**Group (2015-2019)**

**1. (2015/Myanmar /q13a )**

In $\triangle A B C, A D$ and $B E$ are the altitudes. If $\alpha(\triangle D E C)=\frac{3}{4} \alpha^{\prime} \cdot{ }^{\prime} \mathrm{C}$ ), prove that $\angle A C B=30^{\circ}, \quad(5$ marks $)$

**2. (2015/FC /q13a )**

$A B C$ is a triangle. If $B P C, C Q A, A R B$ are equilateral triangles and $\alpha(\triangle B P C)+\alpha(\Delta C Q A)=\alpha(\triangle A R B)$, then prove that $A B C$ is a right triangle. $(5$ marks $)$

**3. (2016/Myanmar /q13a )**

In the figure, $A B / / C D$ and $\alpha(\triangle E C D): \alpha(A B D C)=16: 9$. Find the numerical value of $C D: A B$.

**4. (2016/FC /q13a )**

$A, B, C$ and $D$ are four points in order on a circle $O$, so that $A B$ is a diameter and $\angle C O D=90^{\circ}$. If $A D$ produced and $B C$ produced meet at $E$, prove that $\alpha(\Delta E C D)=\alpha(A B C D)$

**6. (2017/Myanmar /q13a )**

In $\triangle A B C, A D$ and $B E$ are altitudes to the sides $B C$ and $A C$ respectively. If $\angle A C D=45^{\circ}$, prove that $\alpha(\triangle D E C): \alpha(\triangle A B C)=1: 2$

Q13(a) Solution

**7. (2017/FC /q13a )**

In trapezium $\mathrm{ABCD}$ the diagonals $\mathrm{AC}$ and $\mathrm{BD}$ intersect at $\mathrm{O}$. If $\mathrm{AB} / / \mathrm{DC}$ and $16 \alpha(\triangle \mathrm{AOB})=25 \alpha(\Delta \mathrm{COD})$, find the ratios $\mathrm{AB}: \mathrm{CD}$ and $\alpha(\triangle \mathrm{BOC}): \alpha(\Delta \mathrm{COD})$ $(5 \mathrm{marks})$

**8. (2018/Myanmar /q13a )**

$A B C$ is a right triangle with $\angle A$ the right angle. $E$ and $D$ are points on opposiie side of $A C$, with $E$ on the same side of $A C$ as $B$, such that $\triangle A C D$ and $\Delta B C E$ are both equilateral. If $\alpha(\triangle B C E)=2 \alpha(\triangle A C D)$, Prove that $A B C$ is an isosceles right triangle.

Click for Solution

**9. (2018/FC /q13a )**

In the figure $\angle A P Q=\angle C, A P: P B=3: 1$ and $A Q: Q C=1: 2$. If $A Q=2$, find the length of $\mathrm{AP}$ and the ratios of $\alpha(\triangle A P Q): \alpha(\triangle A B C)$ and $\alpha(\triangle A P Q): \alpha(B C Q P)$.

**10. (2019/Myanmar /q11b )**

In the diagram, $\mathrm{P}$ is the point on $\mathrm{AC}$ such that $\mathrm{AP}=3 \mathrm{PC}, \mathrm{R}$ is the point on BP such that $\mathrm{BR}=2 \mathrm{RP}$ and $\mathrm{QR} / / \mathrm{AC}$. Given that $\alpha(\triangle \mathrm{APA})=36 \mathrm{~cm}^{2}$, calculate $\alpha(\Delta \mathrm{BPC})$ and $\alpha(\Delta \mathrm{BRQ}), \quad \mathrm{Q} \quad \mathrm{A} \quad(5 \mathrm{marks})$ Click for Solution

**11. (2019/FC /q11b )**

Given : $\triangle P Q R$ with two medians $P M$ and $\mathrm{QN}$ intersecting at $\mathrm{K}$. $\quad / \mathrm{K} / \mathrm{N} \quad$ (5 marks) Prove $: \alpha(\triangle \mathrm{PNK})=\alpha(\Delta \mathrm{QMK})$ Click for Solution 11(b)

**Answer (2015-2019)**

1. Prove

2. Prove

3. $\alpha(\Delta E A B)=37.5 \mathrm{~cm}^{2}$

4. Prove

6. Prove

7. $\dfrac{5}{4},5:4$

8. Prove

9. $A P=3, \frac{1}{4}, \frac{1}{3}$

10. 16

11. Prove

**Group (2014)**

1. $A, B, C, D$ are four points in order on a circle $O$, so that $A B$ is a diameter. $A D$ and $B C$ meet at $E$. If $\alpha(A B C D)=3 \alpha(\triangle E C D)$, prove that $\angle C O D=60^{\circ}$. (5 marks)

