In a parallelogram PQRS, if ∠QRS = 2x, ∠PQS = 4x and ∠PSQ = 4x, find
the angles of the parallelogram.
Answers
Answer:
In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 Pdf, These solutions are solved subject experts from the latest edition books.
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1
Unless stated otherwise, take π = 227
Question 1.
2 cubes each of volume 64 cm3 are joined end to end. Find the surface area of the resulting cuboid.
Solution:
Volume of each cube = 64 cm3
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 1
Let the edge of each cube = x
∴ x3 = 64 cm3
⇒ x = 4 cm
Now, Length of the resulting cuboid (l) = 2x cm = 8 cm
Breadth of the resulting cuboid (b) = x cm = 4 cm
Height of the resulting cuboid (h) = x cm = 4 cm
∴ Surface area of the cuboid = 2 (lb + bh + hl)
= 2 [(8 × 4) + (4 × 4) + (4 × 8)] cm2
= 2 [32 + 16 + 32] cm2 = 2 [80] cm2
= 160 cm2.
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1
Question 2.
A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.
Solution:
For hemispherical part,
radius (r)= 142 = 7cm
∴ Curved surface area = 2πr2
= 2 × 227 × 7 × 7cm2
= 308cm2 7
Total height of vessel = 13 cm
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 2
∴ Height of cylinder = (13 – 7)cm = 6 cm and radius(r) = 7 cm
∴ Curved surface area of cylinder = 2πrh
= 2 × 227 × 7 × 6cm2 = 264cm2 7
∴ Inner surface area of vessel = (308 + 264)cm2 = 572 cm2
Question 3.
A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.
Solution:
Let h be the height of cone and r be the radius of cone and hemisphere.
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 3
∴ h = [height of toy – radius of hemi sphere]
= (15.5 – 3.5) cm = 12 cm
Also l2 = h2 + r2 = 122 + (3.5)2 = 156.25 cm2
∴ l = 12.5 cm
Curved surface area of the conical part = πrl
Curved surface area of the hemispherical part = 2πr2
∴ Total surface area of the toy = πrl + 2πr2
= πr(l + 2 r)
= 227×3510 (12.5 + 2 × 3.5) cm2
= 11 × (12.5 + 7) cm2 = 11 × 19.5 cm2
= 214.5 cm2
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1
Question 4.
A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.
Solution:
Let side of the block, l = 7 cm
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 4
∴ The greatest diameter of the hemisphere = 7 cm
Surface area of the solid = [Total surface area of the cubical block] + [C.S.A. of the hemisphere] – [Base area of the hemisphere]
= 6 × l2 + 2πr2 – πr2
[where l = 7 cm and r = 72 cm]
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 5
Question 5.
A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter l of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.
Solution:
Let l be the side of the cube.
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 6
∴ Diameter of the hemisphere = l
⇒ Radius of the hemisphere (r) = l2
Curved surface area of hemisphere = 2πr2
= 2 × π × l2×l2=πl22
Base area of the hemisphere = πr2
= π(l2)2=πl24
Surface area of the cube = 6 × l2 = 6l2
∴ Surface area of the remaining solid = [Total surface area of cube + C.S.A. of hemisphere – base area of hemisphere]
MP Board Class 10th Maths Solutions Chapter 13 Surface Areas and Volumes Ex 13.1 7
Answer:
60
Step-by-step explanation:
4x+x+x=180
6x=180
x=180÷6
x=30
4x=4×30
2x=2×30
=60
here is ur answer