NCERT Exemplar Problems Class 9 Maths – Areas of Parallelograms and Triangles

Question 1:
The median of a triangle divides it into two
(a) triangles of equal area (b) congruent triangles
(c) right angled triangles (d) isosceles triangles
Solution:
(a) We know that, a median of a triangle is a line segment joining a vertex to the mid-point of the opposite side. Thus, a median of a triangle divides it into two triangles of equal area.

Question 2:
In which of the following figures, you find two polygons on the same base and between the same parallels?

Solution:
(d) In figures (a), (b) and (c) there are two polygons on the same base but they are not between the same parallels.
In figure (d), there are two polygons (PQRA and BQRS) on the same base and between the same parallels .

Question 3:
The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is
(a) a rectangle of area 24 cm²
(b) a square of area 25 cm²
(c) a trapezium of area 24 cm²
(d) a rhombus of area 24 cm²
Solution:
(d)
Here, length of rectangle ABCD = 8 cm and breadth of rectangle ABCD = 6.cm
Let E, F, G and H are the mid-points of the sides of rectangle ABCD, then EFGH is a rhombus.

Question 4:
In the figure, the area of parallelogram ABCD is
(a) AB x BM (b) BC x BN (c) DC x DL (d) AD x DL

Thinking Process
Use the formula, area of parallelogram =B asex Altitude to get the required result
Solution:
(c) We know that, area of parallelogram is the product of its any side and the corresponding altitude (or height).
Here, when AB is base, then height is DL.
Area of parallelogram = AB x DL and when AD is base, then height is BM.
Area of parallelogram = AD x BM When DC is base, then height is DL.
Area of parallelogram = DC x DL and when BC is base, then height is not given.
Hence, option (c) is correct.

Question 5:
In figure, if parallelogram ABCD and rectangle ABEM are of equal area, then
(a) perimeter of ABCD = perimeter of ABEM
(b) perimeter of ABCD < perimeter of ABEM
(c) perimeter of ABCD > perimeter of ABEM
(d) perimeter of ABCD = ½ (perimeter of ABEM)
Solution:
(c) In rectangle ABEM, AB = EM [sides of rectangle] and in parallelogram ABCD, CD = AB
On adding, both equations, we get
AB + CD = EM + AB …(i)
We know that, the perpendicular distance between two parallel sides of a parallelogram is always less than the length of the other parallel sides.
BE < BC and AM < AD
[since, in a right angled triangle, the hypotenuse is greater than the other side]
On adding both above inequalities, we get
SE + AM <BC + AD or BC + AD> BE + AM
On adding AB + CD both sides, we get
AB + CD + BC + AD> AB + CD + BE + AM
=> AB+BC + CD + AD> AB+BE + EM+ AM [∴ CD = AB = EM]
Perimeter of parallelogram ABCD > perimeter of rectangle ABEM

Question 6:
The mid-point of the sides of a triangle along with any of the vertices as the fourth point make a parallelogram of area equal to
(a) ½ ar (ABC) (b) 1/3 ar (ABC) (c) ¼ ar (ABC) (d) ar (ABC)
Solution:

Question 7:
Two parallelograms are on equal bases and between the same parallels. The ratio of their areas is
(a) 1 : 2 (b) 1 : 1 (c) 2 : 1 (d) 3 : 1
Solution:
(b) We know that, parallelogram on the equal bases and between the same parallels are equal in area. So, ratio of their areas is 1 :1.

Question 8:
ABCD is a quadrilateral whose diagonal AC divides it into two parts, equal in area, then ABCD
(a) is a rectangle
(b) is always a rhombus
(c) is a parallelogram
(d) need not be any of (a), (b) or (c)
Solution:
(d) Here, ABCD need not be any of rectangle, rhombus and parallelogram because if ABCD is a square, then its diagonal AC also divides it into two parts which are equal in area.

Question 9:
If a triangle and a parallelogram are on the same base and between same parallels, then the ratio of the area of the triangle to the area of parallelogram is
(a) 1 : 3 (b) 1:2 (c) 3 : 1 (d) 1 : 4
Solution:
(b) We know that, if a parallelogram and a triangle are on the same base and between the same parallels, then area of the triangle is half the area of the parallelogram.

