A quantity that measures the top shape, planar lamina and plane. The surface area is the area of a two dimensional surface on a three dimensional object. The unit of area is primarily concerned with the side of the length and they are primarily measured in square meters, square kilometers, square yards, square miles etc. Some of the areas of different shapes are:
- Quadrilateral Area
- Triangle Area
- Rectangular Area
- Polygon Area
- Area of a Circle
- Area of an Ellipse
- Surface Area
- Area of Calculus
Learn more about RS Aggarwal Class 9 Solutions Chapter 10 Area below:
RS Aggarwal Class 9 Solutions Chapter 10
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, BM⊥AD
Therefore, Area of parallelogram ABCD= AD x BM
60 = AD x 8cm
AD = 608=7.5cm
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 ΔACD,
OD is its altitude and AC is its base
So, area of ΔACD=1/2 x AC x OD
= 1/2 x 24 x BD/2 = (12×24×8)cm2[Since,BD=16cm] = 96cm2
i.e. Area of ΔABC=12×AC×OB
= (12×24×8) cm2=96cm2
Now, the Area of rhombus = Area of ΔACD + Area of ΔABC = (96 + 96) cm2 = 192 cm2
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 ΔDBC,
DB2 = DC2 – CB2 = 172 – 82 = 289 – 64 = 225 cm2
So, area of ΔDBC = (1/2 x 15 x 8) cm2 = 60 cm2
Again, in right angled ΔDAB,
AB2 = DB2 – AD2 = 152 – 92 = 225 – 81 = 144cm2
Therefore, the area of ΔDAB = (1/2 x 12 x 9) cm2 = 54 cm2
So, area of quadrilateral ABCD = Area of ΔDBC +Area of ΔDAB = (60 + 54) cm2 =114 cm2
Therefore, the area of quadrilateral ABCD = 114 cm2
Q.6: Calculate the area of trap PQRS, given in Fig. (ii)
In right angled ΔRTQ
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 AL⊥BD and CM⊥BD , show that
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)
Area of ΔBAD=12×BD×AL
Area of ΔCBD=12×BD×CM
Therefore, Area of quadrilateral ABCD= Area of ΔABD +Area of ΔCBD = 1/2 x BD x AL + 1/2 x BD x CM
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 AL⊥BD and CM⊥BD such that AL = 8cm and CM = 6cm, find the area of quad ABCD.
Area of ΔBAD=12×BD×AL = (1/2 x 14 x 8) cm2 = 56cm2
Area of ΔCBD=12×BD×CM = (1/2 x 14 x 6) cm2 = 42cm2
Therefore, the Area of quadrilateral ABCD= Area of ΔABD +Area of ΔCBD = (56 + 42) cm2 = 98 cm2
Q.9: In the adjoining figure, ABCD is a trapezium in which AB∥BD and its diagonals AC and BD intersect at O.
Prove that area of triangle AOD = area of the triangle of BOC.
Consider ΔADCandΔDCB.. We find they have the same base CD and lie between two parallel lines DC and AB
Triangles on the same base and between the same parallels are equal in area.
So ΔCDAandΔCDB. are equal in area.
Therefore, Area (ΔAOD)=area(ΔBOC)
Q.10: In the adjoining figure, DE∥BC. Prove that:
(i) Area of triangle ACD = area ABE
(ii) Area of triangle OCE = area of triangle OBD
(i) ΔDBEandΔDCE have the same base DE and lie between parallel lines BC and DE.
So, area (ΔDBE)=area(ΔDCE) . . . . . . . (1)
Adding area (ΔDBE) on both sides ,we get
(ii) Since ar(ΔDBE)=ar(ΔDCE) [From (1)]
Subtracting ar(ΔODE) from both sides we get
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.
Show that: DE∥BC
Given: A ΔABC in which points D and E lie on AB and AC,
Such that ar(ΔBCE)=ar(ΔBCD)
To prove: DE∥AC
As ΔBCEandΔBCD have same base BC, and are equal in area, they have same altitudes.
This means that they lie between two parallel lines. Therefore, DE∥AC
Question 12: In the adjoining figure, O is any point inside a parallelogram ABCD. Prove that:
(i) ar.( △OAB )+ar.( △OCD)= 12(∥gmABCD)
(ii) ar.( △OAD)+ ar.( △OBC)=
Given: A parallelogram ABCD in which O is a point inside it
To prove: (i) ar(ΔOAB)+ar(ΔOCD)=12ar(∥gmABCD)
Construction: Though O draw PQ∥ABandRS∥AD
Proof: (i) ΔAOB and parallelogram ABQP have same base AB and lie between parallel lines AB and PQ.
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.
(ii) ΔAOD and ∥gm ASRD have the same base AD and lie between same parallel lines AD and RS.
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