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# RD Sharma solutions for Class 9 Mathematics chapter 15 - Areas of Parallelograms and Triangles

## Chapter 15 : Areas of Parallelograms and Triangles

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Which of the following figures lie on the same base and between the same parallels. In such a case, write the common base and the two parallels.

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In fig below, ABCD is a parallelogram, AE ⊥ DC and CF ⊥ AD. If AB = 16 cm, AE = 8
cm and CF = 10 cm, find AD.

In Q. No 1, if AD = 6 cm, CF = 10 cm, and AE = 8cm, find AB.

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.

If ABCD is a parallelogram, then prove that
ππ (Δπ΄π΅π·) = ππ (Δπ΅πΆπ·) = ππ (Δπ΄π΅πΆ) = ππ (Δπ΄πΆπ·) = 1/2 ππ (||ππ π΄π΅πΆπ·) .

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In the below figure, compute the area of quadrilateral ABCD.

In the below 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.

Compute the area of trapezium PQRS is Fig. below.

In the below fig. ∠AOB = 90°, AC = BC, OA = 12 cm and OC = 6.5 cm. Find the area of
ΔAOB.

In the below fig. ABCD is a trapezium in which AB = 7 cm, AD = BC = 5 cm, DC = x cm,
and distance between AB and DC is 4cm. Find the value of x and area of trapezium ABCD.

In the below fig. OCDE is a rectangle inscribed in a quadrant of a circle of radius 10 cm. If
OE = 2√5, find the area of the rectangle.

In the below fig. ABCD is a trapezium in which AB || DC. Prove that ar (ΔAOD) =
ar(ΔBOC).

In the given below fig. ABCD, ABFE and CDEF are parallelograms. Prove that ar (ΔADE)
= ar (ΔBCF)

Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that:
ar(ΔAPB) × ar(ΔCPD) = ar(ΔAPD) × ar (ΔBPC)

In the below Fig, ABC and ABD are two triangles on the base AB. If line segment CD is
bisected by AB at O, show that ar (Δ ABC) = ar (Δ ABD)

If P is any point in the interior of a parallelogram ABCD, then prove that area of the
triangle APB is less than half the area of parallelogram.

If AD is a median of a triangle ABC, then prove that triangles ADB and ADC are equal in
area. If G is the mid-point of median AD, prove that ar (Δ BGC) = 2 ar (Δ AGC).

A point D is taken on the side BC of a ΔABC such that BD = 2DC. Prove that ar(Δ ABD) =

ABCD is a parallelogram whose diagonals intersect at O. If P is any point on BO, prove
that:  (1) ar (ΔADO) = ar (ΔCDO)     (2) ar (ΔABP) = ar (ΔCBP)

ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects
CD at F.
(i) Prove that ar (ΔADF) = ar (ΔECF)
(ii) If the area of ΔDFB = 3 cm2, find the area of ||gm ABCD.

ABCD is a parallelogram whose diagonals AC and BD intersect at O. A line through O
intersects AB at P and DC at Q. Prove that ar (Δ POA) = ar (Δ QOC).

ABCD is a parallelogram. E is a point on BA such that BE = 2 EA and F is a point on DC
such that DF = 2 FC. Prove that AE CF is a parallelogram whose area is one third of the
area of parallelogram ABCD.

In a ΔABC, P and Q are respectively the mid-points of AB and BC and R is the mid-point
of AP. Prove that :

(1) ar (Δ PBQ) = ar (Δ ARC)

(2) ar (Δ PRQ) =1/2ar (Δ ARC)

(3) ar (Δ RQC) =3/8 ar (Δ ABC) .

ABCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC
such that CE = 2DE and F is the point of BC such that BF = 2FC. Prove that:

(1)  ar ( ADEG) = ar (GBCD)

(2)  ar (ΔEGB) = 1/6 ar (ABCD)

(3)  ar (ΔEFC) = 1/2 ar (ΔEBF)

(4)  ar (ΔEBG)  = ar (ΔEFC)

(5)ΔFind what portion of the area of parallelogram is the area of οEFG.

