Category: CBSE Class 9

RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2

Other Exercises

• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.1
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS

Question 1.
In the figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD. [NCERT]

Solution:
In ||gm ABCD,
Base AB = 16 cm
and altitude AE = 8 cm

∴ Area = Base x Altitude
= AB x AE
= 16 x 8 = 128 cm²
Now area of ||gm ABCD = 128 cm²
Altitude CF = 10 cm
∴ Base AD = \(\frac { Area }{ Altitude }\) = \(\frac { 128 }{ 10 }\) = 12.8cm

Question 2.
In Q. No. 1, if AD = 6 cm, CF = 10 cm, AE = 8 cm, find AB.
Solution:
Area of ||gm ABCD,

= Base x Altitude
= AD x CF
= 6 x 10 = 60 cm²
Again area of ||gm ABCD = 60 cm²
Altitude AE = 8 cm
∴ Base AB =\(\frac { Area }{ Altitude }\) = \(\frac { 60 }{ 8 }\) = \(\frac { 15 }{ 2 }\) cm = 7.5 cm

Question 3.
Let ABCD be a parallelogram of area 124 cm2. If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AEFD.
Solution:
Area of ||gm ABCD = 124 cm²
E and F are the mid points of sides AB and CD respectively. E, F are joined.

Draw DL ⊥ AB
Now area of ||gm ABCD = Base x Altitude
= AB x DL = 124 cm²
∵ E and F are mid points of sides AB and CD
∴ AEFD is a ||gm
Now area of ||gm AEFD = AE x DL
= \(\frac { 1 }{ 2 }\)AB x DL [∵ E is mid point of AB]
= \(\frac { 1 }{ 2 }\) x area of ||gm ABCD
= \(\frac { 1 }{ 2 }\) x 124 = 62 cm²

Question 4.
If ABCD is a parallelogram, then prove that ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = \(\frac { 1 }{ 2 }\)ar( ||^gm ABCD).
Solution:
Given : In ||gm ABCD, BD and AC are joined

To prove : ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = \(\frac { 1 }{ 2 }\)ar(||gm ABCD)
Proof: ∵ Diagonals of a parallelogram bisect it into two triangles equal in area When BD is the diagonal, then
∴ ar(∆ABD) = ar(∆BCD) = \(\frac { 1 }{ 2 }\)ar(||^gm ABCD) …(i)
Similarly, when AC is the diagonal, then
ar(∆ABC) = ar(∆ADC) = \(\frac { 1 }{ 2 }\)ar(||^gm ABCD) …(ii)
From (i) and (ii),
ar(∆ABD) = ar(∆BCD) = ar(∆ABC) = ar(∆ACD) = \(\frac { 1 }{ 2 }\) ar(||^gm ABCD)

Hope given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4

Other Exercises

• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

Question 1.
In a ∆ABC, D, E and F are respectively, the mid-points of BC, CA and AB. If the lengths of side AB, BC and CA are 7cm, 8cm and 9cm, respectively, find the perimeter of ∆DEF.
Solution:
In ∆ABC, D, E and F are the mid-points of sides,
BC, CA, AB respectively
AB = 7cm, BC = 8cm and CA = 9cm
∵ D and E are the mid points of BC and CA
∴ DE || AB and DE =\(\frac { 1 }{ 2 }\) AB =\(\frac { 1 }{ 2 }\) x 7 = 3.5cm
Similarly,

Question 2.
In a triangle ∠ABC, ∠A = 50°, ∠B = 60° and ∠C = 70°. Find the measures of the angles of the triangle formed by joining the mid-points of the sides of this triangle.
Solution:
In ∆ABC,
∠A = 50°, ∠B = 60° and ∠C = 70°

D, E and F are the mid points of sides BC, CA and AB respectively
DE, EF and ED are joined
∵ D, E and F are the mid points of sides BC, CA and AB respectively
∴ EF || BC
DE || AB and FD || AC
∴ BDEF and CDEF are parallelogram
∴ ∠B = ∠E = 60° and ∠C = ∠F = 70°
Then ∠A = ∠D = 50°
Hence ∠D = 50°, ∠E = 60° and ∠F = 70°

Question 3.
In a triangle, P, Q and R are the mid-points of sides BC, CA and AB respectively. If AC = 21 cm, BC = 29cm and AB = 30cm, find the perimeter of the quadrilateral ARPQ.
Solution:
P, Q, R are the mid points of sides BC, CA and AB respectively
AC = 21 cm, BC = 29 cm and AB = 30°
∵ P, Q, R and the mid points of sides BC, CA and AB respectively.
∴ PQ || AB and PQ = \(\frac { 1 }{ 2 }\) AB

Question 4.
In a ∆ABC median AD is produced to X such that AD = DX. Prove that ABXC is a parallelogram.
Solution:
Given : In ∆ABC, AD is median and AD is produced to X such that DX = AD
To prove : ABXC is a parallelogram
Construction : Join BX and CX
Proof : In ∆ABD and ∆CDX
AD = DX (Given)
BD = DC (D is mid points)
∠ADB = ∠CDX (Vertically opposite angles)
∴ ∆ABD ≅ ∆CDX (SAS criterian)
∴ AB = CX (c.p.c.t.)
and ∠ABD = ∠DCX
But these are alternate angles

∴ AB || CX and AB = CX
∴ ABXC is a parallelogram.

Question 5.
In a ∆ABC, E and F are the mid-points of AC and AB respectively. The altitude AP to BC intersects FE at Q. Prove that AQ = QP.
Solution:
Given : In ∆ABC, E and F are the mid-points of AC and AB respectively.
EF are joined.
AP ⊥ BC is drawn which intersects EF at Q and meets BC at P.
To prove: AQ = QP
proof : In ∆ABC

E and F are the mid points of AC and AB
∴ EF || BC and EF = \(\frac { 1 }{ 2 }\)BC
∴ ∠F = ∠B
In ∆ABP,
F is mid point of AB and Q is the mid point of FE or FQ || BC
∴ Q is mid point of AP,
∴ AQ = QP

Question 6.
In a ∆ABC, BM and CN are perpendiculars from B and C respectively on any line passing through A. If L is the mid-point of BC, prove that ML NL.
Solution:
In ∆ABC,
BM and CN are perpendicular on a line drawn from A. L is the mid point of BC. ML and NL are joined.

Question 7.
In the figure triangle ABC is right-angled at B. Given that AB = 9cm. AC = 15cm and D, E are the mid points of the sides AB and AC respectively, calculate.
(i) The length of BC
(ii) The area of ∆ADC
Solution:
In ∆ABC, ∠B = 90°
AC =15 cm, AB = 9cm
D and E are the mid points of sides AB and AC respectively and D, E are joined.

Question 8.
In the figure, M, N and P are the mid points of AB, AC and BC respectively. If MN = 3 cm, NP = 3.5cm and MP = 2.5cm, calculate BC, AB and AC.

Solution:
In ∆ABC,
M, N and P are the mid points of side, AB, AC and BC respectively.

