36 cm²

Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
\(\therefore B E=E C=\frac{B C}{2}=\frac{a}{2}\)
And, AE = \(\frac{a \sqrt{3}}{2}\)
Given that, BD = \(\frac{1}{3} B C\)
∴ BD = a/3
\(D E=B E-B D=\frac{a}{2}-\frac{a}{3}=\frac{a}{6}\)
Applying Pythagoras theorem in ΔADE, we get
AD² = AE² + DE²
\( A D^{2}=\left(\frac{a \sqrt{3}}{2}\right)^{2}+\left(\frac{a}{6}\right)^{2} \)
\(=\left(\frac{3 a^{2}}{4}\right)+\left(\frac{a^{2}}{36}\right) \)
\(=\frac{28 a^{2}}{36} \)
\(=\frac{7}{9} A B^{2} \)
\(\Rightarrow 9 A D^{2}=7 A B^{2} \)

Yes, ΔABC ∼ ΔPQR [by AAA similarity criterion]

Let the water level raised in cylindrical vessel be h cm
 Volume of Sphere = Volume of water displaced in cylinder 
\(\frac {4}{3}\pi\)(3)3 = \(\pi\)(6)²h
\(\frac {4}{3}\) x 27 = 36h
36=36h
h=1 cm

Let radius be r and the height be h.
Curved surface area of cylinder = 2πrh
Curved surface area of cone =πr\(\sqrt{r^{2}+h^{2}}\)
\(\frac{2\pi{rh}}{\pi{r}\sqrt{r^{2}+h^{2}}}\)=\(\frac{8}{5}\Rightarrow \frac{h}{\sqrt{r^{2}+h^{2}}}=\frac{4}{5}\)
⇒ 25h²=16(r²+h²)
⇒ 25h²=16r²+16h²
⇒ 9h²=16r²
⇒ \(\frac{9}{16}=\frac{r^{2}}{h^{2}}\Rightarrow \frac{r}{h}=\frac{3}{4}\)
=r:h=3:4

In \(\Delta\)ACB,    LM || CB         [given]
\(\Rightarrow \quad \frac{A M}{M B}=\frac{A L}{L C}\)    ....(i)
[ by basic proportionality theorem]
In \(\Delta\)ACD, LN ||CD    [given]
\(\Rightarrow \quad \frac{A N}{N D}=\frac{A L}{L C}\)    ....(ii)
[by basic proportionality theorem]
From Eqs. (i) and (ii), we get
\(\frac{A M}{M B}=\frac{A N}{N D} \Rightarrow \frac{M B}{A M}=\frac{N D}{A N}\)
[on taking reciprocal of the terms]
\(\Rightarrow \quad \frac{M B}{A M}+1=\frac{N D}{A N}\) + 1  [adding 1 on both sides]
\(\begin{aligned} & \Rightarrow \quad \frac{M B+A M}{A M}=\frac{N D+A N}{A N} \\ \end{aligned}\)

\(\begin{aligned} \Rightarrow \quad \frac{A M}{A M+M B}=\frac{A N}{A N+N D} \end{aligned}\)
[on taking reciprocal of the terms]
\(\therefore \quad \frac{A M}{A B}=\frac{A N}{A D} \quad\left[\begin{array}{c} \because A D=A N+N D \\ \text { and } A B=A M+M B \end{array}\right]\)
Hence proved.

Given, in \(\triangle ABC\), \(DE\parallel BC\)
By Thales theorem, we get
\(\frac { AD }{ DB } =\frac { AE }{ EC } \)

\(\Rightarrow \frac { 4x-3 }{ 3x-1 } =\frac { 8x-7 }{ 5x-3 } \)
[\(\because \)  AD = 4x - 3, DB = 3x - 1, AE = 8x - 7, EC = 5x - 3]
\(\Rightarrow \) (4x - 3)(5x - 3) = (8x - 7)(3x - 1)
\(\Rightarrow \)  20x² - 12x + 9 - 15x = 24x² - 21x - 8x + 7
\(\Rightarrow \)  4x² - 2x - 2 = 0
\(\Rightarrow \)  2x² - x - 1 = 0     [dividing both sides by 2]
\(\Rightarrow \)  2x² - 2x + x - 1 = 0   [by splitting the middle term]
\(\Rightarrow \)  2x(x - 1) + 1 (x - 1) = 0
\(\Rightarrow \)  (2x + 1) (x - 1) = 0   \(\therefore x=-\frac { 1 }{ 2 } \) or x = 1
If \(x=-\frac { 1 }{ 2 } \), then AD = \(4\times -\frac { 1 }{ 2 } -3=-5<0\)  [not possible]
Hence, x = 1 is the required value.

Since the mould is in the shape of a frustum of a cone, the quantity (volume) of molasses that can be poured into it \(=\frac{\pi}{3} h\left(r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\right)\)
where r₁ is the radius of the larger base and r₂ is the radius of the smaller base.
\(=\frac{1}{3} \times \frac{22}{7} \times 14\left[\left(\frac{35}{2}\right)^{2}+\left(\frac{30}{2}\right)^{2}+\left(\frac{35}{2} \times \frac{30}{2}\right)\right] \mathrm{cm}^{3}=11641.7 \mathrm{~cm}^{3} .\)
It is given that 1 cm3 of molasses has mass 1.2g. So, the mass of the molasses that can be poured into each mould = (11641.7 × 1.2) g
= 13970.04 g = 13.97 kg
= 14 kg (approx.)

