1
Numerical

JEE Main 2021 (Online) 25th July Evening Shift

If a rectangle is inscribed in an equilateral triangle of side length $$2\sqrt 2 $$ as shown in the figure, then the square of the largest area of such a rectangle is _____________.

Your Input ________

Answer

Correct Answer is 3

Explanation



In $$\Delta$$DBF

$$\tan 60^\circ = {{2b} \over {2\sqrt 2 - l}} \Rightarrow b = {{\sqrt 3 \left( {2\sqrt 2 - l} \right)} \over 2}$$

A = Area of rectangle = l $$\times$$ b

$$A = l \times {{\sqrt 3 } \over 2}\left( {2\sqrt 2 - l} \right)$$

$${{dA} \over {dl}} = {{\sqrt 3 } \over 2}\left( {2\sqrt 2 - l} \right) - {{l.\sqrt 3 } \over 2} = 0$$

$$l = \sqrt 2 $$

$$A = l \times b = \sqrt 2 \times {{\sqrt 3 } \over 2}\left( {\sqrt 2 } \right) = \sqrt 3 $$

$$\Rightarrow$$ A2 = 3
2
Numerical

JEE Main 2021 (Online) 16th March Evening Shift

In $$\Delta$$ABC, the lengths of sides AC and AB are 12 cm and 5 cm, respectively. If the area of $$\Delta$$ABC is 30 cm2 and R and r are respectively the radii of circumcircle and incircle of $$\Delta$$ABC, then the value of 2R + r (in cm) is equal to ___________.
Your Input ________

Answer

Correct Answer is 15

Explanation



Area = $${1 \over 2}(5)(12)\sin \theta = 30$$

$$\sin \theta = 1 \Rightarrow \theta = {\pi \over 2}$$

$$\Delta$$ is right angle $$\Delta$$



$$r = (s - a)\tan {A \over 2}$$

$$r = (s - a)$$$$\tan {{90} \over 2}$$

$$r = (s - a)$$

$$2R + r = s$$, (As $$a = 2R$$)

$$2R + r = {{5 + 12 + 13} \over 2} = 15$$
3
Numerical

JEE Main 2021 (Online) 16th March Morning Shift

Let ABCD be a square of side of unit length. Let a circle C1 centered at A with unit radius is drawn. Another circle C2 which touches C1 and the lines AD and AB are tangent to it, is also drawn. Let a tangent line from the point C to the circle C2 meet the side AB at E. If the length of EB is $$\alpha$$ + $${\sqrt 3 }$$ $$\beta$$, where $$\alpha$$, $$\beta$$ are integers, then $$\alpha$$ + $$\beta$$ is equal to ____________.
Your Input ________

Answer

Correct Answer is 1

Explanation



(i) $$\sqrt 2 r + r = 1$$

$$r = {1 \over {\sqrt 2 + 1}}$$

$$r = \sqrt 2 - 1$$

(ii) $$C{C_2} = 2\sqrt 2 - 2 = 2\left( {\sqrt 2 - 1} \right)$$

From $$\Delta C{C_2}N = \sin \phi = {{\sqrt 2 - 1} \over {2\left( {\sqrt 2 - 1} \right)}}$$

$$\phi = 30^\circ $$

(iii) In $$\Delta$$ACE apply sine law

$${{AE} \over {\sin \phi }} = {{AC} \over {\sin 105^\circ }}$$

$$AE = {1 \over 2} \times {{\sqrt 2 } \over {\sqrt 3 + 1}}.2\sqrt 2 $$

$$AE = {2 \over {\sqrt 3 + 1}} = \sqrt 3 - 1$$

$$ \therefore $$ $$EB = 1 - \left( {\sqrt 3 - 1} \right)$$

= $$2 - \sqrt 3 $$

$$ \therefore $$ $$\alpha$$ = 2, $$\beta$$ = $$-$$1 $$ \Rightarrow $$ $$\alpha$$ + $$\beta$$ = 1
4
Numerical

JEE Main 2020 (Online) 6th September Morning Slot

The angle of elevation of the top of a hill from a point on the horizontal plane passing through the foot of the hill is found to be 45o. After walking a distance of 80 meters towards the top, up a slope inclined at an angle of 30o to the horizontal plane, the angle of elevation of the top of the hill becomes 75o. Then the height of the hill (in meters) is ______.
Your Input ________

Answer

Correct Answer is 80

Explanation



sin 30o = $${x \over {80}}$$ $$ \Rightarrow $$ x = 40

cos 30o = $${y \over {80}}$$ $$ \Rightarrow $$ y = $$40\sqrt 3 $$

Now, In $$\Delta $$AEF

tan 75o = $${{h - x} \over {h - y}}$$

$$ \Rightarrow $$ 2 + $$\sqrt 3 $$ = $${{h - 40} \over {h - 40\sqrt 3 }}$$

$$ \Rightarrow $$ $$\left( {2 + \sqrt 3 } \right)\left( {h - 40\sqrt 3 } \right)$$ = h - 40

$$ \Rightarrow $$ 2h - 80$${\sqrt 3 }$$ + $${\sqrt 3 }$$h - 120 = h - 40

$$ \Rightarrow $$ h + $${\sqrt 3 }$$h = 80 + 80$${\sqrt 3 }$$

$$ \Rightarrow $$ ($${\sqrt 3 }$$ + 1)h = 80($${\sqrt 3 }$$ + 1)

$$ \Rightarrow $$ h = 80 m

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