Class 7 Exam > Class 7 Notes > Mathematics (Maths) Class 7 > RD Sharma Solutions: Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.3)
Mensuration – I (Perimeter and Area of Rectilinear Figures Exercise 20.3) RD Sharma Solutions | Mathematics (Maths) Class 7 PDF Download
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Page 1
1. Find the area of a parallelogram with base 8 cm and altitude 4.5 cm.
Solution:
Given base = 8 cm and altitude = 4.5 cm
We know that area of the parallelogram = Base x Altitude
= 8 cm x 4.5 cm
Therefore, area of parallelogram = 36 cm
2
2. Find the area in square meters of the parallelogram whose base and altitudes are as
under
(i) Base =15 dm, altitude = 6.4 dm
(ii) Base =1 m 40 cm, altitude = 60 cm
Solution:
(i) Given base =15 dm, altitude = 6.4 dm
By converting these to standard form we get,
Base = 15 dm = (15 x 10) cm = 150 cm = 1.5 m
Altitude = 6.4 dm = (6.4 x 10) cm = 64 cm = 0.64 m
We know that area of the parallelogram = Base x Altitude
= 1.5 m x 0.64 m
Area of parallelogram = 0.96 m
2
(ii) Given base = 1 m 40 cm = 1.4 m [Since 100 cm = 1 m]
Altitude = 60 cm = 0.6 m [Since 100 cm = 1 m]
We know that area of the parallelogram = Base x Altitude
= 1.4 m x 0.6 m
= 0.84 m
2
3. Find the altitude of a parallelogram whose area is 54 d m
2
and base is 12 dm.
Solution:
Given area of the given parallelogram = 54 d m
2
Base of the given parallelogram = 12 dm
We know that area of the parallelogram = Base x Altitude
Page 2
1. Find the area of a parallelogram with base 8 cm and altitude 4.5 cm.
Solution:
Given base = 8 cm and altitude = 4.5 cm
We know that area of the parallelogram = Base x Altitude
= 8 cm x 4.5 cm
Therefore, area of parallelogram = 36 cm
2
2. Find the area in square meters of the parallelogram whose base and altitudes are as
under
(i) Base =15 dm, altitude = 6.4 dm
(ii) Base =1 m 40 cm, altitude = 60 cm
Solution:
(i) Given base =15 dm, altitude = 6.4 dm
By converting these to standard form we get,
Base = 15 dm = (15 x 10) cm = 150 cm = 1.5 m
Altitude = 6.4 dm = (6.4 x 10) cm = 64 cm = 0.64 m
We know that area of the parallelogram = Base x Altitude
= 1.5 m x 0.64 m
Area of parallelogram = 0.96 m
2
(ii) Given base = 1 m 40 cm = 1.4 m [Since 100 cm = 1 m]
Altitude = 60 cm = 0.6 m [Since 100 cm = 1 m]
We know that area of the parallelogram = Base x Altitude
= 1.4 m x 0.6 m
= 0.84 m
2
3. Find the altitude of a parallelogram whose area is 54 d m
2
and base is 12 dm.
Solution:
Given area of the given parallelogram = 54 d m
2
Base of the given parallelogram = 12 dm
We know that area of the parallelogram = Base x Altitude
Therefore altitude of the given parallelogram = Area/Base
= 54/12 dm
= 4.5 dm
4. The area of a rhombus is 28 m
2
. If its perimeter be 28 m, find its altitude.
Solution:
Given perimeter of a rhombus = 28 m
But we know that perimeter of a rhombus = 4 (Side)
4(Side) = 28 m
Side = 28/4
Side = 7m
Now, Area of the rhombus = 28 m
2
But we know that area of rhombus = Side x Altitude
(Side x Altitude) = 28 m
2
(7 x Altitude) = 28 m
2
Altitude = 28/7 = 4 m
5. In Fig. 20, ABCD is a parallelogram, DL ? AB and DM ? BC. If AB = 18 cm, BC =12 cm
and DM= 9.3 cm, find DL.