2. $A, B, C, D$ are four points in order on a circle $O$, so that $A B$ is a diameter. $A D$ and $B C$ meet at $\mathrm{E} .$ If $\alpha(\triangle E C D)=\alpha(A B C D)$, prove that $D C=\sqrt{2} A O$. (5 marks)

3. In $\triangle A B C, A D \perp B C$ and $B E \perp A C$. If $2 \alpha(\triangle D E C)=\alpha(\triangle A B C)$, find $\angle A C B$. (5 marks)

4. In $\triangle A B C, \angle B A C=90^{\circ}$ and $\angle A B C=30^{\circ} . D$ and $E$ are points on opposites of $A C$, with $E$ on the same side of $A C$ as $B$, such that both $\triangle A C D$ and $\triangle B C E$ are equilateral. Prove that $\alpha(\triangle B C E)=4 \alpha(\triangle A C D)$. (5 marks)

5. In $\triangle A B C, A D$ and $B E$ are altitudes. If $\angle A C B=45^{\circ}$, prove that $\alpha(\Delta D E C)=\alpha(A B D E)$. (5 marks)

6. $\triangle A B C, A D$ and $B E$ are altitudes. If $\angle A C B=30^{\circ}$, prove that $\alpha(\triangle D E C)=3 \alpha(A B D E)$. (5 marks)

7. $P Q R S$ is a parallelogram. $P S$ is produced to $L$ so that $S L=S R$ and $L R$ produced meets $P Q$ produced at $M$. Prove that $Q M=Q R$. If the area of the parallelogram is $20 \mathrm{~cm}^{2}$ and $P Q=2 P S$, find the area of $\triangle L S R$. (5 marks)

8. In $\triangle A B C, D$ is a point of $A C$ such that $A D=3 C D . \mathrm{E}$ is on $B C$ such that $D E / / A B$. Compare the areas of $\triangle C D E$ and $\triangle A B C$. If $\alpha(A B E D)=30$, what is $\alpha(\Delta A B C) ?$ (5 marks)

9. The area of $\triangle A B C$ is bisected by a line $P Q$ drawn parallel to $B C$ where $P$ lies on the side $A B$ and $Q$ lies on the side $A C$. In what ratio does $P Q$ divide $A B$ and $A C$ ? Find also the ratio $\alpha(\triangle B P Q): \alpha(\triangle B C Q)$. (5 marks)

10. In the diagram, $P$ is the point on $A C$ such that $A P=3 P C, R$ is the point on $B P$ such that $B P=3 R P$ and $Q R \| A C$. Given that $\alpha(\triangle B P A)=36 \mathrm{~cm}^{2}$, calculate the areas of $\triangle B P C$ and $\triangle B R Q .$ (5 marks)

**Answer (2014)**

1. Prove

2. Prove

3. $\angle A C B=45^{\circ}$

4. Prove

5. Prove

6. Prove

7. $20 \mathrm{~cm}^{2}$

8. $1: 16 / 32$

9. $1:(\sqrt{2}-1), 1: \sqrt{2}$

10. $12 \mathrm{~cm}^{2}, 16 \mathrm{~cm}^{2}$

**Group (2013)**

1. In trapezium $A B C D, A B$ is twice $D C$ and $A B \| D C$. If $A C, B D$ intersect at $O$, prove that $\alpha(\triangle A O B)=4 \alpha(\triangle C O D)$. Find the ratio of $A O: C O$. (5 marks)

2. In $\triangle A B C, \angle A=90^{\circ}, A C=5$ and $B C=13 . D$ is a point on $A B$ such that $D E \perp B C$ and $C E=5$. Find $\alpha(\triangle A B C): \alpha(\triangle B D E)$ and find $\alpha(A D E C)$ if $\alpha(\triangle \mathrm{BDE})=\frac{40}{3}$. (5 marks)

3. In $\triangle A B C, \angle B A C=90^{\circ}$ and $A D \perp B C$. If $D C=8 B D$, prove that $B C=3 A B$. (5 marks)

4. In $\triangle A B X, C$ is a point on the segment $B X$ and $D$ is a point on the segment $A X$ such that $\angle B A C=\angle B D C$. Prove that $\alpha(\triangle A B X): \alpha(\Delta C D X)=A

