RD Sharma Class 9 Solutions Maths Chapter 15:
All of the practice problems are addressed in RD Sharma Solutions for Class 9 Maths Chapter 15 Areas of Parallelograms and Triangles. Students will learn about Triangles and Parallelograms in this chapter.A quadrilateral having two pairs of parallel sides that do not self-intersect is called a parallelogram in geometry. A trapezium is a quadrilateral with a single set of parallel sides.
A triangle is a polygon with three vertices and three edges, whereas the diagonals in a parallelogram bisect one another. There are three different types of triangles: scalene, equilateral, and isosceles. A parallelogram is made up of two congruent triangles, each of which is divided by an equal-length diagonal.
RD Sharma Class 9 Solutions Maths Chapter 15 PDF
Expert teachers have created all of the triangle and parallelogram questions in this chapter in accordance with the CBSE syllabus. To achieve high test scores, students are recommended to download the RD Sharma Class 9 Solutions PDF and begin practicing.
The RD Sharma Solutions for the exercise-specific challenges are developed following extensive investigation into every subject. Students are assisted in solving difficulties competently by the solutions' well-organized format.
RD Sharma Class 9 Solutions Maths Chapter 15 PDF
RD Sharma Class 9 Solutions Maths Chapter 15
Exercise 15.1
Question 1: Which of the following figures lie on the same base and between the same parallel. In such a case, write the common base and two parallels:
Solution:
(i)
Triangle APB and trapezium ABCD are on the common base AB and between the same parallels AB and DC.
So,
Common base =
AB
Parallel lines:
AB and DC
(ii)
Parallelograms ABCD and APQD are on the same base AD and between the same parallels AD and BQ.
Common base =
AD
Parallel lines:
AD and BQ
(iii)
Consider, parallelogram ABCD and ΔPQR, lies between the same parallels AD and BC. But not sharing common base.
(iv)
ΔQRT and parallelogram PQRS are on the same base QR and lies between same parallels QR and PS.
Common base =
QR
Parallel lines:
QR and PS
(v)
Parallelograms PQRS and trapezium SMNR share common base SR, but not between the same parallels.
(vi)
Parallelograms: PQRS, AQRD, BCQR are between the same parallels. Also,
Parallelograms: PQRS, BPSC, APSD are between the same parallels.
RD Sharma Class 9 Solutions Maths Chapter 13
RD Sharma Class 9 Solutions Maths Chapter 15 Exercise 15.2
Question 1: If figure, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8 cm and CF = 10 cm, find AD.
Solution
:
In parallelogram ABCD, AB = 16 cm, AE = 8 cm and CF = 10 cm
Since, opposite sides of a parallelogram are equal, then
AB = CD = 16 cm
We know, Area of parallelogram = Base x Corresponding height
Area of parallelogram ABCD:
CD x AE = AD x CF
16 x 18 = AD x 10
AD = 12.8
Measure of AD = 12.8 cm
Question 2: In Q.No. 1, if AD = 6 cm, CF = 10 cm and AE = 8 cm, find AB.
Solution
: Area of a parallelogram ABCD:
From figure:
AD × CF = CD × AE
6 x 10 = CD x 8
CD = 7.5
Since, opposite sides of a parallelogram are equal.
=> AB = DC = 7.5 cm
Question 3: Let ABCD be a parallelogram of area 124 cm
2
. If E and F are the mid-points of sides AB and CD respectively, then find the area of parallelogram AEFD.
Solution:
ABCD be a parallelogram.
Area of parallelogram = 124 cm
2
(Given)
Consider a point P and join AP which is perpendicular to DC.
Now, Area of parallelogram EBCF = FC x AP and
Area of parallelogram AFED = DF x AP
Since F is the mid-point of DC, so DF = FC
From above results, we have
Area of parallelogram AEFD = Area of parallelogram EBCF = 1/2 (Area of parallelogram ABCD)
= 124/2
= 62
Area of parallelogram AEFD is 62 cm
2
.
Question 4: If ABCD is a parallelogram, then prove that
ar(Δ ABD) = ar(Δ BCD) = ar(Δ ABC)=ar(Δ ACD) = 1/2 ar(||
gm
ABCD)
Solution:
ABCD is a parallelogram.