Question 9.
In the figure, AB = AC and CP || BA and AP is the bisector of exterior ∠CAD of ∆ABC. Prove that (i) ∠PAC = ∠BCA (ii) ABCP is a parallelogram.
Solution:
Given : In ABC, AB = AC

nd CP || BA, AP is the bisector of exterior ∠CAD of ∆ABC
To prove :
(i) ∠PAC = ∠BCA
(ii) ABCP is a ||gm
Proof : (i) In ∆ABC,
∵ AB =AC
∴ ∠C = ∠B (Angles opposite to equal sides) and ext.
∠CAD = ∠B + ∠C
= ∠C + ∠C = 2∠C ….(i)
∵ AP is the bisector of ∠CAD
∴ 2∠PAC = ∠CAD …(ii)
From (i) and (ii)
∠C = 2∠PAC
∠C = ∠CAD or ∠BCA = ∠PAC
Hence ∠PAC = ∠BCA
(ii) But there are alternate angles,
∴ AD || BC
But BA || CP
∴ ABCP is a ||gm.

Question 10.
ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the mid-points of the sides, in order, is a rectangle.
Solution:
Given : In fne figure, ABCD is a kite in which AB = AD and BC = CD.
P, Q, R and S are the mid points of the sides AB, BC, CD and DA respectively.
To prove : PQRS is a rectangle.
Construction : Join AC and BD.

Proof: In ∆ABD,
P and S are mid points of AB and AD
∴ PS || BD and PS = \(\frac { 1 }{ 2 }\) BD …(i)
Similarly in ∆BCD,
Q and R the mid points of BC and CD
∴ QR || BD and
QR = \(\frac { 1 }{ 2 }\) BD …(ii)
∴ Similarly, we can prove that PQ || SR and PQ = SR …(iii)
From (i) and (ii) and (iii)
PQRS is a parallelogram,
∵ AC and BD intersect each other at right angles.
∴ PQRS is a rectangle.

Question 11.
Let ABC be an isosceles triangle in which AB = AC. If D, E, F be the mid-points of the sides BC, CA and AB respectively, show that the segment AD and EF bisect each other at right angles.
Solution:
In ∆ABC, AB = AC
D, E and F are the mid points of the sides BC, CA and AB respectively,
AD and EF are joined intersecting at O
To prove : AD and EF bisect each other at right angles.
Construction : Join DE and DF.

Proof : ∵ D, E and F are the mid-points of
the sides BC, CA and AB respectively
∴ AFDE is a ||gm
∴ AF = DE and AE = DF
But AF = AE
(∵ E and F are mid-points of equal sides AB and AC)
∴ AF = DF = DE = AE
∴AFDE is a rhombus
∵ The diagonals of a rhombus bisect each other at right angle.
∴ AO = OD and EO = OF
Hence, AD and EF bisect each other at right angles.

Question 12.
Show that the line segments joining the mid points of the opposite sides of a quadrilateral bisect each other.
Solution:
Given : In quad. ABCD,
P, Q, R and S are the mid points of sides AB, BC, CD and DA respectively.
PR and QS to intersect each other at O
To prove : PO = OR and QO = OS
Construction: Join PQ, QR, RS and SP and also join AC.
Proof: In ∆ABC
P and Q are mid-points of AB and BC
∴ PQ || AC and PQ = \(\frac { 1 }{ 2 }\) AC …(i)
Similarly is ∆ADC,
S and R are the mid-points of AD and CD
∴ SR || AC and SR = \(\frac { 1 }{ 2 }\) AC ..(ii)

from (i) and (ii)
PQ = SQ and PQ || SR
PQRS is a ||gm (∵ opposite sides are equal area parallel)
But the diagonals of a ||gm bisect each other.
∴ PR and QS bisect each other.

Question 13.
Fill in the blanks to make the following statements correct :
(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is …
(ii) The triangle formed by joining the mid-points of the sides of a right triangle is …
(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is …
Solution:
(i) The triangle formed by joining the mid-points of the sides of an isosceles triangle is an isosceles triangle.

(ii) The triangle formed by joining the mid-points of the sides of a right triangle is right triangle.

(iii) The figure formed by joining the mid-points of consecutive sides of a quadrilateral is a parallelogram.

Question 14.
ABC is a triangle and through A, B, C lines are drawn parallel to BC, CA and AB respectively intersecting at P, Q and R. Prove that the perimeter of ∆PQR is double the perimeter of ∆ABC.
Solution:
Given : In ∆ABC,
Through A, B and C, lines are drawn parallel to BC, CA and AB respectively meeting at P, Q and R.

To prove : Perimeter of ∆PQR = 2 x perimeter of ∆ABC
Proof : ∵ PQ || BC and QR || AB
∴ ABCQ is a ||gm
∴ BC = AQ
Similarly, BCAP is a ||gm
∴ BC = AP …(i)
∴ AQ = AP = BL
⇒ PQ = 2BC
Similarly, we can prove that
QR = 2AB and PR = 2AC
Now perimeter of ∆PQR.
= PQ + QR + PR = 2AB + 2BC + 2AC
= 2(AB + BC + AC)
= 2 perimeter of ∆ABC.
Hence proved

Question 15.
In the figure, BE ⊥ AC. AD is any line from A to BC intersecting BE in H. P, Q and R are respectively the mid-points of AH, AB and BC. Prove that PQR = 90°.
Solution:
Given: In ∆ABC, BE ⊥ AC
AD is any line from A to BC meeting BC in D and intersecting BE in H. P, Q and R are respectively mid points of AH, AB and BC. PQ and QR are joined B.

To prove : ∠PQR = 90°
Proof: In ∆ABC,
Q and R the mid points of AB and BC 1
∴ QR || AC and QR = \(\frac { 1 }{ 2 }\) AC
Similarly, in ∆ABH,
Q and P are the mid points of AB and AH
∴ QP || BH or QP || BE
But AC ⊥ BE
∴ QP ⊥ QR
∴ ∠PQR = 90°

Question 16.
ABC is a triangle. D is a point on AB such that AD = \(\frac { 1 }{ 4 }\) AB and E is a point on AC such that AE = \(\frac { 1 }{ 4 }\) AC. Prove that DE = \(\frac { 1 }{ 4 }\) BC.
Solution:
Given : In ∆ABC,
D is a point on AB such that
AD = \(\frac { 1 }{ 4 }\) AB and E is a point on AC such 1
that AE = \(\frac { 1 }{ 4 }\) AC
DE is joined.

To prove : DE = \(\frac { 1 }{ 4 }\) BC
Construction : Take P and Q the mid points of AB and AC and join them
Proof: In ∆ABC,
∵ P and Q are the mid-points of AB and AC

Question 17.
In the figure, ABCD is a parallelogram in which P is the mid-point of DC and Q is a point on AC such that CQ = \(\frac { 1 }{ 4 }\) AC. If PQ produced meets BC at R, prove that R is a mid-point of BC.

Solution:
Given : In ||gm ABCD,
P is the mid-point of DC and Q is a point on AC such that CQ = \(\frac { 1 }{ 4 }\) AC. PQ is produced meets BC at R.

To prove : R is mid point of BC
Construction : Join BD
Proof : ∵ In ||gm ABCD,
∵ Diagonal AC and BD bisect each other at O
∴ AO = OC = \(\frac { 1 }{ 2 }\) AC …(i)
In ∆OCD,
P and Q the mid-points of CD and CO
∴ PQ || OD and PQ = \(\frac { 1 }{ 2 }\) OD
In ∆BCD,
P is mid-poiht of DC and PQ || OD (Proved above)
Or PR || BD
∴ R is mid-point BC.