Solution:
Given DL ? AB and DM ? BC
Taking BC as the base, BC = 12 cm and altitude DM = 9.3 cm
We know that area of parallelogram ABCD = Base x Altitude
= (12 cm x 9.3 cm)
Page 3
1. Find the area of a parallelogram with base 8 cm and altitude 4.5 cm.
Solution:
Given base = 8 cm and altitude = 4.5 cm
We know that area of the parallelogram = Base x Altitude
= 8 cm x 4.5 cm
Therefore, area of parallelogram = 36 cm
2
2. Find the area in square meters of the parallelogram whose base and altitudes are as
under
(i) Base =15 dm, altitude = 6.4 dm
(ii) Base =1 m 40 cm, altitude = 60 cm
Solution:
(i) Given base =15 dm, altitude = 6.4 dm
By converting these to standard form we get,
Base = 15 dm = (15 x 10) cm = 150 cm = 1.5 m
Altitude = 6.4 dm = (6.4 x 10) cm = 64 cm = 0.64 m
We know that area of the parallelogram = Base x Altitude
= 1.5 m x 0.64 m
Area of parallelogram = 0.96 m
2
(ii) Given base = 1 m 40 cm = 1.4 m [Since 100 cm = 1 m]
Altitude = 60 cm = 0.6 m [Since 100 cm = 1 m]
We know that area of the parallelogram = Base x Altitude
= 1.4 m x 0.6 m
= 0.84 m
2
3. Find the altitude of a parallelogram whose area is 54 d m
2
and base is 12 dm.
Solution:
Given area of the given parallelogram = 54 d m
2
Base of the given parallelogram = 12 dm
We know that area of the parallelogram = Base x Altitude
Therefore altitude of the given parallelogram = Area/Base
= 54/12 dm
= 4.5 dm
4. The area of a rhombus is 28 m
2
. If its perimeter be 28 m, find its altitude.
Solution:
Given perimeter of a rhombus = 28 m
But we know that perimeter of a rhombus = 4 (Side)
4(Side) = 28 m
Side = 28/4
Side = 7m
Now, Area of the rhombus = 28 m
2
But we know that area of rhombus = Side x Altitude
(Side x Altitude) = 28 m
2
(7 x Altitude) = 28 m
2
Altitude = 28/7 = 4 m
5. In Fig. 20, ABCD is a parallelogram, DL ? AB and DM ? BC. If AB = 18 cm, BC =12 cm
and DM= 9.3 cm, find DL.
Solution:
Given DL ? AB and DM ? BC
Taking BC as the base, BC = 12 cm and altitude DM = 9.3 cm
We know that area of parallelogram ABCD = Base x Altitude
= (12 cm x 9.3 cm)
= 111.6 c m
2
….. Equation (i)
Now, by taking AB as the base,
We have, Area of the parallelogram ABCD = Base x Altitude
= (18 cm x DL) ….. Equation (ii)
From (i) and (ii), we have
18 cm x DL = 111.6 c m
2
DL = 111.6/18
= 6.2 cm
6. The longer side of a parallelogram is 54 cm and the corresponding altitude is 16 cm.
If the altitude corresponding to the shorter side is 24 cm, find the length of the shorter
side.
Solution:
Let ABCD is a parallelogram with the longer side AB = 54 cm and corresponding altitude
AE = 16 cm.
The shorter side is BC and the corresponding altitude is CF = 24 cm.
We know that area of a parallelogram = base x height.
We have two altitudes and two corresponding bases.
By equating them we get,
½ x BC x CF = ½ x AB x AE
On simplifying, we get
BC x CF = AB x AE
BC x 24 = 54 x 16
BC = (54 × 16)/24
= 36 cm
Hence, the length of the shorter side BC = AD = 36 cm.
7. In Fig. 21, ABCD is a parallelogram, DL ? AB. If AB = 20 cm, AD = 13 cm and area of
Page 4
1. Find the area of a parallelogram with base 8 cm and altitude 4.5 cm.
Solution:
Given base = 8 cm and altitude = 4.5 cm
We know that area of the parallelogram = Base x Altitude
= 8 cm x 4.5 cm
Therefore, area of parallelogram = 36 cm
2
2. Find the area in square meters of the parallelogram whose base and altitudes are as
under
(i) Base =15 dm, altitude = 6.4 dm
(ii) Base =1 m 40 cm, altitude = 60 cm
Solution:
(i) Given base =15 dm, altitude = 6.4 dm
By converting these to standard form we get,
Base = 15 dm = (15 x 10) cm = 150 cm = 1.5 m
Altitude = 6.4 dm = (6.4 x 10) cm = 64 cm = 0.64 m
We know that area of the parallelogram = Base x Altitude
= 1.5 m x 0.64 m
Area of parallelogram = 0.96 m
2
(ii) Given base = 1 m 40 cm = 1.4 m [Since 100 cm = 1 m]
Altitude = 60 cm = 0.6 m [Since 100 cm = 1 m]
We know that area of the parallelogram = Base x Altitude
= 1.4 m x 0.6 m
= 0.84 m
2
3. Find the altitude of a parallelogram whose area is 54 d m
2
and base is 12 dm.