B^{2}: C D^{2}$. (5 marks)

5. $A B C D$ is a segment and $P$ a point outside it such that $\angle P B A=\angle P C D=\angle A P D$. Prove that $\frac{\alpha(\triangle A B P)}{\alpha(\Delta P C D)}=\frac{A B^{2}}{B P^{2}}$. (5 marks)

6. A quadrilateral $A B C D$ is inscribed in a circle. $A B$ and $D C$ are produced to meet at $E$. If $A D=6 \mathrm{~cm}, B C=4 \mathrm{~cm}$ and the area of triangle $B C E$ is $12 \mathrm{~cm}^{2}$, calculate the area of $A B C D$. (5 marks)

7. Two straight lines $A B$ and $C D$ intersect at $E . C M \perp A E, D N \perp B E$ and $\frac{E A}{E C}=\frac{E B}{E D} .$ If $C M=3, D N=4$ and $A B=14$, find $\frac{\alpha(\triangle A C E)}{\alpha(\Delta B D E)}$ and $\alpha(\triangle A C E)$. (5 marks)

8. $\triangle A B C$ is an isosceles right triangle with $A$ the right angle. $E$ and $D$ are points on opposite side of $A C$, with $E$ on the same side of $A C$ as $B$, such that $A C D$ and $B C E$ are both equilateral. Prove that $\alpha(\triangle B C E)=2 \alpha(\triangle A C D) .$ (5 marks)

9. In $\triangle P Q R, Q R=32 \mathrm{~cm}$. The point $Y$ on $P R$ is such that $P Y=6 \mathrm{~cm}, Y R=10 \mathrm{~cm}$. The point $X$ on $P Q$ is such that $X Y / / Q R$. Find the length of $X Y$. If $\alpha(\Delta P X Y)=27 \mathrm{~cm}^{2}$, find $\alpha(Q X Y R)$. (5 marks)

10. In a $\triangle A B C$ the side $A B$ is divided at $X$ so that $\frac{A X}{X B}=\frac{3}{2} .$ A line through $X$ parallel to $B C$, meets $A C$ at $Y$. Find $\alpha(\triangle A B C): \alpha(\triangle B C Y)$. (5 marks)

11. In $\triangle A B C, A D$ and $B E$ are the altitudes. If $\angle A C B=30^{\circ}$, prove that $\alpha(\triangle D E C)=\frac{3}{4} \alpha(\Delta A B C)$. (5 marks)

**Answer (2013)**

1. $AO:CO=2:1$

2. $\alpha(\triangle ABC):\alpha(\triangle BDE)=9:4$

3. Prove

4. Prove

5. Prove

6. Prove

7. Prove

8. Prove

9. Prove

10. Prove

11. Prove

#### ** Group (2012)**

$\qquad$ | $\,$ | |
---|---|---|

1. | $A B C$ is a triangle. If $B P C, C Q A, A R B$ are equilateral triangles, and $\alpha(\triangle B P C)+\alpha(\Delta C Q A)=\alpha(\triangle A R B)$ then prove that $A B C$ is a right triangle. (5 marks) | |

2. | $A, B, C, D$ are four points in order on a circle $O$, so that $A B$ is a diameter and $\angle C O D=90^{\circ} . A D$ and $B C$ meet at $E$, prove that $\alpha(\triangle E C D)=\alpha(A B C D)$. (5 marks) | |

3. | $A B C$ is a triangle such that $B C: C A: A B=3: 4: 5$. If $B P C, C Q A, A R B$ are equilateral triangles, prove that $\alpha(\triangle B P C)+\alpha(\Delta C Q A)=\alpha(\triangle A R B)$. (5 marks) | |

4. | $A, B, C$ and $D$ are four points in order on a circle $O$ so that $A B$ is a diameter and $\angle C O D=90^{\circ}$. If $A D$ produced and $B C$ produced meet at $E$, prove that $\alpha(\Delta E C D)=\alpha(A B C D)$. (5 marks) | |

5. | In $\triangle P Q R, P S$ and $Q T$ are the altitudes. If $\angle P R Q=60^{\circ}$, then prove that $4 \alpha(\Delta S T R)=\alpha(\triangle P Q R)$. (5 marks) | |

6. | (figure) $A D P$ and $B C P$ are two segments such that $\angle B A C=\angle B D C .$ Prove that $\frac{\alpha(\triangle B A P)}{\alpha(\triangle C D P)}=\frac{A B^{2}}{C D^{2}}$. (5 marks) | |