173.25 cm³

Here, a cone and spherical lead shots are given. Since, lead shots are dropped into the vessel, so the water which flows out from the vessel is equal to the volume of lead shots.
Given, height of the vessel, h = 8 cm and radius of the vessel, r = 5 cm.
\(\therefore\) Volume of water filled in a vessel = Volume of cone
\(\begin{aligned} & =\frac{1}{3} \times \pi \times r^2 \times h \\ \end{aligned}\)
\(\begin{aligned} & =\frac{1}{3} \times \frac{22}{7} \times 5 \times 5 \times 8 \\ \end{aligned}\)
\(\begin{aligned} & =\frac{4400}{21} \mathrm{~cm}^3 \end{aligned}\)
Also, radius of a spherical lead shot = 0.5 cm
\(\therefore\) Volume of one spherical lead shot \(=\frac{4}{3} \times \pi \times r^3\)
\(=\frac{4}{3} \times \frac{22}{7} \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}=\frac{11}{21} \mathrm{~cm}^3\)
Let n be the number of lead shots dropped in the vessel, then according to the question,
n \(\times\) Volume of one spherical lead shot
= \(\frac{1}{4}\)\(\times\)Volume of water filled in a vessel
\(\begin{aligned} \Rightarrow & & n \times \frac{11}{21} & =\frac{4400}{21} \times \frac{1}{4} \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow & & n \times 11 & =1100 \\ \end{aligned}\)
\(\begin{aligned} \Rightarrow & & n & =100 \end{aligned}\)
Hence, required number of lead shots is 100.

The total surface area of the cube = 6 × (edge)² = 6 × 5 × 5 cm² = 150 cm².
Note that the part of the cube where the hemisphere is attached is not included in the surface area.
So, the surface area of the block = TSA of cube – base area of hemisphere + CSA of hemisphere
\(=150-\pi r^{2}+2 \pi r^{2}=\left(150+\pi r^{2}\right) \mathrm{cm}^{2} \)
\(=150 \mathrm{~cm}^{2}+\left(\frac{22}{7} \times \frac{4.2}{2} \times \frac{4.2}{2}\right) \mathrm{cm}^{2} \)
\(=(150+13.86) \mathrm{cm}^{2}=163.86 \mathrm{~cm}^{2}\)

We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.

We need to prove that \(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)
Let us join BE and CD and then draw \(\mathrm{DM} \perp \mathrm{AC}\) and \(E N \perp A B\)
Now, area of \(\Delta \mathrm{ADE}\left(=\frac{1}{2} \text { base } \times \text { height }\right)=\frac{1}{2} \mathrm{AD} \times \mathrm{EN}\)
Recall from Class IX, that area of \(\Delta\) ADE is denoted as ar(ADE).
So, \(\operatorname{ar}(\mathrm{ADE})=\frac{1}{2} \mathrm{AD} \times \mathrm{EN}\)
Similarly, \(\operatorname{ar}(\mathrm{BDE})=\frac{1}{2} \mathrm{DB} \times \mathrm{EN}\)
\(\operatorname{ar}(\mathrm{ADE})=\frac{1}{2} \mathrm{AE} \times \mathrm{DM} \text { and } \operatorname{ar}(\mathrm{DEC})=\frac{1}{2} \mathrm{EC} \times \mathrm{DM}\)
Therefore, \(\frac{\operatorname{ar}(\mathrm{ADE})}{\operatorname{ar}(\mathrm{B} \mathrm{DE})}=\frac{\frac{1}{2} \mathrm{AD} \times \mathrm{EN}}{\frac{1}{2} \mathrm{DB} \times \mathrm{EN}}=\frac{\mathrm{AD}}{\mathrm{DB}}\)       (1)
and \(\frac{\operatorname{ar}(\mathrm{ADE})}{\operatorname{ar}(\mathrm{DEC})}=\frac{\frac{1}{2} \mathrm{AE} \times \mathrm{DM}}{\frac{1}{2} \mathrm{EC} \times \mathrm{DM}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)                (2)
Note that \(\Delta\) BDE and DEC are on the same base DE and between the same parallels BC and DE.
So, ar(BDE) = ar(DEC)                                      (3)
Therefore, from (1), (2) and (3), we have:
\(\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)

Let O be the position of airport.

In \(1\frac{1}{2}\) h, the distance travelled by aeroplane when it files due North at a speed of 1000 km/h,
OA = 1000 x \(\frac{3}{2}\) = 1500 km.
[ \(\because\)  distance = speed x time]
In \(1\frac{1}{2}\) h, the distance travelled by aeroplane when it files due West at a speed of 1200 km/h, OB = 1200 x \(\frac{3}{2}\) = 1800 km
[ \(\because\)  distance = speed x time]
Since, North and West directions are perpendicular to each other.
So, using Pythagoras theorem, we get
AB² = OB² + OA² \(\Rightarrow\)  AB² = (1800)² + (1500)²
\(\Rightarrow\)  AB² = 3240000 + 2250000 = 5490000
\(\therefore\)  AB = 300 \(\sqrt{61}\) km    [taking positive square root]
Hence, the distance between two planes after \(1\frac{1}{2}\) h is 300 \(\sqrt{61}\) km.

Total surface area (TSA) of toy = CSA of hemisphere + CSA of cone
This is because the toy is obtained by joining the plane surfaces of hemisphere and cone.

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(i) (c): Pythagoras Theorem
(ii) (a): ADC is a right-angled triangle. By Pythagoras theorem we get AC² = (30)² + (40)²
⇒ AC² = 900 + 1600
⇒ AC² = 2500
⇒ AC = 50m
(iii) (d): (21, 20, 28)
since,\((28)^{2} \neq(20)^{2}+(21)^{2}\)
(iv) (b): AC = 50m and BC = 12m
⇒ AB = AC - BC
⇒ AB = 50 - 12 = 38m
(v) (c): Length of the rope used = 30 + 40 + 12
= 82m