Solution:
Given area of the given parallelogram = 54 d m
2
Base of the given parallelogram = 12 dm
We know that area of the parallelogram = Base x Altitude
Therefore altitude of the given parallelogram = Area/Base
= 54/12 dm
= 4.5 dm
4. The area of a rhombus is 28 m
2
. If its perimeter be 28 m, find its altitude.
Solution:
Given perimeter of a rhombus = 28 m
But we know that perimeter of a rhombus = 4 (Side)
4(Side) = 28 m
Side = 28/4
Side = 7m
Now, Area of the rhombus = 28 m
2
But we know that area of rhombus = Side x Altitude
(Side x Altitude) = 28 m
2
(7 x Altitude) = 28 m
2
Altitude = 28/7 = 4 m
5. In Fig. 20, ABCD is a parallelogram, DL ? AB and DM ? BC. If AB = 18 cm, BC =12 cm
and DM= 9.3 cm, find DL.
Solution:
Given DL ? AB and DM ? BC
Taking BC as the base, BC = 12 cm and altitude DM = 9.3 cm
We know that area of parallelogram ABCD = Base x Altitude
= (12 cm x 9.3 cm)
= 111.6 c m
2
….. Equation (i)
Now, by taking AB as the base,
We have, Area of the parallelogram ABCD = Base x Altitude
= (18 cm x DL) ….. Equation (ii)
From (i) and (ii), we have
18 cm x DL = 111.6 c m
2
DL = 111.6/18
= 6.2 cm
6. The longer side of a parallelogram is 54 cm and the corresponding altitude is 16 cm.
If the altitude corresponding to the shorter side is 24 cm, find the length of the shorter
side.
Solution:
Let ABCD is a parallelogram with the longer side AB = 54 cm and corresponding altitude
AE = 16 cm.
The shorter side is BC and the corresponding altitude is CF = 24 cm.
We know that area of a parallelogram = base x height.
We have two altitudes and two corresponding bases.
By equating them we get,
½ x BC x CF = ½ x AB x AE
On simplifying, we get
BC x CF = AB x AE
BC x 24 = 54 x 16
BC = (54 × 16)/24
= 36 cm
Hence, the length of the shorter side BC = AD = 36 cm.
7. In Fig. 21, ABCD is a parallelogram, DL ? AB. If AB = 20 cm, AD = 13 cm and area of
the parallelogram is 100 c m
2
, find AL.
Solution:
From the figure we have ABCD is a parallelogram with base AB = 20 cm and
corresponding altitude DL.
It is given that the area of the parallelogram ABCD = 100 c m
2
We know that the area of a parallelogram = Base x Height
Therefore,
100 = AB x DL
100 = 20 x DL
DL = 100/20 = 5 cm
By observing the picture it is clear that we have to apply the Pythagoras theorem,
Therefore by Pythagoras theorem, we have,
(AD)
2
= (AL)
2
+ (DL)
2
(13)
2
= (AL)
2
+ (5)
2
(AL)
2
= (13)
2
– (5)
2
(AL)
2
= 169 – 25
= 144
We know that 12
2
= 144
(AL)
2
= (12)
2
AL = 12 cm
Hence, length of AL is 12 cm.
8. In Fig. 21, if AB = 35 cm, AD= 20 cm and area of the parallelogram is 560 cm
2
, find
LB.
Page 5
1. Find the area of a parallelogram with base 8 cm and altitude 4.5 cm.
Solution:
Given base = 8 cm and altitude = 4.5 cm
We know that area of the parallelogram = Base x Altitude
= 8 cm x 4.5 cm
Therefore, area of parallelogram = 36 cm
2
2. Find the area in square meters of the parallelogram whose base and altitudes are as
under
(i) Base =15 dm, altitude = 6.4 dm
(ii) Base =1 m 40 cm, altitude = 60 cm
Solution:
(i) Given base =15 dm, altitude = 6.4 dm
By converting these to standard form we get,
Base = 15 dm = (15 x 10) cm = 150 cm = 1.5 m
Altitude = 6.4 dm = (6.4 x 10) cm = 64 cm = 0.64 m
We know that area of the parallelogram = Base x Altitude
= 1.5 m x 0.64 m
Area of parallelogram = 0.96 m
2
(ii) Given base = 1 m 40 cm = 1.4 m [Since 100 cm = 1 m]
Altitude = 60 cm = 0.6 m [Since 100 cm = 1 m]
We know that area of the parallelogram = Base x Altitude
= 1.4 m x 0.6 m
= 0.84 m
2
3. Find the altitude of a parallelogram whose area is 54 d m
2
and base is 12 dm.