7. | $A B C D$ is a trapezium in which $A B / / C D$ and $\angle A D B=\angle C$. Prove that $A D^{2}: B C^{2}=A B: C D$. (5 marks) | |

8. | Two chords $A C$ and $B D$ of a circle intersect at $O$. Show that$$\alpha(\triangle A O B): \alpha(\Delta C O D)=O A^{2}: O D^{2}$$Show also that$$\alpha(\triangle A O B): \alpha(\triangle A O D)=O B: O D$$. (5 marks) | |

9. | $A B C D$ is a parallelogram and $P Q / / B M$, where $Q$ is midpoint of $C D$ and $P, M$ are points on $B C$ and $A D$ respectively. If $\alpha(\triangle P C Q)=25 \mathrm{~cm}^{2}$, find $\alpha(\triangle A B M)$. (5 marks) | |

10. | $P, Q, R$ and $S$ are four points in order on a circle $O$, such that $P Q$ is a diameter. $P S$ and $Q R$ meet at $T$. If $\alpha(\Delta T R S)=\alpha(P Q R S)$, show that $R S=\sqrt{2} O P$. (5 marks) | |

11. | In $\triangle A B C, \angle A=90^{\circ}$ and $A S \perp B C$. If $2 B C=3 A B$, find the ratio of $B S: C S$. (5 marks) | |

12. | The sides $A B$ and $B C$ of $\triangle A B C$ are $5 \mathrm{~cm}$ and $6 \mathrm{~cm}$ respectively. Points $H$ and $K$ on $A B$ and $A C$ respectively are such that $H K$ and $B C$ are parallel. If the areas of triangles $A H K$ and $A B C$ are in the ratio of $4: 9 ;$ calculate $H K$ and $H B$. (5 marks) |

#### ** Answer (2012)**

$\quad$ | $\,$ | |
---|---|---|

1. | Prove | |

2. | Prove | |

3. | Prove | |

4. | Prove | |

5. | Prove | |

6. | Prove | |

7. | Prove | |

8. | Prove | |

9. | $100 \mathrm{~cm}^{2}$ | |

10. | Prove | |

11. | $4: 5$ | |

12. | $4 \mathrm{~cm} ; \frac{5}{3} \mathrm{~cm}$ |

## ** Group (2011)**

$\quad\;\,$ | $\,$ | |
---|---|---|

1. | Two chords $X B$ and $A Y$ of a circle intersect at $S .$ If $X S=4 \mathrm{~cm}, S A=5 \mathrm{~cm}$, then prove that $\triangle X Y S \sim \triangle A B S$, and hence, find $\alpha(\Delta X Y S): \alpha(\triangle A B S)$. $\mbox{ (5 marks)}$ | |

2. | The chords $B X$ and $A Y$ of a circle intersect at $S .$ If $B S=3 \mathrm{~cm}, A S=5 \mathrm{~cm}$ and $X S=4 \mathrm{~cm}$, find $\alpha(\triangle A B S): \alpha(\Delta X Y S) .$ $\mbox{ (5 marks)}$ | |

3. | In the diagram, $R$ is the point on $B P$ such that $B R=2 R P$ and $Q R / I A C .$ Given that $\alpha(\triangle B P A)=18 \mathrm{~cm}^{2}$ calculate $\alpha(\Delta B R Q), \alpha(P A Q R)$. $\mbox{ (5 marks)}$ | |

4. | $\quad A B C$ is a right triangle with $\angle A$ the right angle.$E$ and $D$ are points onc $+$ posite side of $A C$, with $E$ on the same side of $A C$ as $B$, such that $\triangle A C D$ and $\triangle B C E$ are both equilateral.If $\alpha(\triangle B C E)=2 \alpha(\triangle A C D)$, prove that $A B C$ is an isosceles right triangle. $\mbox{ (5 marks)}$ | |

5. | $\triangle A B C$ is an isosceles right triangle with $\angle A$ the right angle, $E$ and $D$ are points on opposite side of $A C$, with $E$ on the same side of $A C$ as $B$, scuh that $\triangle A C D$ and $\Delta B C E$ are both equilateral.Prove that $\alpha(\triangle B C E)=2 \alpha(\triangle A C D)$. $\mbox{ (5 marks)}$ | |

6. | $A B C$ is a triangle such that $B C: C A: A B=25: 24: 7$.If $B P C, C Q A$ and $A R B$ are equilateral triangles, prove that $\alpha(\triangle B P C)=\alpha(\Delta C Q A)+\alpha(\triangle A R B)$. $\mbox{ (5 marks)}$ | |