Question 10:
ABCD is a trapezium with parallel sides AB = a cm and DC = b cm. E and F are the mid-points of the non-parallel sides. The ratio of ar (ABFE) and ar (EFCD)is
(a) a: b
(b) (3a + b): (a + 3b)
(c) (a + 3b): (3a + b)
(d) (2a +b): (3a + b)
Solution:

Exercise 9.2: Very Short Answer Type Questions

Write whether True or False and justify your answer.
Question 1:
ABCD is a parallelogram and X is the mid-point of AB. If ar (AXCD) = 24 cm², then ar (ABC) = 24 cm².
Solution:
False
Given, ABCD is a parallelogram and ar (AXCD) = 24 cm²
Let area of parallelogram ABCD is 2y cm² and join AC.

Question 2:
PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm and A is any point on PQ. If PS = 5 cm, then ar (ΔPAS) = 30 cm².
Solution:
True
Given, PS = 5 cm
radius of circle = SQ = 13 cm

Question 3:
PQRS is a parallelogram whose area is 180 cm² and A is any point on the diagonal QS. The area of ΔASR = 90 cm².
Solution:
False
Given, area of parallelogram PQRS = 180 cm² and QS is its diagonal which divides it into two triangles of equal area.

Question 4:
ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Then, ar (ΔBDE) = ¼ ar (ΔABC).
Solution:
True

Question 5:
In the figure, ABCD and EFGD are two parallelograms and G is the mid-point of CD. Then, ar (ΔDPC) = ½ ar (EFGD).

Solution:
False
In the given figure, join PG. Since, G is the mid-point of CD.
Thus, PG is a median of ΔDPC and it divides the triangle into parts of equal areas.

Exercise 9.3: Short Answer Type Questions

Question 1:
In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR= RS and PA || QB || RC. Prove that ar (PQE) =ar (CFD).
Thinking Process

Firstly, use the formula, area of parallelogram = Base x Altitude
Further, prove that A PQE = AD CF, by ASA congruent rule.
At the end use the property that congruent figures have same area.

Solution:
Given In a parallelogram PSDA, points 0 and R are on PS such that
PQ = QR = RS and PA || QB || RC.
To prove ar (PQE) = ar (CFD)
Proof In parallelogram PABQ,
and PA||QB [given]
So, PABQ is a parallelogram.
PQ = AB …(i)
Similarly, QBCR is also a parallelogram.
QR = BC …(ii)
and RCDS is a parallelogram.
RS =CD …(iii)
Now, PQ=QR = RS …(iv)
From Eqs. (i), (ii) (iii) and (iv),
PQ || AB [∴ in parallelogram PSDA, PS || AD]
In ΔPQE and ΔDCF, ∠QPE = ∠FDC
[since, PS || AD and PD is transversal, then alternate interior angles are equal] PQ=CD [from Eq. (v)]
and ∠PQE = ∠FCD
[∴ ∠PQE = ∠PRC corresponding angles and ∠PRC = ∠FCD alternate interior angles]
ΔPQE = ΔDCF [by ASA congruence rule]
∴ ar (ΔPQE) = ar (ΔCFD) [since,congruent figures have equal area]
Hence proved.

Question 2:
X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is drawn parallel to LM to meet MN at Z (see figure). Prove that ar (ΔLZY) = ar (MZYX).

Thinking Process
Use the property that the triangles on the same base and between the same two parallel lines are equal in area. Further prove the required result.
Solution:
Given X and Y are points on the side LN such that
LX = XY = YN and XZ || LM To prove ar (ΔLZY) = ar (MZYX)
Proof Since, ΔXMZ and ΔXLZ are on the same base XZ and between the same parallel lines LM and XZ.
Then, ar (ΔXMZ) = ar (ΔXLZ) …(i)
On adding ar (ΔXYZ) both sides of Eq. (i), we get
ar (ΔXMZ) + ar (ΔXXZ) = ar (ΔXLZ) + ar (ΔXYZ)
=> ar (MZYX) = ar (ΔLZY) Hence proved.