Solution:
Given area of the given parallelogram = 54 d m
2
Base of the given parallelogram = 12 dm
We know that area of the parallelogram = Base x Altitude
Therefore altitude of the given parallelogram = Area/Base
= 54/12 dm
= 4.5 dm
4. The area of a rhombus is 28 m
2
. If its perimeter be 28 m, find its altitude.
Solution:
Given perimeter of a rhombus = 28 m
But we know that perimeter of a rhombus = 4 (Side)
4(Side) = 28 m
Side = 28/4
Side = 7m
Now, Area of the rhombus = 28 m
2
But we know that area of rhombus = Side x Altitude
(Side x Altitude) = 28 m
2
(7 x Altitude) = 28 m
2
Altitude = 28/7 = 4 m
5. In Fig. 20, ABCD is a parallelogram, DL ? AB and DM ? BC. If AB = 18 cm, BC =12 cm
and DM= 9.3 cm, find DL.
Solution:
Given DL ? AB and DM ? BC
Taking BC as the base, BC = 12 cm and altitude DM = 9.3 cm
We know that area of parallelogram ABCD = Base x Altitude
= (12 cm x 9.3 cm)
= 111.6 c m
2
….. Equation (i)
Now, by taking AB as the base,
We have, Area of the parallelogram ABCD = Base x Altitude
= (18 cm x DL) ….. Equation (ii)
From (i) and (ii), we have
18 cm x DL = 111.6 c m
2
DL = 111.6/18
= 6.2 cm
6. The longer side of a parallelogram is 54 cm and the corresponding altitude is 16 cm.
If the altitude corresponding to the shorter side is 24 cm, find the length of the shorter
side.
Solution:
Let ABCD is a parallelogram with the longer side AB = 54 cm and corresponding altitude
AE = 16 cm.
The shorter side is BC and the corresponding altitude is CF = 24 cm.
We know that area of a parallelogram = base x height.
We have two altitudes and two corresponding bases.
By equating them we get,
½ x BC x CF = ½ x AB x AE
On simplifying, we get
BC x CF = AB x AE
BC x 24 = 54 x 16
BC = (54 × 16)/24
= 36 cm
Hence, the length of the shorter side BC = AD = 36 cm.
7. In Fig. 21, ABCD is a parallelogram, DL ? AB. If AB = 20 cm, AD = 13 cm and area of
the parallelogram is 100 c m
2
, find AL.
Solution:
From the figure we have ABCD is a parallelogram with base AB = 20 cm and
corresponding altitude DL.
It is given that the area of the parallelogram ABCD = 100 c m
2
We know that the area of a parallelogram = Base x Height
Therefore,
100 = AB x DL
100 = 20 x DL
DL = 100/20 = 5 cm
By observing the picture it is clear that we have to apply the Pythagoras theorem,
Therefore by Pythagoras theorem, we have,
(AD)
2
= (AL)
2
+ (DL)
2
(13)
2
= (AL)
2
+ (5)
2
(AL)
2
= (13)
2
– (5)
2
(AL)
2
= 169 – 25
= 144
We know that 12
2
= 144
(AL)
2
= (12)
2
AL = 12 cm
Hence, length of AL is 12 cm.
8. In Fig. 21, if AB = 35 cm, AD= 20 cm and area of the parallelogram is 560 cm
2
, find
LB.
Solution:
From the figure, ABCD is a parallelogram with base AB = 35 cm and corresponding
altitude DL.
The adjacent side of the parallelogram AD = 20 cm.
It is given that the area of the parallelogram ABCD = 560 cm
2
Now, Area of the parallelogram = Base x Height
560 cm
2
= AB x DL
560 cm
2
= 35 cm x DL
DL = 560/35
= 16 cm
Again by Pythagoras theorem, we have, (AD)
2
= (AL)
2
+ (DL)
2
(20)
2
= (AL)
2
+ (16)
2
(AL)
2
= (20)
2
– (16)
2
= 400 – 256
= 144
(AL)
2
= (12)
2
AL = 12 cm
From the figure, AB = AL + LB
35 = 12 + LB
LB = 35 – 12 = 23 cm
Hence, length of LB is 23 cm.
9. The adjacent sides of a parallelogram are 10 m and 8 m. If the distance between the
longer sides is 4 m, find the distance between the shorter sides
Solution:
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