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**Maharashtra Board Class 10 Solutions for Geometry – Mensuration Ex 6.8 ( Marathi Medium)**

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**Stem**

It is the ascending part of the axis bearing branches, leaves, flowers and fruits. It develops from the plumule of embryo of a germinating seed and bears nodes and internodes. Node is the region, where leaves arise and intemode is the region between two nodes.

**Main functions of stem are as follow:**

- It bears leaves, fruit, flowers and seeds in position.
- Transport water and minerals among leaves, flowers, fruits and roots.
- Stem store food as reserve food material, bears flowers for reproduction and provide support and protection to plant.
- Stem synthesize hormone, e.g. auxin.

**Special features of stem are as follows:**

- In many
**xerophytic**plants like**Opuntia,**stem becomes green, flattened or fleshy, which carries out photosynthesis. It is called**phylloclade**or**cladophyll.** - In some plants, short, green, cylindrical (e.g. Asparagus) or some times flattened (e.g.
**Ruscus**) branches, limited in growth, develop from the node of stem or branch in the**axil**of a leaf, which are reduced to small scales which perform the functions of photosynthesis. It is called**cladode.** - Some stems are thin, weak and lie prostrate on soil.
- Tendril is a thread-like, green, leafless, spirally coiled structure sensitive to touch. These help in the climbing of weak stem.

MathematicsGeneral ScienceMaharashtra Board Solutions

**Solution 1:**

**Solution 2:**

**Solution 3:**

**Solution 1:**

Steps of construction:

- Draw seg PQ of length 4 cm.
- Using a protractor, draw a ray PX ⊥ PQ at point P and a ray QY ⊥ PQ at point Q.
- Placing the point of the compass on point P and taking a radius of 2 cm, draw an arc of circle to cut AX at point S. Taking the same radius and placing the point of the compass on point Q, draw an arc of circle to cut AY at point R
- Draw seg RS.

Thus a rectangle PQRS with PQ = 4 cm and RS = 2 cm is constructed.

**Solution 2:**

Steps of construction:

- Draw seg LM of length 5.5 cm.
- Using a protractor, draw a ray LX ⊥ LM at point L and a ray MY ⊥ LM at point M.
- Placing the point of the compass on point L and taking a radius of 3.5 cm, draw an arc of circle to cut LX at point P. Taking the same radius and placing the point of the compass on point M, draw an arc of circle to cut MY at point N.
- Draw seg NP.

Thus a rectangle LMNP with LM = 5.5 cm and MN = 3.5 cm is constructed.

**Solution 3:**

Steps of construction:

- Draw seg WX of any length, say, 4.5 cm.
- Using a protractor, draw a ray WA ⊥ WX at point W and a ray XB ⊥ WX at point X.
- Placing the point of the compass on point W and taking a radius of 2.5 cm, draw an arc of circle to cut WA at point Z. Taking the same radius and placing the point of the compass on point X, draw an arc of circle to cut XB at point Y.
- Draw seg YZ.

Thus a rectangle WXYZ with WX = 4.5 cm and WZ = 2.5 cm is constructed.

MathematicsGeneral ScienceMaharashtra Board Solutions

**Solution 1:**

- The measure of the place occupied by an object in space is called its
__volume__. - The standard units of length are
__centimeter__and__metre__. - The standard units of area are
__sq. cm__and__sq. m__. - The cube whose length, breadth and height are each 1 cm has a volume of 1
__cu. cm__. - The cube whose length, breadth and height are each 1 m has a volume of 1
__cu. m__.

**Solution 1:**

- Cupboard
- Matchbox
- Cake of soap
- Book
- Trunk

**Solution 2:**

- Each face of a cube is
__square in shape__. - A cuboid has
__12__edges. - A cube has
__6__face altogether. - A cuboid has altogether
__8__vertices. - The standard units of volume are
__cu. cm__and__cu. m__. - All edges of a cube are of
__equal__length.

**Solution 3:**

- Vertices: P, Q, R, S, W, T, U, V
- Edges: seg PQ, seg QR, seg RS, seg SP, seg PW, seg WT, seg TU, seg UV, seg SV, seg RU, seg QT, seg VW
- Face: Square PQRS, square QRTU, square WTVU, square PWSV, square PQWT and square SRVU
- Vertex R is common to seg RQ, seg RS and seg RU.
- Vertex W is common to the seg WT, seg WV and seg WP.
- The edges PW and WT intersect at vertex W.
- Square PQRS is opposite to the square TUVW.

