RS Aggarwal Class 9 Solutions Chapter 10
|Area Exercise 10.1|
Q.1: In the adjoining figure, show that ABCD is parallelogram .Calculate the area parallelogram ABCD.
Since the diagonal BD divides ABCD into two triangles of equal area.
Therefore, Area of parallelogram = 35 cm2
Q.2: In a parallelogram ABCD, it is being given that AB = 10cm and the altitudes corresponding to the sides AB and AD are DL = 6cm and BM = 8cm, respectively. Find AD.
Since, ABCD is a parallelogram and DL Is perpendicular to AB
So, its area =AB x DL = (10 x 6) cm2 = 60cm2
Also in parallelogram ABCD,
Therefore, Area of parallelogram ABCD= AD x BM
60 = AD x 8cm
Therefore, AD = 7.5 cm
Q.3: Find the area of a rhombus, the lengths whose diagonals are 16cm and 24cm respectively.
ABCD is a rhombus in which diagonal Ac = 24cm and BD = 16cm
These diagonals intersect at O:
Since, diagonals of a rhombus are perpendicular to each other. So, in
OD is its altitude and AC is its base
So, area of
= 1/2 x 24 x BD/2 =
i.e. Area of
Now, the Area of rhombus = Area of
Q.4: Find the area of a rhombus, the lengths of whose diagonals are 16 cm and 24 cm respectively.
ABCD is a trapezium in which, AB || CD
AB = 9cm and CD = 6cm
CE is a perpendicular drawn to AB through C and CE = 8cm
Area of trapezium = 1/2 (sum of parallel sides) x distance between them = [1/2 (9+6) x 8] cm2 = [1/2 x 15 x 8] cm2 = 60 cm2
Therefore, the Area of trapezium = 60 cm2
Q.5: Calculate the area of quad ABCD, given in Fig.
ABCD is a quadrilateral:
Now in right angled
DB2 = DC2 – CB2 = 172 – 82 = 289 – 64 = 225 cm2
So, area of
Again, in right angled
AB2 = DB2 – AD2 = 152 – 92 = 225 – 81 = 144cm2
Therefore, the area of
So, area of quadrilateral ABCD = Area of
Therefore, the area of quadrilateral ABCD = 114 cm2
Q.6: Calculate the area of trap PQRS, given in Fig. (ii)
In right angled
RT2 = RQ2 – TQ2 = 172 – 82 = 289 – 64 = 225 cm2
Therefore, Area of trapezium = 1/2(sum of parallel sides) x distance between them
= 1/2 x (PQ + SR) x RT =1/2 x (16 +8) x 15 = (1/2 x 24 x 15) cm2 = 180 cm2
Therefore, the Area of trapezium = 180 cm2
Q.7: BD is one of the diagonals of a diagonal of a quad. ABCD. If
Area (quad. ABCD) = 1/2 * BD * (AL + CM)
Given: ABCD is a quadrilateral and BD is one of its diagonals
To prove: area (quad. ABCD) = 1/2 x BD x (AL + CM)
Therefore, Area of quadrilateral ABCD= Area of
Therefore, the Area of quadrilateral ABCD = 1/2 x BD [AL + CM]
Q.8: In the adjoining figure, ABCD is a quadrilateral in which diag. BD = 14cm. If
Therefore, the Area of quadrilateral ABCD= Area of
Q.9: In the adjoining figure, ABCD is a trapezium in which
Prove that area of triangle AOD = area of the triangle of BOC.
Triangles on the same base and between the same parallels are equal in area.
Q.10: In the adjoining figure,
(i) Area of triangle ACD = area ABE
(ii) Area of triangle OCE = area of triangle OBD
Q.11: In the adjoining figure, D and E are points on the sides AB and AC of triangle ABC such that area of triangle BCE = area of triangle BCD.
This means that they lie between two parallel lines. Therefore,
Question 12: In the adjoining figure, O is any point inside a parallelogram ABCD. Prove that:
Given: A parallelogram ABCD in which O is a point inside it
To prove: (i)
Construction: Though O draw
If a triangle and a parallelogram are on the same base, and between the same parallels, then the area of the triangle is equal to half the area of the parallelogram.
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