When we join the diagonal of parallelogram, it divides it into two quadrilaterals.
Step 1: Let AC is the diagonal, then, Area (ΔABC) = Area (ΔACD) = 1/2(Area of ll
gm
ABCD)
Step 2: Let BD be another diagonal
Area (ΔABD) = Area (ΔBCD) = 1/2( Area of ll
gm
ABCD)
Now,
From Step 1 and step 2, we have
Area (ΔABC) = Area (ΔACD) = Area (ΔABD) = Area (ΔBCD) = 1/2(Area of ll
gm
ABCD)
Hence Proved.
RD Sharma Class 9 Solutions Maths Chapter 15 Exercise 15.3
Question 1: In figure, compute the area of quadrilateral ABCD.
Solution:
A quadrilateral ABCD with DC = 17 cm, AD = 9 cm and BC = 8 cm (Given)
In right ΔABD,
Using Pythagorean Theorem,
AB
2
+ AD
2
= BD
2
15
2
= AB
2
+ 9
2
AB
2
= 225−81=144
AB = 12
Area of ΔABD = 1/2(12×9) cm
2
= 54 cm
2
In right ΔBCD:
Using Pythagorean Theorem,
CD
2
= BD
2
+ BC
2
17
2
= BD
2
+ 8
2
BD
2
= 289 – 64 = 225
or BD = 15
Area of ΔBCD = 1/2(8×17) cm
2
= 68 cm
2
Now, area of quadrilateral ABCD = Area of ΔABD + Area of ΔBCD
= 54 cm
2
+ 68 cm
2
= 112 cm
2
Question 2: In figure, PQRS is a square and T and U are, respectively, the mid-points of PS and QR . Find the area of ΔOTS if PQ = 8 cm.
Solution:
T and U are mid points of PS and QR respectively (Given)
Therefore, TU||PQ => TO||PQ
In ΔPQS ,
T is the mid-point of PS and TO||PQ
So, TO = 1/2 PQ = 4 cm
(PQ = 8 cm given)
Also, TS = 1/2 PS = 4 cm
[PQ = PS, As PQRS is a square)
Now,
Area of ΔOTS = 1/2(TO×TS)
= 1/2(4×4) cm
2
= 8cm
2
Area of ΔOTS is 8 cm
2
.
Question 3: Compute the area of trapezium PQRS in figure.
Solution:
From figure,
Area of trapezium PQRS = Area of rectangle PSRT + Area of ΔQRT
= PT × RT + 1/2 (QT×RT)
= 8 × RT + 1/2(8×RT)
= 12 RT
In right ΔQRT,
Using Pythagorean Theorem,
QR
2
= QT
2
+ RT
2
RT
2
= QR
2
− QT
2
RT
2
= 17
2
−8
2
= 225
or RT = 15
Therefore, Area of trapezium = 12×15 cm
2
= 180 cm
2
Question 4: In figure, ∠AOB = 90
o
, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of ΔAOB.
Solution:
Given: A triangle AOB, with ∠AOB = 90
o
, AC = BC, OA = 12 cm and OC = 6.5 cm
As we know, the midpoint of the hypotenuse of a right triangle is equidistant from the vertices.
So, CB = CA = OC = 6.5 cm
AB = 2 CB = 2 x 6.5 cm = 13 cm
In right ΔOAB:
Using Pythagorean Theorem, we get
AB
2
= OB
2
+ OA
2
13
2
= OB
2
+ 12
2
OB
2
= 169 – 144 = 25
or OB = 5 cm
Now, Area of ΔAOB = ½(Base x height) cm
2
= 1/2(12 x 5) cm
2
= 30cm
2
Question 5: In figure, ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm, and distance between AB and DC is 4 cm. Find the value of x and area of trapezium ABCD.
Solution:
Given: ABCD is a trapezium, where AB = 7 cm, AD = BC = 5 cm, DC = x cm, and
Distance between AB and DC = 4 cm
Consider AL and BM are perpendiculars on DC, then
AL = BM = 4 cm and LM = 7 cm.
