Yaadannoo boqonnaa fi gaaffii shaakalaa
Wanti fuula kana irra jiru hundi boqonnaa 6 kitaaba Herrega kutaa 10 tiif: wanta boqonnaan haguugu, yaadannoo gabaabaa, gaaffii shaakalaa deebii isaanii waliin, fi viidiyoo barnootaa bakka jirutti.
Kaardiiwwan fuula Herrega kutaa 10 irratti boqonnaa kana jala galmaa'an. Tokko tokkoon bani.
Yoom akka itti fayyadamtu
Bal'ina cross-section dalgee fageenya vertex irraa beekamu irratti argamu shallaguuf itti fayyadami.
Of eeggadhu
Fageenyi k vertex irraa shallagama, base irraa miti. Yoo base irraa kenname dura h irraa hir'isi.
Grade 10 Mathematics, fuula 272-327 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e
Yoom akka itti fayyadamtu
Bal'ina fuula cinaacha frustum koonii qajeelaa shallaguuf kanatti fayyadami.
Of eeggadhu
Foormulaan kun lateral area qofa kenna. Total surface area argachuuf bal'ina base koonii lamaanii itti dabali.
Grade 10 Mathematics, fuula 272-327 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e
Yoom akka itti fayyadamtu
Bal'ina fuula waligalaa kan soliidota walitti makamanii shallagdu kana yaadadhu.
Of eeggadhu
Total surface area soliidota dhuunfaa qofa walitti hin dabalin. Fuulotni keessoo wal tuqan waan dhokataniif lakkaa'amuu hin qaban.
Grade 10 Mathematics, fuula 272-327 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e
Yoom akka itti fayyadamtu
Bal'inni base gubbaa fi jalaa akkasumas altitude yoo beekaman volume frustum piiraamiidii shallaguuf fayyadami.
Of eeggadhu
Hasaaba sqrt(A A') bracket keessatti dabaluu hin dagatin.
Grade 10 Mathematics, fuula 272-327 irraa qophaa'e; osoo hin ba'in dura yeroo lama mirkanaa'e
Gaaffii filannoo 10 boqonnaa kana irratti barreeffaman, tokkoon tokkoon isaanii deebii fi sababa qabu. Deebii banuu dura yaali. Gaaffiin kun Ingiliffaa tiin.
Gaaffii 1
A right circular cylinder has a base radius of 3 cm and an altitude of 5 cm. What is its volume?
Deebii: A 45π cm³
Sababa
The volume of a right circular cylinder is V = πr²h. Here r = 3 and h = 5, so V = π × 3² × 5 = π × 9 × 5 = 45π cm³.
Gaaffii 2
A right prism has a base that is a right triangle with legs 3 cm and 4 cm. The altitude of the prism is 6 cm. What is its volume?
Deebii: B 36 cm³
Sababa
First find the base area: BA = ½ × 3 × 4 = 6 cm². The volume of a prism is V = (BA)h, so V = 6 × 6 = 36 cm³.
Gaaffii 3
A sphere has a radius of 10 cm. What is its surface area?
Deebii: B 400π cm²
Sababa
The surface area of a sphere is SA = 4πr². With r = 10, SA = 4π × 10² = 4π × 100 = 400π cm².
Gaaffii 4
A conical tent for a wedding in Gondar is 6 m high. The radius of its base is 8 m. What is the slant height of the tent?
Deebii: D 10 m
Sababa
The slant height of a right circular cone is l = √(h² + r²). Here h = 6 and r = 8, so l = √(6² + 8²) = √(36 + 64) = √100 = 10 m.
Gaaffii 5
A regular square pyramid has a base edge of 4 cm and a slant height of 5 cm. What is its total surface area?
Deebii: C 56 cm²
Sababa
The perimeter of the base is p = 4 × 4 = 16 cm. The lateral surface area is LSA = ½pl = ½ × 16 × 5 = 40 cm². The base area is 4 × 4 = 16 cm². So the total surface area is TSA = LSA + BA = 40 + 16 = 56 cm².
Gaaffii 6
A regular square pyramid has a base edge of 4 cm and an altitude of 7 cm. What is its volume?
Deebii: A 112/3 cm³
Sababa
The base is a square with area BA = 4 × 4 = 16 cm². The volume of a pyramid is one third of the base area times the altitude: V = ⅓ × 16 × 7 = 112/3 cm³.
Gaaffii 7
The base area of a pyramid is 270 cm² and its altitude is 6 cm. A horizontal cross-section is taken 4 cm from the vertex. What is the area of this cross-section?
Deebii: B 120 cm²
Sababa
For a cross-section at distance k from the vertex, the area of the cross-section divided by the base area equals k²/h². Here k = 4 and h = 6, so the ratio is 4²/6² = 16/36 = 4/9. The cross-section area is 270 × 4/9 = 120 cm².
Gaaffii 8
A frustum of a right circular cone has top and bottom diameters of 6 m and 10 m. Its height is 12 m. What is its volume? Give your answer in terms of π.
Deebii: D 196π m³
Sababa
The radii are r = 5 m (bottom) and r′ = 3 m (top), since the radius is half the diameter. The frustum volume is V = ⅓πh(r² + r′² + rr′) = ⅓π × 12 × (25 + 9 + 15) = ⅓π × 12 × 49 = 196π m³.
Gaaffii 9
The areas of the upper and lower bases of a frustum of a pyramid are 36 cm² and 81 cm². The altitude of the frustum is 5 cm. What is the altitude of the whole pyramid?
Deebii: C 15 cm
Sababa
Let k be the distance from the vertex to the upper base. The area ratio gives 36/81 = k²/(5 + k)², so 6/9 = k/(5 + k). Cross-multiplying: 6(5 + k) = 9k, so 30 = 3k and k = 10 cm. The altitude of the whole pyramid is 5 + k = 5 + 10 = 15 cm.
Gaaffii 10
A solid is made of a cylinder with radius 4 cm and height 7 cm, with a cone of altitude 3 cm on top. The cone has the same radius of 4 cm. The overall height of the solid is 10 cm. What is the total volume of the solid? Give your answer in terms of π.
Deebii: A 128π cm³
Sababa
The volume of the solid is the cylinder volume plus the cone volume. Cylinder: πr²h = π × 16 × 7 = 112π cm³. Cone: ⅓πr²h = ⅓π × 16 × 3 = 16π cm³. The total volume is 112π + 16π = 128π cm³.
Appiin boqonnaa kana irratti gaaffii 12 qaba; utuu hojjettuu ni sirreessa. Kan asii jiran 10 jalqabaa ti.
Boqonnaan 6 Maxxansa Ministeera Barnootaa keessatti fuula 271 hanga fuula 326 ti.
Maxxansa Ministeera Barnootaa, Ingiliffaa, Fuula 373
Temari boqonnaa kana gara shaakala yeroon daangeffametti jijjiira; utuu hojjettuu ni sirreessa; wanti si hin galinis barsiisaa AI tiin ni ibsama.
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