**Solution 1:**

- Volume of cube = (l)
^{3}= (26)^{3}= 17576 cu. cm - Volume of cube = (l)
^{3}= (2.6)^{3}= 17.576 cu. cm - Volume of cube = (l)
^{3}= (3.9)^{3}= 59.319 cu. cm - Volume of cube = (l)
^{3}= (12.5)^{3}= 1953.125 cu. cm - Volume of cube = (l)
^{3}= (13.2)^{3}= 2299.968 cu. cm - Volume of cube = (l)
^{3}= (24.3)^{3}= 14348.907 cu. cm - Volume of cube = (l)
^{3}= (9.7)^{3}= 912.673 cu. cm - Volume of cube = (l)
^{3}= (10.3)^{3}= 1092.727 cu.cm

**Solution 2:**

Side of the cube (l) = 2.5 cm.

Volume of the cubic die = Volume of a die = (l)^{3}

= (2.5)^{3}

= 15.625 cu. cm

**Solution 3:**

Side of the cube (l) = 6 m

Volume of water in a cube-shaped tank = Volume of cube = (l)^{3}

= (6)^{3}

= 216 cu. m

Thus, the cube-shaped tank can hold 216 cu. m of water.

**Solution 4:**

Side of the cube-shaped box (l) = 1.9 cm

Volume of the cube-shaped box = (l)^{3}

= (1.9)^{3}

= 6.859 cu. cm

Thus, the volume of the cube-shaped box is 6.859 cu. cm.

**Solution 1:**

1. Total surface area of a cube = 6l^{2}

= 6(3)^{2}

= 6 × 3 × 3

= 54 sq. cm

2. Total surface area of a cube = 6l^{2}

= 6(5)^{2}

= 6 × 5 × 5

= 150 sq. cm

3. Total surface area of a cube = 6l^{2}

= 6(7.2)^{2}

= 6 × 7.2 × 7.2

= 311.04 sq. m

4. Total surface area of a cube = 6l^{2}

= 6(6.8)^{2}

= 6 × 6.8 × 6.8

= 277.44 sq. m

5. Total surface area of a cube = 6l^{2}

= 6(9.3)^{2}

= 6 × 9.3 × 9.3

= 518.94 sq. cm

6. Total surface area of a cube = 6l^{2}

= 6(5.8)^{2}

= 6 × 5.8 × 5.8

= 201.84 sq. cm

7. Total surface area of a cube = 6l^{2}

= 6(8.6)^{2}

= 6 × 8.6 × 8.6

= 443.76 sq. cm

**Solution 2:**

Side of the cube (l) = 5.5 cm

Total surface area = 6l^{2}

= 6(5.5)^{2}

= 6 × 5.5 × 5.5

= 181.5 sq. cm

**Solution 3:**

Side of the safe (l) = 0.5 m

Total surface area = 6l^{2}

= 6(0.5)^{2}

= 6 × 0.5 × 0.5

= 1.5 sq. m

Cost of painting 1 sq. m = Rs. 60

∴ Cost of painting 1.5 sq. m = Rs. (60 × 1.5)

= Rs. 90

Thus, the cost to paint all sides of the safe is Rs. 90.

**Solution 4:**

**Solution 5:**

**Solution 6:**

**Solution 7:**

**Solution 1:**

1. l = 14 cm, b = 12 cm, h = 8 cm

Volume of a cuboid = l × b × h

= 14 × 12 × 8

= 1344 cu. cm

2. l = 20.5 cm, b = 16 cm, h = 10 cm

Volume of a cuboid = l × b × h

= 20.5 × 16 × 10

= 3280 cu. cm

3. l = 7.5 cm, b = 5.2 cm, h = 4.5 cm

Volume of a cuboid = l × b × h

= 7.5 × 5.2 × 4.5

= 175.5 cu. cm

4. l = 1.4 cm, b = 1.1 cm, h = 0.6 cm

Volume of a cuboid = l × b × h

= 1.4 × 1.1 × 0.6

= 0.924 cu. cm

5. l = 2.2 cm, b = 1.5 cm, h = 0.9 cm

Volume of a cuboid = l × b × h

= 2.2 × 1.5 × 0.9

= 2.97 cu. cm

**Solution 2:**

**Solution 3:**

**Solution 4:**

**Solution 5:**

l = 1.8 km = 1.8 × 1000 = 1800 m, b = 8 m,

h = 15 cm = (15 ÷ 100) m = 0.15 m

Volume of the required metal = l × b × h

= 1800 × 8 × 0.15

= 2160 cu. m

Thus, the required volume of the metal is 2160 cu. m.

**Solution 6:**

l = 7.5 m, b = 2.4 m, h = 3 m

Volume of the tank = l × b × h

= 7.5 × 2.4 × 3

= 54 cu. m

Thus, the tank will hold 54 cu. m of water.