Question 3:
The area of the parallelogram ABCD is 90 cm². Find

ar (ABEF)
ar (ΔABD)
ar (ΔBEF)

Solution:
Given, area of parallelogram, ABCD = 90 cm².

We know that, parallelograms on the same base and between the same parallel are equal in areas.
Here, parallelograms ABCD and ABEF are on same base AB and between the same parallels AB and CF.
So, ar (ΔBEF) = ar (ABCD) = 90 cm²
We know that, if a triangle and a parallelogram are on the same base and between the same parallels, then area of triangle is equal to half of the area of the parallelogram.
Here, ΔABD and parallelogram ABCD are on the same base AB and between the same parallels AB and CD.
So, ar (ΔABD) = ½ ar (ABCD)
= ½ x 90 = 45 cm² [∴ ar (ABCD) = 90 cm²]
Here, ABEF and parallelogram ABEF are on the same base EF and between the same parallels AB and EF.
ar (ΔBEF) = ½ ar (ABEF)
= ½ x 90 = 45 cm² [∴ ar (ABEF) = 90 cm², from part (i)]

Question 4:
In ΔABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (shown in figure), then prove that ar (ΔBPQ) = ½ ar (ΔABC).

Solution:

Question 5:
ABCD is a square. E and F are respectively the mid-points of BC and CD. If R is the mid-point of EF, prove that ar (ΔAER) = ar (ΔAFR).

Solution:

Question 6:
O is any point on the diagonal PR of a parallelogram PQRS (figure). Prove that ar(ΔPSO) = ar(ΔPQO).

Solution:

Question 7:
ABCD is a parallelogram in which BC is produced to E such that CE = BC. AE intersects CD at F

If ar (ΔDFB) = 3 cm², then find the area of the parallelogram ABCD.
Solution:

Question 8:
In trapezium ABCD, AB || DC and L is the mid-point of BC. Through L, a line PQ || AD has been drawn which meets AB in P and DC produced in Q. Prove that ar (ABCD) = ar (APQD).

∴ BL = CL
Solution:

Question 9:
If the mid-points of the sides of a quadrilateral are joined in order, prove that the area of the parallelogram, so formed will be half of the area of the given quadrilateral (figure).

Solution:

Exercise 9.4: Long Answer Type Questions

Question 1:
A point E is taken on the side BC of a parallelogram ABCD. AE and DC are produced to meet at F. Prove that ar (ΔADF) = ar (ΔBFC).
Solution:
Given ABCD is a parallelogram and E is a point on BC. AE and DC are produced to meet at F.
AB||CD anti BC||AD ,..(i)

Question 2:
The diagonals of a parallelogram ABCD intersect at a point O. Through O, a line is drawn to intersect AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal area.
Solution:

Question 3:
The median BE and CF of a triangle ABC intersect at G. Prove that the area of ΔGBC = area of the quadrilateral AFGE.
Solution:

Question 4:
In figure, CD||AE and CY||BA. Prove that ar (ΔCBX) = ar (ΔAXY).

Solution:
Given In figure, CD||AE
and CY || BA
To prove ar (ΔCBX) = ar (ΔAXY) .
Proof We know that, triangles on the same base and between the same parallels are equal . in areas.
Here, ΔABY and ΔABC both lie on the same base AB and between the same parallels CY and BA.
ar (ΔABY) = ar (ΔABC)
=> ar (ABX) + ar (AXY) = ar (ABX) + ar (CBX)
=> ar (AXY) = ar (CBX) [eliminating ar (ABX) from both sides]

Question 5:
ABCD is trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = 7/9 ar (XYBA).
Solution:

Question 6:
In ΔABC, if L and M are the points on AB and AC, respectively such that LM || BC. Prove that ar (ΔLOB) = ar (ΔMOC).
Solution:
Given In ΔABC, L and M are points on AB and AC respectively such that LM || BC.

Question 7:
In figure, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn parallel to AD meets CD produced at Q. Prove that ar (ABCDE) = ar (ΔAPQ).

