB || DC
⇒ AB || PR || DC
(iii) PQ, AB and DC are the three lines cut by transversal AD at P such that AP = PD.
These lines PQ, AB, DC are also cut by transversal BC at Q such that BQ = QC.
Similarly, lines PQ, AB and DC are also cut by AC at R.
∴ AR = RC (By intercept theorem)
Question 8:
In the adjoining figure, AD is a median of ∆ABC and DE || BA. Show that BE is also a median of ∆ABC.
Answer 8:
AD is a median of ∆ABC.
∴ BD = DC
We know that the line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side.
Here, in ∆ABC, D is the mid point of BC and DE || BA (given). Then DE bisects AC.
i.e., AE = EC
∴ E is the midpoint of AC.
⇒ BE is the median of ∆ABC.
Question 9:
In the adjoining figure, AD and BE are the medians of ∆ABC and DF || BE. Show that .
Answer 9:
In ∆ABC, we have:
AC = AE + EC ...(i)
AE = EC ...(ii) [BE is the median of ∆ABC]
∴ AC = 2EC ...(iii)
In ∆BEC, DF || BE.
∴ EF = CF (By midpoint theorem, as D is the midpoint of BC)
But EC = EF + CF
⇒ EC = 2 ⨯ CF ...(iv)
From (iii) and (iv), we get:
AC = 2 ⨯ (2 ⨯ CF)
∴ CF =
Question 10:
Prove that the line segments joining the middle points of the sides of a triangle divide it into four congruent triangles.
Answer 10:
∆ABC is shown below. D, E and F are the midpoints of sides AB, BC and CA, respectively.
As, D and E are the mid points of sides AB, and BC of ∆ ABC.
∴ DE ∣∣ AC (By midpoint theorem)
Similarly, DF ∣∣ BC and EF ∣∣ AB.
Therefore, ADEF, BDFE and DFCE are all parallelograms.
Now, DE is the diagonal of the parallelogram BDFE.
∴ ∆BDE ≅ ∆FED
Similarly, DF is the diagonal of the parallelogram ADEF.
And, EF is the diagonal of the parallelogram DFCE.
So, all the four triangles are congruent.
Question 11:
In the adjoining figure, D, E, F are the midpoints of the sides BC, CA and AB respectively, of ∆ABC. Show that ∠EDF = ∠A, ∠DEF = ∠B and ∠DEF = ∠C.
Answer 11:
∆ ABC is shown below. D, E and F are the midpoints of sides BC, CA and AB, respectively.
As F and E are the mid points of sides AB and AC of ∆ ABC.
∴ FE ∣∣ BC (By mid point theorem)
Similarly, DE ∣∣ FB and FD ∣∣ AC.
Therefore, AFDE, BDEF and DCEF are all parallelograms.
In parallelogram AFDE, we have:
∠A = ∠EDF (Opposite angles are equal)
In parallelogram BDEF, we have:
∠B = ∠DEF (Opposite angles are equal)
In parallelogram DCEF, we have:
∠ C = ∠ DFE (Opposite angles are equal)
Question 12:
Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rectangle is a rhombus.
Answer 12:
Let ABCD be the rectangle and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join AC, a diagonal of the rectangle.
In ∆ ABC, we have:
∴ PQ ∣∣ AC and PQ$=\frac{1}{2}$AC [By midpoint theorem]
Again, in ∆ DAC, the points S and R are the mid points of AD and DC, respectively.
∴ SR ∣∣ AC and SR$=\frac{1}{2}$AC [By midpoint theorem]
Now, PQ ∣∣ AC and SR ∣∣ AC
⇒ PQ ∣∣ SR
So, PQRS is a parallelogram.
Now, in ∆SAP and ∆QBP, we have:
AS = BQ
∠A = ∠B = 90o
AP = BP
i.e., ∆SAP ≅ ∆QBP
∴ PS = PQ ...(ii)
Similarly, ∆SDR ≅ ∆QCR
∴ SR = RQ ...(iii)
From (i), (ii) and (iii), we have:
PQ = PQ = SR = RQ
Hence, PQRS is a rhombus.