7. | In the figure, $P A$ is a tangent segment and $P B C$ is a secant segment.Prove that $\frac{A B^{2}}{C A^{2}}=\frac{P B}{P C}$. $\mbox{ (5 marks)}$ |

#### ** Answer (2011)**

$\quad\;\,$ | $\,$ | |
---|---|---|

1. | $\alpha(\Delta X Y S): \alpha(\triangle A B S)=16: 25$ | |

2. | $\alpha(\triangle A B S): \alpha(\Delta X Y S)=25: 16$ | |

3. | $\alpha(\triangle B R Q)=8 \mathrm{~cm}^{2} / \alpha(P A Q R)=10 \mathrm{~cm}^{2}$ | |

4. | Prove | |

5. | Prove | |

6. | Prove | |

7. | Prove |

## ** Group (2010)**

$\quad\;\,$ | $\,$ | |
---|---|---|

1. | In $\triangle A B X, C$ is a point on the segment $B X$ and $D$ is a point on the segment $A X$ such that $\angle B A C=\angle B D C.$ Prove that $\alpha(\triangle A B X): \alpha(\Delta C D X)=A B^{2}: C D^{2}$.$\text{ (5 marks)}$ | |

2. | In $\triangle A B C, D$ is a point of $A C$ such that $A D=C D.E$ is on $B C$ such that $D E / / A B.$ Compare the areas of $\triangle A B C$ and $\triangle C D E.$ If $\alpha(A B E D)=30$, what is $\alpha(\triangle A B C) ?$ $\text{ (5 marks)}$ | |

3. | In $\triangle A B C, D$ is a point of $A C$ such that $A D=C D.E$ is on $B C$ such that $D E / / A B.$ Compare the areas of $\triangle C D E$ and $\triangle A B C.$ If $\alpha(A B E D)=30$, what is $\alpha(\triangle A B C) ?$ $\text{ (5 marks)}$ | |

4. | In a trapezium $A B C D, A B$ is twice $D C$ and $A B / / D C.$ If $A C$ and $B D$ intersect at $P$, prove that $\alpha(\triangle A P B)=4 \alpha(\Delta C P D).$ $\text{ (5 marks)}$ | |

5. | $P Q R S$ is a trapezium in which $P Q / / R S$ and $\angle P S Q=\angle R.$ Prove that $P S^{2}: Q R^{2}=P Q: R S$.$\text{ (5 marks)}$ | |

6. | In trapezium $A B C D$ the diagonals $A C$ and $B D$ intersect at $O.$ If $A B / / D C$ and $9 \alpha(\triangle A \mathrm{O} B)=16 \alpha(\triangle C \mathrm{OD})$, find the ratios $A B: C D$ and $\alpha(\triangle A O D): \alpha(\Delta C O D)$.$\text{ (5 marks)}$ | |

7. | In $\triangle P Q R, S$ and $T$ are the points on the sides $P Q$ and $P R$ respectively, and $S T \| Q R.$ If $P S=5, S Q=10$ and $\alpha(S Q R T)=104$, find $\alpha(\Delta P Q R)$.$\text{ (5 marks)}$ | |

8. | In $\triangle P Q R, P Q=6, P R=9$, and $S$ is a point on $P R$ such that $\angle P Q S=\angle R.$ Given that $\alpha(\triangle P Q S)=20$, calculate $\alpha(\Delta Q R S).$ $\text{ (5 marks)}$ | |

9. | In $\triangle A B C, A D$ and $B E$ are the altitudes.If $4 \alpha(\triangle D E C)=3 \alpha(\triangle A B C)$, find $\angle A C B.$ $\text{ (5 marks)}$ |

#### ** Answer (2010)**

$\quad\;\,$ | ||
---|---|---|

1. | Prove | |

2. | $\alpha(\triangle A B C): \alpha(\triangle C D E)=4: 1 ; \alpha(\triangle A B C)=40$ | |

3. | $\alpha(\triangle C D E): \alpha(\triangle A B C)=1: 4 ; \alpha(\triangle A B C)=40$ | |

4. | Prove | |

5. | $A B: C D=4: 3 ; \alpha(\triangle AOD): \alpha(\triangle COD)=4: 3$ | |

6. | Prove | |

7. | $117 $ | |

8. | $25$ | |

9. | $30^{\circ}$ |

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