Question 18.
In the figure, ABCD and PQRC are rectangles and Q is the mid-point of AC.
Prove that (i) DP = PC (ii) PR = \(\frac { 1 }{ 2 }\) AC.

Solution:
Given : ABCD are PQRC are rectangles and Q is the mid-point of AC.
To prove : (i) DP = PC (ii) PR = \(\frac { 1 }{ 2 }\) AC
Construction : Join diagonal AC which passes through Q and join PR.

Proof : (i) In ∆ACD,
Q is mid-point of AC and QP || AD (Sides of rectangles)
∴ P is mid-point of CD
∴ DP = PC
(ii) ∵PR and QC are the diagonals of rectangle PQRC
∴ PR = QC
But Q is the mid-point of AC
∴ QC = \(\frac { 1 }{ 2 }\) AC
Hence PR = \(\frac { 1 }{ 2 }\) AC

Question 19.
ABCD is a parallelogram, E and F are the mid points AB and CD respectively. GFI is any line intersecting AD, EF and BC at Q P and H respectively. Prove that GP = PH.
Solution:
Given : In ||gm ABCD,
E and F are mid-points of AB and CD
GH is any line intersecting AD, EF and BC at GP and H respectively

To prove : GP = PH
Proof: ∵ E and F are the mid-points of AB and CD

Question 20.
BM and CN are perpendiculars to a line passing, through the vertex A of a triangle ABC. If L is the mid-point of BC, prove that LM = LN.
Solution:
In ∆ABC,
BM and CN are perpendicular on a line drawn from A.
L is the mid point of BC.
ML and NL are joined.

Hope given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

Question 1.
In a parallelogram ABCD, write the sum of angles A and B.
Solution:
In ||gm ABCD,
∠A + ∠B = 180°
(Sum of consecutive angles of a ||gm)

Question 2.
In a parallelogram ABCD, if ∠D = 115°, then write the measure of ∠A.
Solution:
In ||gm ABCD,
∠D = 115°
But ∠A + ∠D = 180°
(Sum of consecutive angles of a ||gm)

⇒ ∠A + 115°= 180° ∠A = 180°- 115°
∴ ∠A = 65°

Question 3.
PQRS is a square such that PR and SQ intersect at O. State the measure of ∠POQ.
Solution:
In a square PQRS,
Diagonals PR and QS intersects each other at O.

∵ The diagonals of a square bisect each other at right angles.
∴ ∠POQ = 90°

Question 4.
If PQRS is a square then write the measure of ∠SRP.
Solution:
In square PQRS,

Join PR,
∵Diagonals of a square bisect are opposite angles
∴∠SRP = \(\frac { 1 }{ 2 }\)x ∠SRQ
= \(\frac { 1 }{ 2 }\) x 90° = 45°

Question 5.
If ABCD is a rhombus with ∠ABC = 56°, find the measure of ∠ACD.
Solution:
In rhombus ABCD,
Diagonals bisect each other at 0 at right angles.
∠ABC = 56°
But ∠ABC + ∠BCD = 180° (Sum of consecutive angles)
⇒ 56° + ∠BCD = 180°
⇒ ∠BCD = 180° – 56° = 124°
∵ Diagonals of a rhombus bisect the opposite angle
∴ ∠ACD = \(\frac { 1 }{ 2 }\) ∠BCD = \(\frac { 1 }{ 2 }\) x 124°
= 62°

Question 6.
The perimeter of a parallelogram is 22 cm. If the longer side measures 6.5 cm, what is the measure of the shorter side.
Solution:
Perimeter of a ||gm ABCD = 22cm
∴ Sum of two consecutive sides = \(\frac { 22 }{ 2 }\)
= 11cm
i.e. AB + BC = 11 cm
AB = 6.5 cm and let BC = x cm
∴ 6.5 + x = 11 cm
x = 11 – 6.5 = 4.5
∴ Shorter side = 4.5 cm

Question 7.
If the angles of a quadrilateral are in the ratio 3 : 5 : 9 : 13. Then find the measure of the smallest angle.
Solution:
Ratio in the angles of a quadrilateral = 3 : 5 : 9 : 13
Let first angle = 3x
Second angle = 5x
Third angle = 9x
and fourth angle = 13x
∵ The sum of angles of a quadrilateral = 360°
∴ 3x + 5x + 9x + 13x = 360°
⇒ 30x = 360° ⇒ x = \(\frac { { 360 }^{ \circ } }{ 30 }\)  = 12
∴ Smallest angle = 3x = 3 x 12° = 36°

Question 8.
In parallelogram ABCD if ∠A = (3x – 20°), ∠B = (y + 15)°, ∠C = (x + 40°), then find the value of x and y.
Solution:
In a ||gm ABCD,
∠A = (3x – 20°), ∠B = y + 15°,
∠C = x + 40°
Now, ∠A = ∠C (Opposite angles of a ||gm)
⇒ 3x – 20 = x + 40°
⇒ 3x – x = 40° + 20° ⇒ 2x = 60°
⇒ x = \(\frac { { 60 }^{ \circ } }{ 2 }\)  = 30°
and ∠A + ∠B = 180° (Sum of the consecutive angles)
⇒ 3x-20° + y + 15° = 180°
⇒ 3x + y – 5° = 180°
⇒ 3 x 30° +y- 5° = 180°
⇒ 90° – 5° + y = 180
y = 180° – 90° + 5 = 95°
∴ x = 30°, y = 95°

Question 9.
If measures opposite angles of a parallelogram are (60 – x)° and (3x – 4)°, then find the measures of angles of the parallelogram.
Solution:
Opposite angles of a ||gm ABCD are (60 – x)° and (3x – 4°)
But opposite angles of a ||gm are equal, the
60° – x° = 3x – 4° ⇒ 60° + 4° = 3x + x
⇒ 4x = 64° ⇒ x = \(\frac { { 64 }^{ \circ } }{{ 4 }^{ \circ } }\)  = 16°
∴ ∠A = 60° – x = 60° – 16° = 44°
But ∠A + ∠B = 180° (sum of consecutive angle)
⇒ 44° + ∠B = 180°
⇒ ∠B = 180° – 44°
⇒ ∠B = 136°
But ∠A = ∠C and ∠B = ∠D (Opposite angles)
∴ Angles are 44°, 136°, 44°, 136°

Question 10.
In a parallelogram ABCD, the bisectors of ∠A also bisect BC at x, find AB : AD.
Solution:
In ||gm ABCD,
Bisectors of ∠A meets BC at X and BX = XC
Draw XY ||gm AB meeting AD at Y

Question 11.
In the figure, PQRS in an isosceles trapezium find x and y.
Solution:
∵ PQRS is an isosceles trapezium in which
SP = RQ and SR || PQ
∴ ∠P + ∠S = 180° (Sum of co-interior angles)
3x + 2x = 180° ⇒ 5x = 180°
⇒ x = \(\frac { { 180 }^{ \circ } }{ 5 }\)  = 36°

But ∠P = ∠Qm (Base angles of isosceles trapezium)
y = 2x = 2 x 36° = 12°
∴ y = 12°
Hence x = 36°, y = 12°

Question 12.
In the figure ABCD is a trapezium. Find the values of x and y.

