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R.D. Sharma solutions Mathematics for Class 9 by R D Sharma (2018-19 Session) chapter 15 Areas of Parallelograms and Triangles

Chapters

R.D. Sharma Mathematics Class 9 by R D Sharma (2018-19 Session)

Mathematics for Class 9 by R D Sharma (2018-19 Session)

Chapter 15 - Areas of Parallelograms and Triangles

Page 0

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‘Žπ‘Ÿ (Δ𝐴𝐡𝐷) = π‘Žπ‘Ÿ (Δ𝐡𝐢𝐷) = π‘Žπ‘Ÿ (Δ𝐴𝐡𝐢) = π‘Žπ‘Ÿ (Δ𝐴𝐢𝐷) = `1/2` π‘Žπ‘Ÿ (||π‘”π‘š 𝐴𝐡𝐢𝐷) .

Page 0

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) =
2ar (ΔADC).

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) =`1/2`ar (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 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) =`9/11`ar (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.

R.D. Sharma Mathematics Class 9 by R D Sharma (2018-19 Session)

Mathematics for Class 9 by R D Sharma (2018-19 Session)
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