Question 13:
Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle.
Answer 13:
Let ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join the diagonals, AC and BD.
In ∆ ABC, we have:
PQ ∣∣ AC and PQ$=\frac{1}{2}$AC [By midpoint theorem]
Again, in ∆DAC, the points S and R are the midpoints of AD and DC, respectively.
∴ SR ∣∣ AC and SR$=\frac{1}{2}$AC [By midpoint theorem]
Now, PQ ∣∣ AC and SR ∣∣ AC ⇒ PQ ∣∣ SR
So, PQRS is a parallelogram.
We know that the diagonals of a rhombus bisect each other at right angles.
∴ ∠EOF = 90o
Now, RQ∣∣ DB
⇒RE ∣∣ FO
Also, SR∣∣ AC
⇒FR ∣∣ OE
∴ OERF is a parallelogram.
So, ∠FRE = ∠EOF = 90o (Opposite angles are equal)
Thus, PQRS is a parallelogram with ∠R = 90o.
∴ PQRS is a rectangle.
Question 14:
Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is a square.
Answer 14:
Let ABCD be a square and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join the diagonals AC and BD. Let BD cut SR at F and AC cut RQ at E. Let O be the intersection point of AC and BD.
In ∆ ABC, we have:
∴ PQ ∣∣ AC and PQ$=\frac{1}{2}$AC [By midpoint theorem]
Again, in ∆DAC, the points S and R are the midpoints of AD and DC, respectively.
∴ SR ∣∣ AC and SR$=\frac{1}{2}$AC [By midpoint theorem]
Now, PQ ∣∣ AC and SR ∣∣ AC
⇒ PQ ∣∣ SR
So, PQRS is a parallelogram.
Now, in ∆SAP and ∆QBP, we have:
AS = BQ
∠A = ∠B = 90o
AP = BP
i.e., ∆SAP ≅ ∆QBP
∴ PS = PQ ...(ii)
Similarly, ∆SDR ≅ ∆RCQ
∴ SR = RQ ...(iii)
From (i), (ii) and (iii), we have:
PQ = PS = SR = RQ ...(iv)
We know that the diagonals of a square bisect each other at right angles.
∴ ∠EOF = 90o
Now, RQ ∣∣ DB
⇒RE ∣∣ FO
Also, SR ∣∣ AC
⇒FR ∣∣ OE
∴ OERF is a parallelogram.
So, ∠FRE = ∠EOF = 90o (Opposite angles are equal)
Thus, PQRS is a parallelogram with ∠R = 90o and PQ = PS = SR = RQ.
∴ PQRS is a square.
Question 15:
Prove that the line segments joining the midpoints of opposite sides of a quadrilateral bisect each other.
Answer 15:
Let ABCD be the quadrilateral in which P, Q, R, and S are the midpoints of sides AB, BC, CD, and DA, respectively.
Join PQ, QR, RS, SP and BD. BD is a diagonal of ABCD.
In ΔABD, S and P are the midpoints of AD and AB, respectively.
∴ SP || BD and SP = $\dfrac{1}{2}$ BD ... (i) (By midpoint theorem)
Similarly in Δ BCD, we have:
QR || BD and QR = $\dfrac{1}{2}$ BD ... (ii) (By midpoint theorem)
From equations (i) and (ii), we get:
SP || BD || QR
∴ SP || QR and SP = QR [Each equal to $\dfrac{1}{2}$ BD]
In quadrilateral SPQR, one pair of the opposite sides is equal and parallel to each other.
∴ SPQR is a parallelogram.
We know that the diagonals of a parallelogram bisect each other.
∴ PR and QS bisect each other.