Solution:
In trapezium ABCD,
AB || CD

∴ ∠A + ∠D = 180° (Sum of cointerior angles)
x + 20° + 2x + 10° = 180°
3x + 30° = 180°
⇒ 3x= 180° – 30°
3x = 150°
x = \(\frac { { 150 }^{ \circ } }{ 3 }\)  = 50°
Similarly, ∠B + ∠C = 180°
⇒ y + 92° = 180°
⇒ y = 180° – 92° = 88°
∴ x = 50°, y = 88°

Question 13.
In the figure, ABCD and AEFG are two parallelograms. If ∠C = 58°, find ∠F.

Solution:
In the figure, ABCD and AEFG are two parallelograms ∠C = 58°

∵ DC || GF and CB || FE (Sides of ||gms)
∴ ∠C = ∠F
But ∠C = 58°
∴ ∠F = 58°

Question 14.
Complete each of the following statements by means of one of those given in brackets against each:
(i) If one pair of opposite sides are equal and parallel, then the figure is ……… (parallelogram, rectangle, trapezium)
(ii) If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is …….. (square, rectangle, trapezium)
(iii) A line drawn from the mid-point of one side of a triangle ………. another side intersects the third side at its mid-point, (perpendicular to, parallel to, to meet)
(iv) If one angle of a parallelogram is a right angle, then it is necessarily a …….. (rectangle, square, rhombus)
(v) Consecutive angle of a parallelogram are ……… (supplementary, complementary)
(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a ……… (rectangle, parallelogram, rhombus)
(vii) If opposite angles of a quadrilateral are equal, then it is necessarily a ………. (parallelogram, rhombus, rectangle)
(viii)If consecutive sides of a parallelogram are equal, then it is necessarily a …….. (kite, rhombus, square)
Solution:
(i) If one pair of opposite sides are equal and parallel, then the figure is parallelogram.
(ii) If in a quadrilateral only one pair of opposite sides are parallel, the quadrilateral is trapezium.
(iii) A line drawn from the mid-point of one side of a triangle parallel to another side intersects the third side at its mid-point,
(iv) If one angle of a parallelogram is a right angle, then it is necessarily a rectangle.
(v) Consecutive angle of a parallelogram are supplementary.
(vi) If both pairs of opposite sides of a quadrilateral are equal, then it is necessarily a parallelogram.
(vii) If opposite angles of a quadrilateral are equal, then it is necessarily a parallelogram.
(viii) If consecutive sides of a parallelogram are equal, then it is necessarily a rhombus.

Question 15.
In a quadrilateral ABCD, bisectors of A and B intersect at O such that ∠AOB = 75°, then write the value of ∠C + ∠D.
Solution:
In quadrilateral ABCD,
Bisectors of ∠A and ∠B meet at O and ∠AOB = 75°

In AOB, ∠AOB = 75°
∴ ∠OAB + ∠OBA = 180° – 75° = 105°
But OA and OB are the bisectors of ∠A and ∠B.
∴ ∠A + ∠B = 2 x 105° = 210°
But ∠A + ∠B + ∠C + ∠D = 360° (Sum of angles of a quad.)
∴ 210° + ∠C + ∠D = 360°
⇒ ∠C + ∠D = 360° – 210° = 150°
Hence ∠C + ∠D = 150°

Question 16.
The diagonals of a rectangle ABCD meet at O. If ∠BOC = 44° find ∠OAD.
Solution:
In rectangle ABCD,
Diagonals AC and BD intersect each other at O and ∠BOC = 44°

But ∠AOD = ∠BOC (Vertically opposite angles)
∴ ∠AOD = 44°
In ∆AOD,
∠AOD + ∠OAD + ∠ODA = 180° (Sum of angles of a triangle)
⇒ 44° + ∠OAD + ∠OAD = 180° [∵ OA = OD, ∠OAD = ∠ODA]
⇒ 2∠OAD = 180° – 44° = 136°
∴ ∠OAD = \(\frac { { 136 }^{ \circ } }{ 2 }\)  = 68°

Question 17.
If ABCD is a rectangle with ∠BAC = 32°, find the measure if ∠DBC.
Solution:
In rectangle ABCD,

Diagonals bisect each other at O
∠BAC = 32°
∵ OA = OB
∴ ∠OBA Or ∠DBA = ∠BAC = 32°
But ∠ABC = 90° (Angle of a rectangles)
∴ ∠DBC = ∠ABC – ∠DBA
= 90° – 32° = 58°

Question 18.
If the bisectors of two adjacent angles A and B of a quadrilateral ABCD intersect at a point O. Such that ∠C + ∠D = k(∠AOB), then find the value of k.
Solution:
In quadrilateral ABCD,
Bisectors of ∠A and ∠B meet at O
Such that ∠C + ∠D = k (∠AOB)

Question 19.
In the figure, PQRS is a rhombus in which the diagonal PR is produced to T. If ∠SRT = 152°, find x, y and z.

Solution:
In rhombus PQRS,
Diagonal PR and SQ bisect each other at right angles and PR is produced to T such that ∠SRT = 152°

But ∠SRT + ∠SRP = 180° (Linear pair)
⇒ 152° +∠SRP = 180°
⇒ ∠SRP =180°- 152° = 28°
But ∠SPR = ∠SRP (∵ PR bisects ∠P and ∠R)
⇒ z = 28°
y = 90° (∵ Diagonals bisect each other at right angles)
∠RPQ = z = 28°
∴ In ∆POQ,
z + x = 90° ⇒ 28° + x = 90°
⇒ x = 90° – 28° = 62°
∴ x = 62°, y = 90°, z = 28°

Question 20.
In the figure, ABCD is a rectangle in which diagonal AC is produced to E. If ∠ECD = 146°, find ∠AOB.

Solution:
In rectangle ABCD,
Diagonals AC and BD bisect each other at O
AC is produced to E and ∠DCE = 146°

∠DCE + ∠DCA = 180° (Linear pair)
⇒ 146°+ ∠DCA= 180°
⇒ ∠DCA = 180°- 146°
⇒ ∠DCA = 34°
∴ ∠CAB = ∠DCA (Alternate angles)
= 34°
Now in ∆AOB,
∠AOB = 180° – (∠DAB + ∠OBA)
= 180° – (34° + 34°)
= 1803 – 68° = 112°

Hope given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.1

Other Exercises

• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.1
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS

Question 1.
Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and two parallels: [NCERT]

Solution:
(i) ΔPCD and trapezium ABCD are on the same base CD and between the same parallels AB and DC.
(ii) Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ.
(iii) Parallelogram ABCD and ΔPQR are between the same parallels AD and BC but they are not on the same base.
(iv) ΔQRT and parallelogram PQRS are on the same base QR and between the same parallels QR and PS.
(v) Parallelogram PQRS and trapezium SMNR on tire same base SR but they are not between the same parallels.
(vi) Parallelograms PQRS, AQRD, BCQR are between the same parallels. Also, parallelograms PQRS, BPSC, and APSD are between the same parallels.

