**1. In △ABC, AB = 7 cm, BC = 3.5 cm, CA = 2√2 cm, ∠A = 75° and ∠B = 47° and in △PQR, PQ = 21 cm, QR = 10.5 cm, RP = 6√2 cm. What is the value of ∠P?**

a) 75°

b) 47°

c) 58°

d) 68°

**Answer:** c) 58°

**Explanation:** We are given that in △ABC, AB = 7 cm, BC = 3.5 cm, CA = 2√2 cm, ∠A = 75° and ∠B = 47°.

In △PQR, PQ = 21 cm, QR = 10.5 cm, RP = 6√2 cm.

The figure is shown below:

According to SSS (Side-Side-Side) Similarity Criterion,

△ABC ~ △PQR

Thus, ∠C = ∠P [Corresponding angles of the similar triangles]

We know that the sum of all angles of a triangle is 180°.

In △ABC,

∠A + ∠B + ∠C = 180°

75° + 47° + ∠C = 180°

122° + ∠C = 180°

∠C = 180° − 122°

∠C = 58°

Since ∠C = ∠P

∠P = 58°

**2. Consider the following statements:**

**(i) If A and B are the points on the sides PQ and PR of △PQR such that ****PQ = 21 cm, PA = 6 cm, BR = 20 cm and PB = 8 cm, then AB || QR.**

**(ii) If in △ABC and △PQR, ∠A = 38°, ∠B = 57°, ∠P = 85° and ∠Q = 57°, then △BAC ~ △RQP.**

**Out of the following, which statement is TRUE?**

a) Only (i)

b) Only (ii)

c) Both (i) and (ii)

d) Neither (i) nor (ii)

**Answer: **a) Only (i)

**Explanation:**

**Consider Statement (i): **We are given PQ = 21 cm, PA = 6 cm, BR = 20 cm and PB = 8 cm

Thus, AQ = PQ − PA

AQ = 21 − 6

AQ = 15 cm

Now, PA/AQ = 6/15 = 2/5

Also, PB/BR = 8/20 = 2/5

PA/AQ = PB/BR

The converse of basic proportionality theorem states that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Thus, AB || QR

**Thus, statement (i) is TRUE.**

**Consider Statement (ii):**

We are given that in △ABC and △PQR, ∠A = 38°, ∠B = 57°, ∠P = 85° and ∠Q = 57°.

We know that the sum of all angles of a triangle is 180°.

Thus, in △ABC,

∠A + ∠B + ∠C = 180°

38° + 57° + ∠C = 180°

95° + ∠C = 180°

∠C = 180° − 95°

∠C = 85°

Now in △PQR,

∠P + ∠Q + ∠R = 180°

85° + 57° + ∠R = 180°

142° + ∠R = 180°

∠R = 180° − 142°

∠R = 38°

Thus, in △BAC and △QRP,

∠A = ∠R = 38°

∠B = ∠Q = 57°

∠C = ∠P = 85°

Thus, by AAA similarity

△BAC ~ △QRP

**Thus, statement (ii) is FALSE.**

Therefore, Only (i) is TRUE.

**3. Consider the figure given below:**

**ABC is a triangle with ∠B = 50° and AB = 12 cm. ADE is another triangle with ∠AED = 50°, AD = 7 cm and AE = 5 cm. What is the length of EC?**

b) 11.75 cm

c) 11.6 cm

d) 11.8 cm

**Answer: **d) 11.8 cm

**Explanation:** The given figure is shown as:

Let ∠A = θ

In △ABC and △AED,

∠BAC = ∠DAE (Common)

∠ABC = ∠AED = 50° (Given)

Thus, by AA similarity criterion,

△ABC ~ △AED

AC / AD = AB / AE = BC / ED

Taking AC / AD = AB / AE

AC / 7 = 12 / 5

AC = (12 × 7) / 5

AC = 84 / 5

Now, AC = AE + EC

EC = AC − AE

= (84 / 5) − 5

= (84 − 25) / 5

= 59 / 5

= 11.8 cm

**∴**** EC = 11.8 cm**

**4. In a trapezium ABCD, AB || DC and DC = 3AB. EF drawn parallel to AB cuts AD at F and BC at E such that BE/EC = $\frac{\frac{4}{}}{\frac{5}{}}$. If diagonal DB intersects EF at G, then which of the following is correct?**

a) EF = 2AB

b) 9EF = 17AB

c) 9EF = 16AB

d) 9EF = 19AB

**Answer:** b) 9EF = 17AB

**Explanation: **The figure is shown below:

We are given that DC = 3AB and $\frac{\frac{\mathrm{BE}}{}}{\frac{\mathrm{EC}}{}}$**= ****$\frac{\frac{4}{}}{\frac{5}{}}$**

In △DFG and △DAB,

∠DFG = ∠DAB [Corresponding angles since AB || FG]

∠FDG = ∠ADB [Common]

We know that if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar by AA similarity criteria.