Solution:
Given ABCDE is a pentagon.
BP || AC and EQ|| AD.
To prove ar (ABCDE) = ar (APQ)
Proof We know that, triangles on the same base and between the same parallels are equal in area.
Here, ΔADQ and ΔADE lie on the same base AD and between the same parallels AD and EQ.
So, ar (ΔADQ) = ar (ΔADE) ...(i)
Similarly, ΔACP and ΔACB lie on the same base AC and between the same parallels AC and BP.
So, ar (ΔACP) = ar (ΔACB) …(ii)
On adding Eqs. (i) and (ii), we get
ar (ΔADQ) + ar (ΔACP) = ar (ΔADE) + ar (ΔACB)
On adding ar (ΔACD) both sides, we get
ar (ΔADQ) + ar (ΔACP) + ar (ΔACD) = ar (ΔADE) + ar (ΔACB) + ar (ΔACD)
=> ar (ΔAPQ) = ar (ABCDE) Hence proved.

Question 8:
If the medians of a AABC intersect at G, then show that ar (ΔAGB) = ar (ΔAGC) = ar (ΔBGC) = 1/3 ar(ΔABC).
Thinking Process
Use the property that median of a triangle divides it into two triangles of equal area.
Further, apply above property by considering different triangles and prove the required result.
Solution:
Given In ΔABC, AD, BE and CF are medians and intersect at G.

Question 9:
In figure X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are straight lines. Prove that ar (ΔABP) = ar (ΔACQ).

Thinking Process

Firstly, use the theorem that joining the mid-points of the two sides of a triangle is parallel to the third side.
Further, use the theorem that triangles on the same base and between the same
parallels are equal in area. Use this theorem by considering different triangles and prove the required result

Solution:
Given X and Y are the mid-points of AC and AB respectively. Also, QP|| BC and CYQ, BXP are straight lines.
To prove ar (ΔABP) = ar (ΔACQ)
Proof Since, X and Y are the mid-points”of AC and AB respectively.
So, XY || BC
We know that, triangles on the same base and between the same parallels are equal in area. Here, ΔBYC and ΔBXC lie on same base BC and between the same parallels BC and XY.
So, ar (ΔBYC) = ar (ΔBXC)
On subtracting ar (ΔBOC)from both sides, we get
ar (ΔBYC) – ar (ΔBOC) = ar (ΔBXC) – ar (ΔBOC)
=» ar (ΔBOY) = ar (ΔCOX)
On adding ar (ΔXOY) both sides, we get
ar (ΔSOY) + ar (ΔXOY) = ar (ΔCOX) + ar (ΔXOY)
=> ar (ΔBYX) = ar (ΔCXY) …(i)
Hence, we observe that quadrilaterals XYAP and YXAQ are on the same base XY and between the same parallels XY and PQ.
ar (XYAP) = ar (YXAQ) …(ii)
On adding Eqs. (i) and (ii), we get
ar (ΔBYX) + ar (XYAP) = ar (ΔCXY) + ar (YXAQ)
=> ar (ΔABP) = ar (ΔACQ) Hence proved.

Question 10:
In figure, ABCD and AEFD are two parallelograms. Prove that ar (APEA) = ar(AQFO).

Solution:
Given, ABCD and AEFD are two parallelograms.
To prove ar (APEA) = ar (AQFD)
Proof In quadrilateral PQDA,
AP || DQ [since, in parallelogram ABCD, AB || CD ] and PQ || AD [since, in parallelogram AEFD, FE || AD]
Then, quadrilateral PQDA is a parallelogram.
Also, parallelogram PQDA and AEFD are on the same base AD and between the same parallels AD and EQ.
ar (parallelogram PQDA) = ar (parallelogram AEFD)
On subtracting ar (quadrilateral APFD) from both sides, we get
ar (parallelogram PQDA)- ar (quadrilateral APFD)
= ar (parallelogram AEFD) – ar (quadrilateral APFD) => ar (AQFD) = ar (APEA) Hence proved.

NCERT Exemplar Problems Class 9 Maths – Areas of Parallelograms and Triangles