in ch the last leaf
Find hard words from the chapter (at least 30 words) and write two
synonyms for each.
Answers
Answer:
GIVEN :-
A solid toy is in the form of a hemisphere surmounted by a right circular cone.
The height of the cone is 4 cm.
The diameter of the base is 8cm.
TO FIND :-
The volume of the toy.
If a cube circumscribes the toy, then find the difference of the volumes of the cube and the toy.
The total surface area of the toy.
Formula Used :-
\longmapsto \boxed{ \green{ \bf \: Volume_{(cone)} = \dfrac{1}{3} \pi \: {r}^{2}h }}⟼
Volume
(cone)
=
3
1
πr
2
h
\longmapsto \boxed{ \green{ \bf \:CSA_{(cone)} = \pi \: rl }}⟼
CSA
(cone)
=πrl
\longmapsto \boxed{ \green{ \bf \: Volume_{(hemisphere)} = \dfrac{2}{3} \pi \: {r}^{3} }}⟼
Volume
(hemisphere)
=
3
2
πr
3
\longmapsto \boxed{ \green{ \bf \: CSA_{(hemisphere)} = 2\pi \: {r}^{2} }}⟼
CSA
(hemisphere)
=2πr
2
\longmapsto \boxed{ \green{ \bf \:Volume_{(cube)} = {(side)}^{3} }}⟼
Volume
(cube)
=(side)
3
where,
r = radius
h = height of cone
l = slant height of cone
CALCULATION :-
1. VOLUME OF TOY
Given that
Radius of cone, r = 4 cm
Height of cone, h = 4 cm
Radius of hemisphere, r = 4 cm
So,
\pink{\rm :\implies\:Volume_{(toy)} = Volume_{(cone)} + Volume_{(hemisphere)}}:⟹Volume
(toy)
=Volume
(cone)
+Volume
(hemisphere)
\rm :\implies\:Volume_{(toy)} = \dfrac{1}{3} \pi \: {r}^{2}h + \dfrac{2}{3} \pi \: {r}^{3}:⟹Volume
(toy)
=
3
1
πr
2
h+
3
2
πr
3
\rm :\implies\:Volume_{(toy)} = \dfrac{1}{3} \pi \: {r}^{2}(h + 2r):⟹Volume
(toy)
=
3
1
πr
2
(h+2r)
\rm :\implies\:Volume_{(toy)} = \dfrac{1}{3} \times \dfrac{22}{7} \times {4}^{2} \times (4 + 8):⟹Volume
(toy)
=
3
1
×
7
22
×4
2
×(4+8)
\rm :\implies\:Volume_{(toy)} = \dfrac{1}{ \cancel3} \times \dfrac{22}{7} \times 16 \times \cancel{12} \: \: \: ^{4} :⟹Volume
(toy)
=
3
1
×
7
22
×16×
12
4
\rm :\implies\: \boxed{ \red{ \bf \: Volume_{(toy)} = \dfrac{1408}{7} \: {cm}^{3} }}:⟹
Volume
(toy)
=
7
1408
cm
3
2. SURFACE AREA OF TOY
Given that
Radius of cone, r = 4 cm
Height of cone, h = 4 cm
Radius of hemisphere, r = 4 cm
So,
Slant (l) height of cone is given by
\longmapsto \boxed{ \green{ \bf \: {l}^{2} = {h}^{2} + {r}^{2} }}⟼
l
2
=h
2
+r
2
\rm :\implies\: {l}^{2} = {4}^{2} + {4}^{2} :⟹l
2
=4
2
+4
2
\rm :\implies\: {l}^{2} = 16 + 16:⟹l
2
=16+16
\rm :\implies\: {l}^{2} = 32:⟹l
2
=32
\rm :\implies\:l = \sqrt{32} = 4 \sqrt{2} \: cm:⟹l=
32
=4
2
cm
\rm :\implies\:l \: = \: 4 \times 1.414:⟹l=4×1.414
\rm :\implies\:l \: = \: 5.656 \: cm:⟹l=5.656cm
Now,
Surface Area of toy is given by
\rm :\implies\: \pink{ \bf \: Surface Area_{(toy)} =CSA_{(cone)} + CSA_{(hemisphere)} }:⟹SurfaceArea
(toy)
=CSA
(cone)
+CSA
(hemisphere)
\rm :\implies\:Surface Area_{(toy)} = \pi \: rl \: + \: 2\pi \: {r}^{2} :⟹SurfaceArea
(toy)
=πrl+2πr
2
\rm :\implies\:Surface Area_{(toy)} = \pi \: r(l \: + \: 2r):⟹SurfaceArea
(toy)
=πr(l+2r)
\rm :\implies\:Surface Area_{(toy)} = \dfrac{22}{7} \times 4 \times (5.656 + 8):⟹SurfaceArea
(toy)
=
7
22
×4×(5.656+8)
\rm :\implies\:Surface Area_{(toy)} = \dfrac{22}{7} \times 4 \times 13.656:⟹SurfaceArea
(toy)
=
7
22
×4×13.656
\longmapsto \boxed{ \green{ \bf \:Surface Area_{(toy)} = 171.68 \: {cm}^{2} }}⟼
SurfaceArea
(toy)
=171.68cm
2
Now,
According to statement,
The toy is circumscribes by a cube.
So,
Edge of the cube = height of cone + height of hemisphere
So,
Edge of cube, x = 4 + 4 = 8 cm
\rm :\implies\: \pink{ \bf \: Volume_{(cube)} = {x}^{3} }:⟹Volume
(cube)
=x
3
\rm :\implies\:Volume_{(cube)} = {8}^{3} :⟹Volume
(cube)
=8
3
\longmapsto \boxed{ \green{ \bf \: Volume_{(cube)} = 512 \: {cm}^{3} }}⟼
Volume
(cube)
=512cm
3
Now,
Difference in the volume of cube and volume of toy is
\bull\purple{ \bf \: Volume_{(difference)} = Volume_{(cube)} - Volume_{(toy)}}∙Volume
(difference)
=Volume
(cube)
−Volume
(toy)
\rm :\implies\:Volume_{(difference)} =512 - \dfrac{1408}{7} :⟹Volume
(difference)
=512−
7
1408
\rm :\implies\:Volume_{(difference)} =\dfrac{3584 - 1408}{7} :⟹Volume
(difference)
=
7
3584−1408
\rm :\implies\:Volume_{(difference)} =\dfrac{2176}{7} :⟹Volume
(difference)
=
7
2176
\longmapsto \boxed{ \green{ \bf \:Volume_{(difference)} =310.86 \: {cm}^{3} }}⟼
Volume
(difference)
=310.86cm
3