△DFG ~ △DAB [By AA similarity]

Thus, ^{DF}⁄_{DA} = ^{FG}⁄_{AB} … (1)

In trapezium ABCD, AB || EF || CD

^{AF}⁄_{FD} = ^{BE}⁄_{EC}

But ^{BE}⁄_{EC}** = **^{4}⁄_{5}

^{AF}⁄_{FD}** = **^{4}⁄_{5}

Adding 1 on both the sides

^{AF}⁄_{FD} + 1 = ^{4}⁄_{5} + 1

$\frac{\frac{\mathrm{AF\; +\; FD}}{}}{\frac{\mathrm{FD}}{}}$ = $\frac{\frac{\mathrm{4\; +\; 5}}{}}{\frac{5}{}}$

^{AD}⁄_{FD} = ^{9}⁄_{5}

^{FD}⁄_{AD} = ^{5}⁄_{9} … (2)

From equation (1) and (2),

^{FG}⁄_{AB} = ^{5}⁄_{9}

FG = ^{5}⁄_{9}AB … (3)

In △BEG and △BCD,

∠BEG = ∠BCD [Corresponding angles since EG || CD]

∠GBE = ∠DBC [Common]

We know that if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar by AA similarity criteria.

△BEG ~ △BCD [By AA similarity]

Thus, ^{BE}⁄_{BC} = ^{EG}⁄_{CD}

We know that ^{BE}⁄_{EC}** = **^{4}⁄_{5}

^{EC}⁄_{BE} = ^{5}⁄_{4}

Adding 1 on both sides

$\frac{\frac{\mathrm{EC}}{}}{\frac{\mathrm{BE}}{}}$ + 1 = $\frac{\frac{5}{}}{\frac{4}{}}$ + 1

$\frac{\frac{\mathrm{EC\; +\; BE}}{}}{\frac{\mathrm{BE}}{}}$ = $\frac{\frac{\mathrm{5\; +\; 4}}{}}{\frac{4}{}}$

^{EC}⁄_{BE} = ^{9}⁄_{4}

^{BE}⁄_{EC} = ^{4}⁄_{9}

But ^{BE}⁄_{BC} = ^{EG}⁄_{CD}

^{EG}⁄_{CD} = ^{4}⁄_{9}

EG = ^{4}⁄_{9}CD

We know that CD = 3AB

EG = ^{4}⁄_{9}(3AB)

EG = ^{12}⁄_{9}AB … (4)

Adding equation (3) and (4),

FG + EG = ^{5}⁄_{9}AB + ^{12}⁄_{9}AB

EF = ^{17}⁄_{9}AB

9EF = 17AB

**5. ABCD is a parallelogram in the given figure. AB is divided at P and CD at Q such that AP : PB = 5 : 2 and CQ : QD = 3 : 1. If PQ meets AC at R, then which of the following is true?**

a) AR = ^{21}⁄_{40}AC

b) AR = ^{21}⁄_{41}AC

c) AR = ^{20}⁄_{41}AC

d) AR = ^{23}⁄_{40}AC

**Answer:** c) AR = ^{20}⁄_{41}AC

**Explanation:** In △APR and △CQR,

∠PAR = ∠QCR [Alternate interior angles since AB || CQ]

∠ARP = ∠CRQ [Vertically opposite angles]

We know that if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar by AA similarity criteria.

△APR ~ △CQR [By AA similarity]

^{AP}⁄_{CQ} = ^{PR}⁄_{QR} = ^{AR}⁄_{CR} … (1)

We are given that AP : PB = 5 : 2

$\frac{\frac{\mathrm{AP}}{}}{\frac{\mathrm{AB}}{}}$ = $\frac{\frac{\mathrm{AP}}{}}{\frac{\mathrm{AP\; +\; PB}}{}}$

= $\frac{\frac{5}{}}{\frac{\mathrm{5\; +\; 2}}{}}$

= $\frac{\frac{5}{}}{\frac{7}{}}$

AP = ^{5}⁄_{7}AB

We are given that CQ : QD = 3 : 1

$\frac{\frac{\mathrm{CQ}}{}}{\frac{\mathrm{CD}}{}}$ = $\frac{\frac{\mathrm{CQ}}{}}{\frac{\mathrm{CQ\; +\; QD}}{}}$

= ^{3}⁄_{3 + 1} = 3⁄_{4}

CQ = 3⁄_{4}CD

Since ABCD is a parallelogram, its opposite sides are equal

AB = CD

CQ = 3⁄_{4}AB

From equation (1), we have

^{AP}⁄_{CQ} = ^{AR}⁄_{CR}

Substituting AP =$\frac{\frac{5}{}}{\frac{7}{}}$AB and CQ = $\frac{\frac{3}{}}{\frac{4}{}}$AB in the above equation

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