In Fig. below, CD || AE and CY || BA.
(i) Name a triangle equal in area of ΔCBX
(ii) Prove that ar (Δ ZDE) = ar (ΔCZA)
(iii) Prove that ar (BCZY) = ar (Δ EDZ)

In below fig., PSDA is a parallelogram in which PQ = QR = RS and AP || BQ || CR. Prove
that ar (Δ PQE) = ar (ΔCFD).

In the below fig. ABCD is a trapezium in which AB || DC and DC = 40 cm and AB = 60
cm. If X and Y are respectively, the mid-points of AD and BC, prove that:
(i) XY = 50 cm
(ii) DCYX is a trapezium
(iii) ar (trap. DCYX) =9/11ar (trap. (XYBA))

In Fig. below, ABC and BDE are two equilateral triangles such that D is the mid-point of
BC. AE intersects BC in F. Prove that

(1)  ar (Δ BDE) = 1/2 ar (ΔABC)

(2) Area ( ΔBDE) = 1/2  ar (ΔBAE)

(3)  ar (BEF) = ar (ΔAFD)

(4) area (Δ ABC) = 2 area (ΔBEC)

(5) ar (ΔFED) = 1/8 ar (ΔAFC)

(6) ar (Δ BFE) = 2 ar (ΔEFD)

D is the mid-point of side BC of ΔABC and E is the mid-point of BD. if O is the mid-point
of AE, prove that ar (ΔBOE) = 1/8 ar (Δ ABC).

In the below fig. X and Y are the mid-points of AC and AB respectively, QP || BC and
CYQ and BXP are straight lines. Prove that ar (Δ ABP) = ar (ΔACQ).

In the below fig. ABCD and AEFD are two parallelograms. Prove that
(1) PE = FQ
(2) ar (Δ APE) : ar (ΔPFA) = ar Δ(QFD) : ar (Δ PFD)
(3) ar (ΔPEA) = ar (ΔQFD)

In the below figure, ABCD is parallelogram. O is any point on AC. PQ || AB and LM ||
AD. Prove that ar (||gm DLOP) = ar (||gm BMOQ)

In a ΔABC, if L and M are points on AB and AC respectively such that LM || BC. Prove
that:

(1) ar (ΔLCM ) = ar (ΔLBM )
(2) ar (ΔLBC) = ar (ΔMBC)
(3) ar (ΔABM) ar (ΔACL)
(4) ar (ΔLOB) ar (ΔMOC)

In the below fig. D and E are two points on BC such that BD = DE = EC. Show that ar
(ΔABD) = ar (ΔADE) = ar (ΔAEC).

In the following figure, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:-

(i) ΔMBC ≅ ΔABD

(ii) ar (BYXD) = 2 ar(MBC)

(iii) ar (BYXD) = ar(ABMN)

(iv) ΔFCB ≅ ΔACE

(v) ar(CYXE) = 2 ar(FCB)

(vi) ar (CYXE) = ar(ACFG)

(vii) ar (BCED) = ar(ABMN) + ar(ACFG)

Note : Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in Class X.

## RD Sharma solutions for Class 9 Mathematics chapter 15 - Areas of Parallelograms and Triangles

RD Sharma solutions for Class 9 Maths chapter 15 (Areas of Parallelograms and Triangles) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Mathematics for Class 9 by R D Sharma (2018-19 Session) solutions in a manner that help students grasp basic concepts better and faster.

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Concepts covered in Class 9 Mathematics chapter 15 Areas of Parallelograms and Triangles are Triangles on the Same Base and Between the Same Parallels, Parallelograms on the Same Base and Between the Same Parallels, Figures on the Same Base and Between the Same Parallels, Introduction to Areas of Parallelograms and Triangles.

Using RD Sharma Class 9 solutions Areas of Parallelograms and Triangles exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in RD Sharma Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 9 prefer RD Sharma Textbook Solutions to score more in exam.

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