The correct answer is: 2 Explanation: Given, ​DE∥AB ∴ $ \displaystyle \dfrac{{\text{CD}}}{{\text{AD}}}=\dfrac{{\text{CE}}}{{\text{BE}}}$​​ [by basic proportionality theorem] ⇒ $ \displaystyle \dfrac{{x+3}}{{3x+19}}=\dfrac{x}{{3x+4}}$ ⇒(x + 3)(3x + 4) = x(3x + 19) ⇒ 3$ \displaystyle {{x}^{2}}$ + 4x + 9x + 12 = 3$ \displaystyle {{x}^{2}}$ + 19x 19x − 13x = 12 6x = 12 x = $ […]

The correct answer is: 40 m Explanation: Also, let the distance between the tower (BD) and the cliff (AC) be d m. Thus, in △ABC, tan x = $\displaystyle \dfrac{{20}}{d}$ ​……(i) Also, in △CDE, tan x = $\displaystyle \dfrac{h}{d}$ ​………(ii) From equations (i) and (ii) $\displaystyle \dfrac{h}{d}$ ​= $\displaystyle \dfrac{{20}}{d}$​ ⇒h  = 20 m ⇒ Height of the

In △ADC and △BAC, ∠ADC = ∠BAC (Already given) ∠ACD = ∠BCA (Common angles) ∴ △ADC ∼ △BAC (AA similarity criterion) We know that corresponding sides of similar triangles are in proportion. $ \displaystyle \dfrac{{\text{CA}}}{{\text{CB}}}=\dfrac{{\text{CD}}}{{\text{CA}}}$ ​∴ $ \displaystyle {\Rightarrow \text{C}{{\text{A}}^{2}}=\text{CB}\text{.CD}\text{.}}$  Hence, proved.

The correct answer is: 21.7 m Explanation: $\displaystyle \dfrac{x}{{20\sqrt{3}}}$= tan30° = $\displaystyle \dfrac{1}{√3}$  ​⇒ x = 20 m ∴AB = 21.7 m

The correct answer is: ₹1925 Explanation: Length of the fence (in meters) = $\displaystyle \dfrac{Total cost}{Rate}$ = $\displaystyle \dfrac{5820}{24}$ = 220 So, circumference of the the field  = 220 m Therefore, if r meters is the radius of the field , then 2πr = 220 Or, 2 × $\displaystyle \dfrac{22}{7}$​× r = 220  or, r =

The correct answer is: $\displaystyle \dfrac{P}{720}$​ 2πR² Explanation: Sector angle is p in degrees Radius of the Circle = R Area of the sector = $\displaystyle \dfrac{πR²}{360°}$ = ​$\displaystyle \dfrac{(πR²p)²}{720°}$​ = $\displaystyle \dfrac{P}{720}$​ 2πR²

The correct answer is: Sublimation Explanation: Radius of bigger circle $ \displaystyle {{{r}_{1}}}$​ = OA = 7 Radius of smaller circle $ \displaystyle {{{r}_{2}}}$ = $ \displaystyle \dfrac{{{{r}_{1}}}}{2}$ = $ \displaystyle \dfrac{7}{2}$​ Area of the bigger circle $ \displaystyle {{{C}_{1}}}$​ = $ \displaystyle {\pi r_{1}^{2}}$​ ​= $ \displaystyle \dfrac{22}{7}$ ​× 7 × 7 = 154 $ \displaystyle \text{c}{{\text{m}}^{2}}$​

Sublimation Explanation: Given, in △ABC, circle touch the triangle at point D, F and E respectively and let the lengths of the segment AF be x. So, BF = BD = 6 cm [Tangent drawn from same point are equal] CE = CD = 9 cm [Tangent drawn from same point are equal ] and AE =

The correct answer is: 22 cm, 231 cm² Explanation: Let AB be the given arc subtending an angle of 60° at the centre. Here, r = 21cm and θ = 60° $\displaystyle \begin{array}{l}\text{Length of the arc ABC}=\dfrac{{2\pi r\theta {}^\circ }}{{360{}^\circ }}cm\\=\left( {2\times \dfrac{{22}}{7}\times 21\times \dfrac{{60{}^\circ }}{{360{}^\circ }}} \right)cm=\text{ }22cm\\\text{Area of the sector OACBO}=\dfrac{{\pi {{r}^{2}}\theta {}^\circ }}{{360{}^\circ

3293.5 $ \displaystyle {{\text{m}}^{2}}$ Explanation: Length of rectangular field = L = 70 m Breadth of rectangular field = B = 52 m So, Area of the field = L × B = 70 × 52 = 3640 Area of the field is 3640 $ \displaystyle {{\text{m}}^{2}}$ And, Area it can graze = Area of quadrant of