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RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3

Other Exercises

• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

Question 1.
In a parallelogram ABCD, determine the sum of angles ZC and ZD.
Solution:
In a ||gm ABCD,
∠C + ∠D = 180°
(Sum of consecutive angles)

Question 2.
In a parallelogram ABCD, if ∠B = 135°, determine the measures of its other angles.
Solution:
In a ||gm ABCD, ∠B = 135°

∴ ∠D = ∠B = 135° (Opposite angles of a ||gm)
But ∠A + ∠B = 180° (Sum of consecutive angles)
⇒ ∠B + 135° = 180°
∴ ∠A = 180° – 135° = 45°
But∠C = ∠B = 45° (Opposite angles of a ||gm)
∴ Angles are 45°, 135°, 45°, 135°.

Question 3.
ABCD is a square, AC and BD intersect at O. State the measure of ∠AOB.
Solution:
In a square ABCD,
Diagonal AC and BD intersect each other at O
∵ Diagonals of a square bisect each other at right angle
∵∠AOB = 90°

Question 4.
ABCD is a rectangle with ∠ABD = 40°. Determine ∠DBC.
Solution:
In rectangle ABCD,

∠B = 90°, BD is its diagonal
But ∠ABD = 40°
and ∠ABD + ∠DBC = 90°
⇒ 40° + ∠DBC = 90°
⇒ ∠DBC = 90° – 40° = 50°
Hence ∠DBC = 50°

Question 5.
The sides AB and CD of a parallelogram ABCD are bisected at E and F. Prove that EBFD is a parallelogram.
Solution:
Given : In ||gm ABCD, E and F are the mid points of the side AB and CD respectively
DE and BF are joined
To prove : EBFD is a ||gm
Construction : Join EF

Proof : ∵ ABCD is a ||gm
∴ AB = CD and AB || CD
(Opposite sides of a ||gm are equal and parallel)
∴ EB || DF and EB = DF (∵ E and F are mid points of AB and CD)
∴ EBFD is a ||gm.

Question 6.
P and Q are the points of trisection of the diagonal BD of the parallelogram ABCD. Prove that CQ is parallel to AP. Prove also that AC bisects PQ.
Solution:
Given : In ||gm, ABCD. P and Q are the points of trisection of the diagonal BD

To prove : (i) CQ || AP
AC bisects PQ
Proof: ∵ Diagonals of a parallelogram bisect each other
∴ AO = OC and BO = OD
∴ P and Q are point of trisection of BD
∴ BP = PQ = QD …(i)
∵ BO = OD and BP = QD …(ii)
Subtracting, (ii) from (i) we get
OB – BP = OD – QD
⇒ OP = OQ
In quadrilateral APCQ,
OA = OC and OP = OQ (proved)
Diagonals AC and PQ bisect each other at O
∴ APCQ is a parallelogram
Hence AP || CQ.

Question 7.
ABCD is a square. E, F, G and H are points on AB, BC, CD and DA respectively, such that AE = BF = CG = DH. Prove that EFGH is a square.
Solution:
Given : In square ABCD
E, F, G and H are the points on AB, BC, CD and DA respectively such that AE = BF = CG = DH
To prove : EFGH is a square
Proof : E, F, G and H are points on the sides AB, BC, CA and DA respectively such that
AE = BF = CG = DH = x (suppose)
Then BE = CF = DG = AH = y (suppose)
Now in ∆AEH and ∆BFE

AE = BF (given)
∠A = ∠B (each 90°)
AH = BE (proved)
∴ ∆AEH ≅ ∆BFE (SAS criterion)
∴ ∠1 = ∠2 and ∠3 = ∠4 (c.p.c.t.)
But ∠1 + ∠3 = 90° and ∠2 + ∠4 = 90° (∠A = ∠B = 90°)
⇒ ∠1 + ∠2 + ∠3 + ∠4 = 90° + 90° = 180°
⇒ ∠1 + ∠4 + ∠1 + ∠4 = 180°
⇒ 2(∠1 + ∠4) = 180°
⇒ ∠1 + ∠4 = \(\frac { { 180 }^{ \circ } }{ 2 }\)  = 90°
∴ ∠HEF = 180° – 90° = 90°
Similarly, we can prove that
∠F = ∠G = ∠H = 90°
Since sides of quad. EFGH is are equal and each angle is of 90°
∴ EFGH is a square.

Question 8.
ABCD is a rhombus, EABF is a straight line such that EA = AB = BF. Prove that ED and FC when produced meet at right angles.
Solution:
Given : ABCD is a rhombus, EABF is a straight line such that
EA = AB = BF
ED and FC are joined
Which meet at G on producing

To prove: ∠EGF = 90°
Proof : ∵ Diagonals of a rhombus bisect
each other at right angles
AO = OC, BO = OD
∠AOD = ∠COD = 90°
and ∠AOB = ∠BOC = 90°
In ∆BDE,
A and O are the mid-points of BE and BD respectively.
∴ AO || ED
Similarly, OC || DG
In ∆ CFA, B and O are the mid-points of AF and AC respectively
∴ OB || CF and OD || GC
Now in quad. DOCG
OC || DG and OD || CG
∴ DOCG is a parallelogram.
∴ ∠DGC = ∠DOC (opposite angles of ||gm)
∴ ∠DGC = 90° (∵ ∠DOC = 90°)
Hence proved.

Question 9.
ABCD is a parallelogram, AD is produced to E so that DE = DC = AD and EC produced meets AB produced in F. Prove that BF = BC.
Solution:
Given : In ||gm ABCD,
AB is produced to E so that
DE = DA and EC produced meets AB produced in F.
To prove : BF = BC
Proof: In ∆ACE,

O and D are the mid points of sides AC and AE
∴ DO || EC and DB || FC
⇒ BD || EF
∴ AB = BF
But AB = DC (Opposite sides of ||gm)
∴ DC = BF
Now in ∆EDC and ∆CBF,
DC = BF (proved)
∠EDC = ∠CBF
(∵∠EDC = ∠DAB corresponding angles)
∠DAB = ∠CBF (corresponding angles)
∠ECD = ∠CFB (corresponding angles)
∴ AEDC ≅ ACBF (ASA criterion)
∴ DE = BC (c.p.c.t.)
⇒ DC = BC
⇒ AB = BC
⇒ BF = BC (∵AB = BF proved)
Hence proved.

Hope given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3 are helpful to complete your math homework.

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RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.1
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.2
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.3
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables Ex 13.4
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables VSAQS
• RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS

Mark the correct alternative in each of the following:
Question 1.
The opposite sides of a quadrilateral have
(a) no common point
(b) one common point
(c) two common points
(d) infinitely many common points
Solution:
The opposite sides of a quadrilateral have no common point. (a)

Question 2.
The consecutive sides of a quadrilateral have
(a) no common point
(b) one common point
(c) two common points
(d) infinitely many common points
Solution:
The consecutive sides of a quadrilateral have one common point. (b)

Question 3.
PQRS is a quadrilateral, PR and QS intersect each other at O. In which of the following cases, PQRS is a parallelogram?
(a) ∠P = 100°, ∠Q = 80°, ∠R = 100°
(b) ∠P = 85°, ∠Q = 85°, ∠R = 95°
(c) PQ = 7 cm, QR = 7 cm, RS = 8 cm, SP = 8 cm
(d) OP = 6.5 cm, OQ = 6.5 cm, OR = 5.2 cm, OS = 5.2 cm
Solution:
PQRS is a quadrilateral, PR and QS intersect each other at O. PQRS is a parallelogram if ∠P = 100°, ∠Q = 80°, ∠R = 100° (a)

