Sol. c)18°

Since ∠Q = 36° & PQ = QR,  ∠QPR= ∠QRP = 72°   { Represented with blue colour}

∠Q = ∠SRP = 36°{represented with green colour}

∠TPR = 54°{represented with red colour}

∠QRP= ∠PRS + ∠SRT  = 72°

Thus, ∠SRT  = 36°

Let us name the point of intersection of PT and SR as O.

Then ∠POR  = 54°also, ∠TOR  =∠POS  =∠TOS  =90°

Now, In Δ ROT, ∠OTR  =54°

Now, in ΔTPR, ∠PTR  =∠TPR  and OR is teh angle bisector of ∠R, thus OT=OP.

Thus,  ΔSOT ≅ ΔSOP (By SAS)

So,  ∠STO  =∠SPO =  18°

Sol. c) (s-a)(s-b)(s-c)≤ abc/8

Let us assume the sides a,b,c be 2, 2,2

Then s=(a+b+c)/2=3

Now the LHS =(s-a)(s-b)(s-c) = 1

And RHS = abc/8 = 1

Thus option a or b cannot be the answer.

Now, let us take the sides a,b,c as 3,4,5

Then s = 6

Now the LHS =(s-a)(s-b)(s-c) = 3*2*1 =6

And RHS = abc/8 = 15/2 = 7.5

Thus LHS≤ RHS

Sol. c)5

We know that a right triangle with two perpendicular sides as 6 and 8 will have its hypotenuse as 10. To check the in-radius of the triangle, we use the formula

Area of triangle = semi-perimeter * in-radius

=> 6*8/2 = 12*r

Thus, r=2

Now, we know that this triangle will have the hypotenuse of 10 units and thus the Circum-radius will be 5 units.

Sol. d) 1:37

As AC = 10 cm and CP = 30cm thus AP=20 and AC:AP=1:2

Similarly, AB:BR = 2:5 and BC:CQ = 1:3

Now, let us join PB and RC and area of ΔABC be x.

Area ΔABP: area ΔABC = AP:AC  (base area ratio)

Thus Area ΔABP = 2x.

Also, Area ΔABP: area ΔBRP = AB:BR =   2:5

area ΔBRP = 5x

Area ΔBCR: area ΔABC = BR:AB =2:5

 Area ΔBCR = 2.5x

Similarly other areas are calculated as shown in the figure.

The final ratio of area of ΔABC to area of ΔPQR = x: 37x =1:37

Sol. b)10

Let us construct the remaining median PN, which would also be the altitude (equilateral triangle)

If the length of PN is h, PM = h/2 (ΔPST ~ ΔPQR, with each side half  of ΔPQR)

Thus area ΔPST = ΔPQR/4 = 30 sq units

Now, we know that the intersection point of the medians, divides it in the ratio of 2:1

Hence, ON = h/3

Thus MO = h-h/2-h/3 = h/6

area ΔPST :area ΔSMO = (h/2)/(h/6) {Triangle on same base, with different heights}

Thus, area ΔSMO = area ΔPST/3 = 30/3 = 10 sq units.

Sol. h)

As the radius of in-circle is 6, the length of SQ and QT would also be 6, as can be seen in the image.

Also, as PS is 14 units and tangents to a circle from a point are equal in length, thus PU would be 14 too. And UR=RT=x.

Now, applying Pythagoras theorem,

20² + (6+x)2 = (14+x)2

On solving, we get x=15. Thus hypotenuse = 14+x = 29 units.

Sol a) 2:1

Referring to the figure for the question,

Since QT:TU = 2:3,

Lets suppose area of  ΔQTS = 2a, area of Δ PST would be 3a {base area  ratio}

Let area of ΔVTS= x  and  area of ΔUVS = y

Then, area of Δ PVT = 3a-x.

Similarly, lets suppose the area of area of ΔUSR = 7b, area of Δ PUV = 3b-y

Now, since S is the midpoint of QR, the area of Δ PSQ = area of Δ PSR

Thus, 5a= 10b; a=2b

Now, area of Δ PVT: area of Δ PVU = TV:VU

Also area of Δ STV: area of Δ TUV = TV: VU

Equating both, x/y = 3a-x/3b-y = 6b-x/3b-y

On solving, we get x/y = 2/1

Sol. c)

Equal sides of a triangle will subtend equal angle.

Since QV=VS, let us name the equal angle as x, as shown in the figure. Also, QS=SP so the ∠VPS = x.

For ΔQVS, ∠PVS is external angle, thus ∠PVS = ∠VQS+∠VSQ = 2x

But, VP=VS thus ∠VPS = ∠VSP = x

Now in ΔSVP, applying sum of angles equals 180°, we get x= 45°

Now, this means ∠PSQ = 90° and also ∠PSU = 90°

Since PSUT is cyclic, ∠PUT =180°-∠PSU = 90°

This means ∠PTR = 90° {Supplementary angle with ∠PU}

Also, UT=TR, thus ∠TUR = ∠TRU = 45°

Now, statement  i) is true as PS is the altitude of the isosceles right triangle ΔPQR.

Statement ii) is true as ΔSQV and ΔSPV are congruent {by RHS}

Statement iii) is also true as ortho-center of a right triangle lies on the vertex with right angle.

Hence all three are definitely true.

Sol.  d) 5√2

Now, in right ΔMBO, since MB = AB/2 = 7, BN and CM are the medians of the triangle, intersecting at O, thus O will divide

Them in the ratio of 2:1

Thus, if OM=a, OC=2a; on=b, OB=2b.

a²+4b² = 49

Similarly in right ΔNCO, since CN = AC/2 = √37/2

4a²+b² = 54/4

Adding both, we get

5a²+5b² = 250/4

Or a²+b²=50/4

BC = √(4a²+4b²) = √( 4*50/4) = √(50) = 5√2

Sol. a)

As sum of internal angles of each triangle is 180° and the unmarked angles of the figure equal the angles of the quadrilateral in the figure (Vertically opposite angles) required sum of eight angles = 4*180°-360° = 360°