**Solution 7:**

**Solution 1:**

l = 1.5 m, b = 1.2 m, h = 1.3 m

Total surface area of the trunk

= 2(l × b + b × h + h × l)

= 2(1.5 × 1.2 + 1.2 × 1.3 + 1.3 × 1.5)

= 2(1.80 + 1.56 + 1.95)

= 2 × 5.31

= 10.62 sq. m

**Solution 2:**

l = 12 cm, b = 10 cm, h = 5 cm

Metal sheet required

= 2(l × b + b × h + h × l)

= 2(12× 10 + 10 × 5 + 5 × 12)

= 2(120 + 50 + 60)

= 2 × 230

= 460 sq. cm

**Solution 3:**

l = 4 cm, b = 2.5 cm, h = 1.5 cm

Paper required = Total surface area of the matchbox

= 2(l × b + b × h + h × l)

= 2(4 × 2.5 + 2.5 × 1.5 + 1.5 × 4)

= 2(10.00 + 3.75 + 6.0)

= 2 × 19.75

= 39.5 sq. cm

**Solution 4:**

l = 2.5 m, b = 2 m, h = 2.4 m

Metal sheet required

= 2(l × b + b × h + h × l)

= 2(2.5 × 2 + 2 × 2.4 + 2.4 × 2.5)

= 2(5 + 4.8 + 6)

= 2 × 15.8

= 31.6 sq. m

Cost of constructing 1 sq. m = Rs. 10

∴ Cost of constructing 31.6 sq. m = Rs. (31.6 × 10)

= Rs. 316

Volume of the tank = l × b × h

= 2.5 × 2 × 2.4

= 12 cu. m

MathematicsGeneral ScienceMaharashtra Board Solutions

**Solution 1:**

- Side (l) = 2 m

Volume of the cube = l^{3}

= 2^{3}

= 2 × 2 × 2

= 8 cu. m

Side (l) = 5 m

Volume of the cube = l^{3}

= 5^{3}

= 5 × 5 × 5

= 125 cu. m

Side (l) = 8 cm

Volume of the cube = l^{3}

= 8^{3}

= 8 × 8 × 8

= 512 cu. cm

Side (l) = 4 cm

Volume of the cube = l^{3}

= 4^{3}

= 4 × 4 × 4

= 64 cu. cm

Side (l) = 10 m

Volume of the cube = l^{3}

= 10^{3}

= 10 × 10 × 10

= 1000 cu. cm

**Solution 2:**

Side of the room (l) = 4 m

Volume of the room = l^{3}

= 4^{3}

= 4 × 4 × 4

= 64 cu. m

The volume of the room is 64 cu. m,

∴ The room can hold 64 cu. m of air.

**Solution 3:**

**Solution 1:**

- Volume of a cuboid =
__Length__×__breadth__×__height__ - Forty cubes just fit into a cuboid box. Each side of the cube is 1 cm. Hence, the volume of the box is
__40 cc__.

**Solution 2:**

Length of the box (l) = 5 cm

Breadth of the box (b) = 3 cm

Height of the box (h) = 1 cm

Volume of the matchbox = l × b × h

= 5 × 3 × 1

= 15 cu. cm

Thus, the volume of the matchbox is 15 cu. cm.

**Solution 3:**

Length of the water tank (l) = 5 m

Breadth of the water tank (b) = 3 m

Height of the water tank (h) = 1 m

Volume of the water tank = l × b × h

= 5 × 3 × 1

= 15 cu. m

Thus, the volume of the water tank is 15 cu. m

**Solution 4:**

Length of the ditch (l) = 5 m

Breadth of the ditch (b) = 2.5 m

Depth of the ditch (h) = 1.5 m (Depth is same as height)

Volume of the ditch = l × b × h

= 5 × 2.5 × 1.5

= 18.75 cu. m

Volume of soil taken out = Volume of the ditch

∴ Volume of soil taken out = 18.75 cu. m

Thus, 18.75 cu. m of soil will be taken out while digging the ditch.

**Solution 5:**

1 cubic metre = 1 metre × 1 metre × 1 metre

1 metre = 100 cm

∴ 1 cubic metre = (100 × 100 × 100) cubic centimetres

= 10,00,000 cubic centimetres1 cubic metre = 1 metre × 1 metre × 1 metre

**Solution 6:**

Length of the tank (l) = 2.5 m

Breadth of the tank (b) = 2 m

Height of the tank (h) = 3 m

Volume of the tank = l × b × h

= 2.5 × 2 × 3

= 15 cu. m

Volume of water which the tank can hold = Volume of the tank.

Thus, the volume of water which the tank can hold is 15 cu. m.

**Solution 7:**

Length of the storage bin (l) = 2 m

Breadth of the storage bin (b) = 1.2 m

Height of the storage bin (h) = 1.8 m

Volume of the storage bin = l × b × h

= 2 × 1.2 × 1.8

= 4.32 cu. m

Thus, the volume of the storage bin is 4.32 cu. m.

**Solution 8:**