Question 16:
The diagonals of a quadrilateral ABCD are equal. Prove that the quadrilateral formed by joining the midpoints of its sides is a rhombus.
Answer 16:
To prove: KLMN is a rhombus.
Proof:
In ∆ADC,
Since, K and L are the mid-points of sides AD and CD, respectively.
So, KL || AC and KL$=\frac{1}{2}$AC ...(1)
Similarly, in ∆ABC,
Since, M and N are the mid-points of sides BC and AB, respectively.
So, NM || AC and NM $=\frac{1}{2}$AC ...(2)
From (1) and (2), we get
But this a pair of opposite sides of the quadrilateral KLMN.
So, KLMN is a parallelogram.
Now, in ∆ABD,
Since, K and N are the mid-points of sides AD and AB, respectively.
So, KN || BD and KN$=\frac{1}{2}$BD ...(3)
But BD = AC (Given)
$=\frac{1}{2}$BD$=\frac{1}{2}$AC
KN = NM [From (2) and (3)]
But these are a pair of adjacent sides of the parallelogram KLMN.
Hence, KLMN is a rhombus.
Question 17:
The diagonals of a quadrilateral ABCD are perpendicular to each other. Prove that the quadrilateral formed by joining the midpoints of its sides is a rectangle.
Answer 17:
Given: In quadrilateral ABCD, AC BD. P, Q, R and S are the mid-points of AB, BC, CD and AD, respectively.
To prove: PQRS is a rectangle.
Proof:
In ΔABC, P and Q are mid-points of AB and BC, respectively.
∴ PQ || AC and PQ$=\frac{1}{2}$AC (Mid-point theorem) ...(1)
Similarly, in ΔACD,
So, R and S are mid-points of sides CD and AD, respectively.
∴ SR || AC and SR$=\frac{1}{2}$AC (Mid-point theorem) ...(2)
From (1) and (2), we get
PQ || SR and PQ = SR
But this is a pair of opposite sides of the quadrilateral PQRS,
So, PQRS is parallelogram.
Now, in ΔBCD, Q and R are mid-points of BC and CD, respectively.
∴ QR || BD and QR$=\frac{1}{2}$BD (Mid-point theorem) ...(3)
From (2) and (3), we get
SR || AC and QR || BD
But, AC ⊥ BD (Given)
∴ RS ⊥ QR
Hence, PQRS is a rectangle.
Question 18:
The midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD are joined to form a quadrilateral. If AC = BD and AC ⊥ BD then prove that the quadrilateral formed is a square.
Answer 18:
Given: In quadrilateral ABCD,
AC = BD and AC ⊥ BD. P, Q, R and S are the mid-points of AB, BC, CD and AD, respectively.
To prove: PQRS is a square.
Construction: Join AC and BD.
Proof:
In ΔABC,
P and Q are mid-points of AB and BC, respectively.
PQ || AC and PQ$=\frac{1}{2}$AC (Mid-point theorem) ...(1)
Similarly, in ΔACD,
R and S are mid-points of sides CD and AD, respectively.
SR || AC and SR$=\frac{1}{2}$AC (Mid-point theorem) ...(2)
From (1) and (2), we get
PQ || SR and PQ = SR
But this a pair of opposite sides of the quadrilateral PQRS.
So, PQRS is parallelogram.
Now, in ΔBCD,
Q and R are mid-points of sides BC and CD, respectively.
QR || BD and QR$=\frac{1}{2}$BD (Mid-point theorem) ...(3)
From (2) and (3), we get
RS || AC and QR || BD
But, AC ⊥ BD (Given)
∴ RS ⊥ QR
But this a pair of adjacent sides of the parallelogram PQRS.
So, PQRS is a rectangle.
Again, AC = BD (Given)
$=\frac{1}{2}$AC$=\frac{1}{2}$BD
RS = QR [From (2) and (3)]
But this a pair of adjacent sides of the rectangle PQRS.
Hence, PQRS is a square.
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