Question 4.
Which of the following quadrilateral is not a rhombus?
(a) All four sides are equal
(b) Diagonals bisect each other
(c) Diagonals bisect opposite angles
(d) One angle between the diagonals is 60°
Solution:
A quadrilateral is not a rhombus if one angle between the diagonals is 60°. (d)

Question 5.
Diagonals necessarily bisect opposite angles in a
(a) rectangle
(b) parallelogram
(c) isosceles trapezium
(d) square
Solution:
Diagonals necessarily bisect opposite angles in a square. (d)

Question 6.
The two diagonals are equal in a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) trapezium
Solution:
The two diagonals are equal in a rectangle. (c)

Question 7.
We get a rhombus by joining the mid-points of the sides of a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) triangle
Solution:
We get a rhombus by joining the mid points of the sides of a rectangle. (c)

Question 8.
The bisectors of any two adjacent angles of a parallelogram intersect at
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Solution:
The bisectors of any two adjacent angles of a parallelogram intersect at 90°. (d)

Question 9.
The bisectors of the angle of a parallelogram enclose a
(a) parallelogram
(b) rhombus
(c) rectangle
(d) square
Solution:
The bisectors of the angles of a parallelogram enclose a rectangle. (c)

Question 10.
The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a
(a) parallelogram
(b) rectangle
(c) square
(d) rhombus
Solution:
The figure formed by joining the mid-points of the adjacent sides of a quadrilateral is a parallelogram. (a)

Question 11.
The figure formed by joining the mid-points of the adjacent sides of a rectangle is a
(a) square
(b) rhombus
(c) trapezium
(d) none of these
Solution:
The figure formed by joining the mid-points of the adjacent sides of a rectangle is a rhombus. (b)

Question 12.
The figure formed by joining the mid-points of the adjacent sides of a rhombus is a
(a) square
(b) rectangle
(c) trapezium
(d) none of these
Solution:
The figure formed by the joining the mid-points of the adjacent sides of a rhombus is a rectangle. (b)

Question 13.
The figure formed by joining the mid-points of the adjacent sides of a square is a
(a) rhombus
(b) square
(c) rectangle
(d) parallelogram
Solution:
Tire figure formed by joining the mid-points of the adjacent sides of a square is a square. (b)

Question 14.
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a
(a) rectangle
(b) parallelogram
(b) rhombus
(d) square
Solution:
The figure formed by joining the mid-points of the adjacent sides of a parallelogram is a parallelogram. (b)

Question 15.
If one angle of a parallelogram is 24° less than twice the smallest angle, then the measure of the largest angle of the parallelogram is
(a) 176°
(b) 68°
(c) 112°
(d) 102°
Solution:
Let the smallest angle be x
The largest angle = 2x – 24°
But sum of two adjacent angles = 180°

Question 16.
In a parallelogram ABCD, If ∠DAB = 75° and ∠DBC = 60°, then ∠BDC =
(a) 75°
(b) 60°
(c) 45°
(d) 55°
Solution:
In ||gm ABC,

∠A = 75°, ∠DBC = 60°
But ∠A + ∠B = 180° (Sum of two consecutive angles)
⇒ 75° + ∠B = 180°
⇒ ∠B = 180°- 75“= 105°
But ∠DBC = 60°
∴ ∠DBA = 105°-60° = 45°
But ∠BDC = ∠DBA (Alternate angles)
∴ ∠BDC = 45° (c)

Question 17.
ABCD is a parallelogram and E and F are the centroids of triangles ABD and BCD respectively, then EF =
(a) AE
(b) BE
(c) CE
(d) DE
Solution:
In ||gm ABCD, BD is joined forming two triangles ABD and BCD

E and F are the centroid of ∆ABD and ∆BCD
Now E and F trisect AC
i.e. AE = EF = FC
∴ EF = AE (a)

Question 18.
ABCD is a parallelogram, M is the mid¬point of BD and BM bisects ∠B. Then, ∠AMB =
(a) 45°
(b) 60°
(c) 90°
(d) 75°
Solution:
In ||gm ABCD, M is mid-point of BD and
BM bisects ∠B
AM is joined

∴AM bisects ∠A
But ∠A + ∠B = 180° (Sum of two consecutive angles)
∴ ∠AMB = 90° (c)

Question 19.
If an angle of a parallelogram is two-third of its adjacent angle, the smallest angle of the parallelogram is
(a) 108°
(b) 54°
(c) 12°
(d) 81°
Solution:
Let adjacent angle of a ||gm = x
Then second angle = \(\frac { 2 }{ 3 }\) x
∴ x+ \(\frac { 2 }{ 3 }\) x= 180°
(Sum of two adjacent angles of a ||gm is 180°)

Question 20.
If the degree measures of the angles of quadrilateral are Ax, lx, 9x and 10JC, what is the sum of the measures of the smallest angle and largest angle?
(a) 140°
(b) 150°
(c) 168°
(d) 180°
Solution:
Sum of the angles of a quadrilateral = 360°
∴ 4x + 1x + 9x + 10x = 360°
⇒ 30x = 360°
⇒ x = \(\frac { { 360 }^{ \circ } }{ 30 }\)  = 12°
Now sum of smallest and largest angle = 4 x 12° + 10 x 12°
= 48° + 120° = 168° (c)

Question 21.
If the diagonals of a rhombus are 18 cm and 24 cm respectively, then its side is equal to
(a) 16 cm
(b) 15 cm
(c) 20 cm
(d) 17 cm
Solution:
Diagonals of a rhombus are 18 cm and 24 cm But diagonals of a rhombus bisect each other at right angles

Question 22.
ABCD is a parallelogram in which diagonal AC bisects ∠BAD. If ∠BAC = 35°, then ∠ABC =
(a) 70°
(b) 110°
(c) 90°
(d) 120°
Solution:
In ||gm ABCD, AC is its diagonal which bisect ∠BAD

∠BAD = 35°
∴ ∠BAD = 2 x 35° = 70°
But ∠A + ∠B = 180° (Sum of consecutive angles)
⇒ 70° + ∠B = 180°⇒ ∠B = 180° – 70°
∴ ∠B = 110°
⇒ ABC = 110° (b)

Question 23.
In a rhombus ABCD, if ∠ACB = 40°, then ∠ADB =
(a) 70°
(b) 45°
(c) 50°
(d) 60°
Solution:
In rhombus ABCD, ∠ACB = 40°

∴ ∠BCD = 2 x ∠ACB
= 2 x 40° = 80°
But ∠BCD + ∠ADC = 180° (Sum of consecutive angles of ||gm)
⇒ 80° + ∠ADC = 180°
⇒ ∠ADC = 180° – 80° = 100°
∴ ∠ADB = \(\frac { 1 }{ 2 }\)∠ADC = \(\frac { 1 }{ 2 }\)x 100° = 50° (c)

Question 24.
In ∆ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°. The angles of the triangle formed by joining the mid-points of the sides of this triangle are
(a) 70°, 70°, 40°
(b) 60°, 40°, 80°
(c) 30°, 40°, 110°
(d) 60°, 70°, 50°
Solution:
In ∆ABC,

