1.

In the figure above, PQ=QR and ∠Q = ∠SRP = 36°. If ∠TPR = 54°, find ∠STP.

a) 24° b)27° c)18° d)36°

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 **Δ**TPR, **∠**PTR =**∠**TPR and OR is teh angle bisector of **∠**R, thus OT=OP.

Thus, **Δ**SOT ≅ **Δ**SOP (By SAS)

So, **∠**STO =**∠**SPO = 18°

2.If s is the semiperimeter of a triangle with sides as a, b and c. Then which of the following is true?

a) (s-a)(s-b)(s-c)>abc/8

b) (s-a)(s-b)(s-c)<abc/8

c) (s-a)(s-b)(s-c)≤ abc/8

d) (s-a)(s-b)(s-c)≥ abc/8

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

3.Two sides of a right triangle are 6 and 8 and the in-radius is 2, What is the circum-radius?

a) 10 b) 4 c)5 d)6

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.

4.

The ΔPQR is formed by extending the lines AB, BC and CA of the ΔABC. If AC is 10 cm and CP is 30cm, AB is 14cm and AR is 35 cm also, BC is 12 cm while QB is 48 cm.

What is the ratio of area of ΔABC to area of ΔPQR?

a)2:31 b)2:29 c)1:20 d) 1:37

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

5. In an equilateral triangle ΔPQR has the area of 120 sq units. Medians RS and QT are drawn intersecting at O. What is the area of triangle ΔSTO in sq units?

a)12 b)10 c)30 d)None of these

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.

6. PQR is a right triangle, right angled at Q. The circle inscribed in the triangle, with radius of 6 units, touches the side PQ at S. If PS=14 units, find the length of PR.

a)27 units b)29 units c)35 units d)23 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,

20^{2 }+ (6+x)2 = (14+x)2

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

7.In ΔPQR, S is the midpoint of QR. Point T is on PQ such that PT:TQ is 3:2 and point U is on PR such that PU:UR = 3:7. It TU and PS intersect at V, what is the ratio of area of ΔVTS to area of ΔUVS?

a)2:1 b)3:5 c)2:1 d)5:3

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

8.

In the figure, PSTU is a cyclic quadrilateral and QS=SP, TU=TR , SV=PV=QV. How many of the statements would always be true?

i) PS is the median of ΔPQR

ii) ΔSQV and ΔSPV are congruent

iii)The orthocenter of ΔPQR lies at point P.

a) 1 b)2 c)3 d) Cannot be determined

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.

9.In a ΔABC, M is the midpoint of AB and N is the midpoint of AC. CM and BN are perpendicular to each other and meet at point O. If AB 14 and AC is √54 , find BC.

a) 5√7 b) 4√2 c) 2√7 d) 5√2

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.

a^{2}+4b^{2} = 49

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

4a^{2}+b^{2} = 54/4

Adding both, we get

5a^{2}+5b^{2} = 250/4

Or a^{2}+b^{2}=50/4

BC = √(4a^{2}+4b^{2}) = √( 4*50/4) = √(50) = 5√2

10.

Find the sum of all angles shown. **∠**P + **∠**Q + **∠**R+**∠**S+**∠**T +**∠**U+**∠**V +**∠**W

a)360° b) 1080° c) 1000° d) 720°

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°

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