Option (b)

Assume a 45-45-90 triangle

sin $\alpha$= P = $\frac{1}{\sqrt{2}}$ . (Sin= opposite side/ hypotenuse)

tan $\alpha$= Q = 1 (Tan = opposite side/ adjacent side)

$\frac{1}{P^2}-\frac{1}{Q^2}$ = 2-1 =1

Shortcut

$cose{c}^2\alpha$ – $co{t}^2\alpha$ = 1 is a trigonometric identity

If you are not aware of this identity, substitute $\alpha$ = 45.

Option (b)

Perimeter = 3a

Area = $\frac{\sqrt{3}}{4a^2}$

10x$\frac{\sqrt{3}}{4a^2}$= 25x3a

a=$\frac{30}{\sqrt{3}}$ = 17.32 m

Option (a)

Take DBF = a${}^{\circ }$ and ECF = b${}^{\circ }$.
40 + a${}^{\circ }$ + b${}^{\circ }$ = 180 $\Rightarrow$ a${}^{\circ }$ + b${}^{\circ }$= 140
DFB + EFC + DFE = 180 $\Rightarrow$ 180 – 2a${}^{\circ }$ + 180 – 2b${}^{\circ }$ + DFE = 180 $\Rightarrow$ 180${}^{\circ }$ – 2(a + b) = - DFE
DFE = 280 – 180 = 100${}^{\circ }$

Option (d)
AC has to less than 8 as the diameter will be of 8 units. So the correct option is “D” as all are greater than 8

Option b
It’s important that we note that it is a Pythagoras triplet

7${}^2$ = 25${}^2$+24${}^2$ 

Now,

In a right-angled triangle, the median to the hypotenuse is half the hypotenuse and is also the circum radius of the triangle.

As the hypotenuse is 25, the circum radius is 12.5

Draw perpendicular from D to GF meeting at X.

Also, let GN = GP = k

Since, DE = XF = 3 cm

NX = 3 – 2 = 1 = DM = DP

Therefore, GX = k – 1

$\Delta$DXG is right angled triangle.

Hence, (k + 1)${}^2$ – (k – 1)${}^2$ = 42

Hence, k = 4 and so GF = 6 cm

Area of trapezium DEFG = $\frac{(3+6)}{2}$ * 4 = 18 cm${}^2$

$\angle$PTQ= 90 (angle in a semi circle)

, hence $\angle$RTS = 45.

$\angle$ROS = 90 ( Inscribed angle= 2 *Angle at circumference)

OR=OS=OR (radius)

$\angle$RSO = $\angle$SOQ= 45${}^{\circ }$ (interior alternate angles)

Option (a)

Let the side of the equilateral triangle ΔABC be 3 units. Clearly, BD = 2, CD = 1 and CE = 2.

CD/CE = $\frac{1}{2}$ So $\angle$DEC=30${}^{\circ }$

Hence, ΔCDE is a right angled triangle.

Similarly, ΔBDE is also a right angled triangle.

BE = $\sqrt{7}$ and ED = $\sqrt{3}$ hence BE:ED = $\sqrt{7}$ : $\sqrt{3}$

Consider any triangle ABC, AM is a line connecting A to mid point of BC. Drop a perpendicular from A to BC touching BC at D. The angle ADM is a right angled triangle with 45 – 45 – 90 angles. Hence DAM = 45 and BAM > 45. BAM= 45${}^{\circ }$ If $\angle$ABC =90${}^{\circ }$. hence, answer is option (d)

Option c

In a quadrilateral, product of area of diagonally opposite triangles is same.

So, the product of other set of opposite triangles will be 12*27

Now, we have to find minimum a and b such that a*b=12*27 and a+b is minimum.

So a = b = 18 (as 12*27 = 324)

Hence the area of the quadrilateral = 18 + 18 + 12 + 27 = 75