∠A = 30°, ∠B = 40°, ∠C = 110°
D, E and F are mid-points of the sides of the triangle. By joining them in order,
DEF is a triangle formed
Now BDEF, CDFE and AFDE are ||gms
∴ ∠A = ∠D = 30°
∠B = ∠E = 40°
∠C = ∠F= 110°
∴ Angles are 30°, 40°, 110° (c)

Question 25.
The diagonals of a parallelogram ABCD intersect at O. If ∠BOC = 90° and ∠BDC = 50°, then ∠OAB =
(a) 40°
(b) 50°
(c) 10°
(d) 90°
Solution:
In ||gm ABCD, diagonals AC and BD intersect each other at O

BOC = 90°, ∠BDC = 50°
∵ ∠BOC = 90°
∴ Diagonals of ||gm bisect each other at 90°
∴∠COD = 90°
In ∆COD,
∠OCD = 90° – 50° = 40°
But ∠OAB = ∠OCD (Alternate angles)
∴∠OAB = 40° (a)

Question 26.
ABCD is a trapezium in which AB || DC. M and N are the mid-points of AD and BC respectively. If AB = 12 cm, MN = 14 cm, then CD =
(a) 10 cm
(b) 12 cm
(c) 14 cm
(d) 16 cm
Solution:
In trapezium AB || DC
M and N are mid-points of sides AD and BC and MN are joined
AB = 12 cm, MN = 14 cm

∵ MN = \(\frac { 1 }{ 2 }\)(AB + CD)
⇒ 2MN = AB + CD
⇒ 2 x 14 = 12 + CD
CD = 2 x 14 – 12 = 28 – 12 = 16 cm (d)

Question 27.
Diagonals of a quadrilateral ABCD bisect each other. If ∠A = 45°, then ∠B =
(a) 115°
(b) 120°
(c) 125°
(d) 135°
Solution:
Diagonals AC and BD of quadrilateral ABCD bisect each other at O

∴ AO = OC, BO = OD
∴ ABCD is a ||gm ∠A = 45°
But ∠A + ∠B = 180° (Sum of consecutive angles)
∴ ∠B = 180° – ∠A = 180° – 45°
= 135° (d)

Question 28.
P is the mid-point of side BC of a paralleogram ABCD such that ∠BAP = ∠DAP. If AD = 10 cm, then CD =
(a) 5 cm
(b) 6 cm
(c) 8 cm
(d) 10 cm
Solution:
In ||gm ABCD, P is mid-point of BC
AD = 10cm

∠BAP = ∠DAP
Through P, draw PQ || AB
∴ ABPQ is rhombus
∴ AB = BP = AQ
= \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) x 10 = 5 cm
But CD = AB (Opposite sides of ||gm)
∴ CD = 5 cm (a)

Question 29.
In ∆ABC, E is the mid-point of median AD such that BE produced meets AC at E If AC = 10.5 cm, then AF =
(a) 3 cm
(b) 3.5 cm
(c) 2.5 cm
(d) 5 cm
Solution:
In ∆ABC, E is the mid-point of median AD
Such that BE produced meets AC at F
AC = 10.5 cm
Draw DG || AF

In ∆ADG
E is mid-point of AD and EF || DG
∴ F is mid-point of AG
⇒ AF = FG …(i)
In ∆BCF
D is mid-point of BC and DG || BF
∴ G is mid-point of FC
∴ FG = GC …(i)
From (i) and (ii)
AF = FG = GC = \(\frac { 1 }{ 3 }\) AC
But AC = 10.5 cm
∴ AF = \(\frac { 1 }{ 3 }\) AC = \(\frac { 1 }{ 3 }\) x 10.5 = 3.5 cm (b)

Question 30.
ABCD is a parallelogram and E is the mid-point of BC. DE and AB when produced meet at F. Then, AF =

Solution:
In ||gm ABCD, E is mid-point of BC DE and AB are produced to meet at F

∵ E is mid point of BC
∴ BE = EC
In ∆BEF and ∆CDE
BE = EC
∠BEF = ∠CED (Vertically opposite angle)
and ∠EBF = ∠ECD (Alternate angles)
∴ ∆BEF ≅ ∆CDE (ASA criterian)
∴ DC = BF
But DC = AB
∴ AB = BF
AF = AB + BF = AB + AB
= 2AB (b)

Question 31.
In a quadrilateral ABCD, ∠A + ∠C is 2 times ∠B + ∠D. If ∠A = 140° and ∠D = 60°, then ∠B =
(a) 60°
(b) 80°
(c) 120°
(d) None of these
Solution:
In quadrilateral ABCD
⇒ ∠A + ∠C = 2(∠B + ∠D)
⇒ ∠A + ∠C = 2∠B + 2∠D
Adding 2∠A + 2∠C both sides
2∠A + 2∠C + ∠A + ∠C = 2∠A + 2∠B + 2∠C + 2∠D
⇒ 3∠A + 3∠C = 2(∠A + ∠B + ∠C + ∠D)
⇒ 3(∠A + ∠C) = 2 x 360° = 720°
∴ ∠A + ∠C = \(\frac { { 720 }^{ \circ } }{ 3 }\)  = 240°
⇒ 40° + ∠C = 240° (∵ ∠A = 40°)
∠C = 240° – 40° = 200°
Now 2(∠B + ∠D) = ∠A + ∠C = 240°
∠B + ∠D = \(\frac { { 240 }^{ \circ } }{ 2 }\)  = 120°
∴ ∠B = 60° = 120°
∴ ∠B = 60° (a)

Question 32.
The diagonals AC and BD of a rectangle ABCD intersect each other at P. If ∠ABD = 50°, then ∠DPC =
(a) 70°
(b) 90°
(c) 80°
(d) 100°
Solution:
In rectangle ABCD, diagonals AC and BD intersect each other at P

∠ABD = 50°
∴ ∠CAB = ∠ABD = 50° (∵ AP = BP)
Now in ∆APB
∠CAB + ∠ABD + ∠APB = 180° (Angles of a triangle)
⇒ ∠PAB + ∠PBA + ∠APB = 180°
⇒ 50° + 50° + ∠APB = 180°
⇒ ∠APB = 180° – 50° – 50° = 80°
But ∠DPC = ADB (Vertically opposite angles)
∴ ∠DPC = 80° (c)

Hope given RD Sharma Class 9 Solutions Chapter 13 Linear Equations in Two Variables MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS

Other Exercises

• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.1
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.2
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals Ex 14.3
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS
• RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals MCQS

Question 1.
If ABC and BDE are two equilateral triangles such that D is the mid-ponit of BC, then find ar(∆ABC) : ar(∆BDE).
Solution:
ABC and BDE are two equilateral triangles and D is the mid-point of BC

Let each side of AABC = a
Then BD = \(\frac { a }{ 2 }\)
∴ Each side of triangle BDE will be \(\frac { a }{ 2 }\)

Question 2.
In the figure, ABCD is a rectangle in which CD = 6 cm, AD = 8 cm. Find the area of parallelogram CDEF.

Solution:
In rectangle ABCD,
CD = 6 cm, AD = 8 cm
∴ Area of rectangle ABCD = CD x AD
= 6 x 8 = 48 cm²
∵ DC || AB and AB is produced to F and DC is produced to G
∴ DG || AF
∵ Rectangle ABCD and ||gm CDEF are on the same base CD and between the same parallels
∴ ar(||gm CDEF) = ar(rect. ABCD)
= 48 cm²

Question 3.
In the figure of Q. No. 2, find the area of ∆GEF.