In right ΔBMC :
Using Pythagorean Theorem, we get
BC
2
= BM
2
+ MC
2
25 = 16 + MC
2
MC
2
= 25 – 16 = 9
or MC = 3
Again,
In right Δ ADL :
Using Pythagorean Theorem, we get
AD
2
= AL
2
+ DL
2
25 = 16 + DL
2
DL
2
= 25 – 16 = 9
or DL = 3
Therefore, x = DC = DL + LM + MC = 3 + 7 + 3 = 13
=> x = 13 cm
Now,
Area of trapezium ABCD = 1/2(AB + CD) AL
= 1/2(7+13)4
= 40
Area of trapezium ABCD is 40 cm
2
.
Question 6: In figure, OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If OE = 2√5 cm, find the area of the rectangle.
Solution
:
From given:
Radius = OD = 10 cm and OE = 2√5 cm
In right ΔDEO,
By Pythagoras theorem
OD
2
= OE
2
+ DE
2
(10)
2
= (2√5 )
2
+ DE
2
100 – 20 = DE
2
DE = 4√5
Now,
Area of rectangle OCDE = Length x Breadth = OE x DE = 2√5 x 4√5 = 40
Area of rectangle is 40 cm
2
.
Question 7: In figure, ABCD is a trapezium in which AB || DC. Prove that ar(ΔAOD) = ar(ΔBOC)
Solution:
ABCD is a trapezium in which AB || DC (Given)
To Prove: ar(ΔAOD) = ar(ΔBOC)
Proof:
From figure, we can observe that ΔADC and ΔBDC are sharing common base i.e. DC and between same parallels AB and DC.
Then, ar(ΔADC) = ar(ΔBDC) ……(1)
ΔADC is the combination of triangles, ΔAOD and ΔDOC. Similarly, ΔBDC is the combination of ΔBOC and ΔDOC triangles.
Equation (1) => ar(ΔAOD) + ar(ΔDOC) = ar(ΔBOC) + ar(ΔDOC)
or
ar(ΔAOD) = ar(ΔBOC)
Hence Proved.
Question 8: In figure, ABCD, ABFE and CDEF are parallelograms. Prove that ar(ΔADE) = ar(ΔBCF).
Solution:
Here, ABCD, CDEF and ABFE are parallelograms:
Which implies:
AD = BC
DE = CF and
AE = BF
Again, from triangles ADE and BCF:
AD = BC, DE = CF and AE = BF
By SSS criterion of congruence, we have
ΔADE ≅ ΔBCF
Since both the triangles are congruent, then ar(ΔADE) = ar(ΔBCF).
Hence Proved,
Question 9: Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that: ar(ΔAPB) x ar(ΔCPD) = ar(ΔAPD) x ar(ΔBPC).
Solution:
Consider: BQ and DR are two perpendiculars on AC.
To prove: ar(ΔAPB) x ar(ΔCPD) = ar(ΔAPD) x ar(ΔBPC).
Now,
L.H.S. = ar(ΔAPB) x ar(ΔCDP)
= (1/2 x AP × BQ) × (1/2 × PC × DR)
= (1/2 x PC × BQ) × (1/2 × AP × DR)
=
ar(ΔAPD) x ar(ΔBPC)
= R.H.S.
Hence proved.
Question 10: In figure, ABC and ABD are two triangles on the base AB. If line segment CD is bisected by AB at O, show that ar(ΔABC) = ar(ΔABD).
Solution:
Draw two perpendiculars CP and DQ on AB.
Now,
ar(ΔABC) = 1/2×AB×CP ⋅⋅⋅⋅⋅⋅⋅(1)
ar(ΔABD) = 1/2×AB×DQ ⋅⋅⋅⋅⋅⋅⋅(2)
To prove the result, ar(ΔABC) = ar(ΔABD), we have to show that CP = DQ.
In right angled triangles, ΔCPO and ΔDQO
∠CPO = ∠DQO = 90
o
CO = OD (Given)
∠COP = ∠DOQ (Vertically opposite angles)
By AAS condition: ΔCP0 ≅ ΔDQO
So, CP = DQ …………..(3)
(By CPCT)
From equations (1), (2) and (3), we have
ar(ΔABC) = ar(ΔABD)
Hence proved.