Solution:

Question 4.
In the figure, ABCD is a rectangle with sides AB = 10 cm and AD = 5 cm. Find the area of ∆EFG.

Solution:
ABCD is a rectangle in which
AB = 10 cm, AD = 5 cm

∵ ABCD is a rectangle
∴DC || AB,
DC is produced to E and AB is produced to G
∴DE || AG
∵ Rectangle ABCD and ||gm ABEF are on the same base AB and between the same parallels
∴ ar(rect. ABCD) = ar(||gm ABEF)
= AB x AD = 10 x 5 = 50 cm²
Now ||gm ABEF and AEFG are on the same
base EF and between the same parallels
∴ area ∆EFG = \(\frac { 1 }{ 2 }\) ar(||gm ABEF)
= \(\frac { 1 }{ 2 }\) x 50 = 25 cm²

Question 5.
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then find or(∆RAS).
Solution:
In quadrant PLRM, rectangle PQRS is in scribed

Radius of the circle = 13 cm
A is any point on PQ
AR and AS are joined, PS = 5 cm
In right ∆PRS,
PR² = PS² + SR²
⇒ (13² = (5)²+ SR²
⇒ 169 = 25 + SR²
⇒ SR² = 169 – 25 = 144 = (12)²
∴ SR = 12 cm
Area of rect. PQRS = PS x SR = 5x 12 = 60 cm²
∵ Rectangle PQRS and ARAS are on the same
base SR and between the same parallels
∴ Area ARAS = \(\frac { 1 }{ 2 }\) area rect. PQRS 1
= \(\frac { 1 }{ 2 }\) x 60 = 30 cm²

Question 6.
In square ABCD, P and Q are mid-point of AB and CD respectively. If AB = 8 cm and PQ and BD intersect at O, then find area of ∆OPB.
Solution:
In sq. ABCD, P and Q are the mid points of sides AB and CD respectively PQ and BD are joined which intersect each other at O
Side of square AB = 8 cm

∴ Area of square ABCD = (side)²
∵ Diagonal BD bisects the square into two triangle equal in area
∴ Area ∆ABD = \(\frac { 1 }{ 2 }\) x area of square ABCD
= \(\frac { 1 }{ 2 }\) x 64 = 32 cm²
∵ P is mid point of AB of AABD, and PQ || AD
∴ O is the mid point of BD
∴ OP = \(\frac { 1 }{ 2 }\)AD = \(\frac { 1 }{ 2 }\) x 8 = 4 cm
and PB = \(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\) x 8 = 4 cm
∴ Area ∆OPB = \(\frac { 1 }{ 2 }\)PB x OP
= \(\frac { 1 }{ 2 }\) x4x4 = 8 cm²

Question 7.
ABC is a triangle in which D is the mid-point of BC. E and F are mid-points of DC and AE respectively. If area of ∆ABC is 16 cm², find the area of ∆DEF.
Solution:
In ∆ABC, D is mid point of BC. E and F are the mid points of DC and AE respectively area of ∆ABC = 16 cm²
FD is joined

∵ D is mid point of BC
∴ AD is the median and median divides the triangle into two triangles equal in area
area ∆ADC = \(\frac { 1 }{ 2 }\) ar(∆ABC)
= \(\frac { 1 }{ 2 }\) x 16 = 8 cm²
Similarly, E is mid point of DC
∴ area (∆ADE) = \(\frac { 1 }{ 2 }\) ar(∆ADC)
= \(\frac { 1 }{ 2 }\) x 8 = 4 cm²
∵ F is mid point of AE of ∆ADE
∴ ar(∆DEF) = \(\frac { 1 }{ 2 }\)area (∆ADE)
= \(\frac { 1 }{ 2 }\) x 4 = 2 cm²

Question 8.
PQRS is a trapezium having PS and QR as parallel sides. A is any point on PQ and B is a point on SR such that AB || QR. If area of ∆PBQ is 17 cm2, find the area of ∆ASR.
Solution:
In trapezium PQRS,
PS || QR
A and B are points on sides PQ and SR
Such that AB || QR
area of ∆PBQ = 17 cm²

∆ABQ and ∆ABR are on the same base AB and between the same parallels
∴ ar(∆ABQ) = ar(∆ABR) …(i)
Similarly, ∆ABP and ∆ABS are on the same base and between the same parallels
∴ ar(ABP) = ar(∆ABS) …(ii)
Adding (i) and (ii)
ar( ∆ABQ) + ar( ∆ABP)
= ar(∆ABR) + ar(∆ABS)
⇒ ar(∆PBQ) = ar(∆ASR)
Put ar(PBQ) = 17 cm²
∴ ar(∆ASR) = 17 cm²

Question 9.
ABCD is a parallelogram. P is the mid-point of AB. BD and CP intersect at Q such that CQ : QP = 3 : 1. If ar(∆PBQ) = 10 cm2, find the area of parallelogram ABCD.
Solution:
In ||gm ABCD, P is mid point on AB,
PC and BD intersect each other at Q

CQ : QP = 3 : 1
ar(∆PBQ) = 10 cm²
In ||gm ABCD,
BD is its diagonal
∴ ar(∆ABD) = ar(∆BCD) = \(\frac { 1 }{ 2 }\) ar ||gm ABCD
∴ ar(||gm ABCD) = 2ar(∆ABD) …(i)
In ∆PBC CQ : QP = 3 : 1
∵ ∆PBQ and ∆CQB have same vertice B
∴ 3 x area ∆PBQ = ar(∆CBQ)
⇒ area(∆CBQ) = 3 x 10 = 30 cm²
∴ ar(∆PBC) = 30 + 10 = 40 cm²
Now ∆ABD and ∆PBC are between the
same parallel but base PB = \(\frac { 1 }{ 2 }\) AB
∴ ar(∆ABD) = 2ar(∆PBC)
= 2 x 40 = 80 cm²
But ar(||gm ABCD) = 2ar(∆ABD)
= 2 x 80 = 160 cm²

Question 10.
P is any point on base BC of ∆ABC and D is the mid-point of BC. DE is drawn parallel to PA to meet AC at E. If ar(∆ABC) = 12 cm², then find area of ∆EPC.
Solution:
P is any point on base of ∆ABC
D is mid point of BC
DE || PA drawn which meet AC at E
ar(∆ABC) = 12 cm²
AD and PE are joined

∵ D is mid point of BC
∴ AD is median
∴ ar(∆ABD) = ar(∆ACD)
= \(\frac { 1 }{ 2 }\) (∆ABC) = \(\frac { 1 }{ 2 }\) x 12 = 6 cm² …(i)
∵ ∆PED and ∆ADE are on the same base DE and between the same parallels
∴ ar(∆PED) = ar(∆ADE)
Adding ar(∆DCE) to both sides,
ar(∆PED) + ar(∆DCE) = ar(∆ADE) + ar(∆DCE)
ar(∆EPC) = ar(∆ACD)
⇒ ar(∆EPC) = ar(∆ABD) = 6 cm2 [From (i)]
∴ ar(∆EPC) = 6 cm²

Hope given RD Sharma Class 9 Solutions Chapter 14 Quadrilaterals VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.