90686
Choose the correct answer from the given four options: Its is given that \(\triangle\text{ABC}\sim\triangle\text{PQR},\text{with}\frac{\text{BC}}{\text{QR}}=\frac{1}{3}. \)\(\text{Then},\frac{\text{ar}(\text{PQR})}{\text{ar}(\text{BCA})} \) is equal to:
1 \(9 \)
2 \(3 \)
3 \(\frac{1}{3} \)
4 \(\frac{1}{9} \)
Explanation:
A\(9 \) Given, \(\triangle\text{ABC}\sim\triangle\text{PQR},\text{and}\frac{\text{BC}}{\text{QR}}=\frac{1}{3}. \) We know that the ratio of the areas of two similar triangles is equal to squal of the ratio of their corresponding sides. \(\therefore\frac{\text{ar}(\text{PQR})}{\text{ar}(\text{BCA})}=\frac{(\text{QR})^2}{(\text{BC})^2} \) \(\Big(\frac{\text{QR}}{\text{BC}}\Big)^2=\Big(\frac{3}{1}\Big)^2=\frac{9}{1}=9 \)
TRIANGLES
90688
In \(\triangle\text{ABC, }\text{AB}=6\sqrt{3 }\text{cm},\text{AC}=12\text{cm} \) and \(\text{BC}=6\text{cm}. \) Then, \(\angle\text{B} \) is: 45º 60º 90º 120º
1 45º
2 60º
3 90º
4 120º
Explanation:
D120º In \(\triangle\text{ABC}, \) \(\text{AB}=6\sqrt{3 }\text{cm},\text{AC}=12\text{cm} \) and \(\text{BC}=6\text{cm} \) \(\text{AB}^2+\text{BC}^2=3\sqrt{3}^2+6^2=108+36=144 \) \(\text{AC}^2=12^2=144 \) \(\Rightarrow\text{AB}^2+\text{BC}^2=\text{AC}^2 \) So, by the Converse of Pythagoras theorem, \(\triangle\text{ABC} \) is a right angled triangle and since AC is the hypotenuse, \(\angle\text{B} \) which is opposite \(\text{AC} = 90^\circ. \)
TRIANGLES
90689
The lenght of the hypotenuse of an isosceles right triangle whose one side is \(4\sqrt{2}\text{cm} \) is:
1 \(12\text{cm}. \)
2 \(8\text{cm}. \)
3 \(8\sqrt{2}\text{ cm}. \)
4 \(12\sqrt{2}\text{ cm}. \)
Explanation:
B\(8\text{cm}. \) Het ABC be an isosceles right triangle. We have,
\(\text{AB}=\text{BC}=4\sqrt{2}\text{ cm} \) \(\Rightarrow\text{AC}^2=\text{AB}^2+\text{BC}^2 \) \(\Rightarrow\text{AC}^2=(4\sqrt{2})^2+(4\sqrt{2})^2 \) \(\Rightarrow\text{AC}^2=32+32 \) \(\Rightarrow\text{AC}^2=64 \) \(\Rightarrow\text{AC}=8\text{cm} \) Thus, the length of hypotenuse is 8cm.
TRIANGLES
90691
Tick the correct answer and justify: ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is:
1 2 : 1
2 1 : 2
3 4 : 1
4 1 : 4
Explanation:
C4 : 1 \(\triangle\text{ABC}\ \text{and}\ \triangle\text{BDE} \) are both equilateral triangles. \(\therefore\ \triangle\text{ABC}\sim\triangle\text{BDE} \) [Using AAA similar condition]. We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. \(\therefore\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac{\text{AB}^2}{\text{BD}^2} \) \(\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\Big(\frac{\text{BC}}{\text{BD}}\Big)^2 \) [\(\therefore \) AB = BC = CA] \(\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}= \Big(\frac{2\text{BD}}{\text{BD}}\Big)^2 \) \(\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac{4}{1} \) Hard **[Triangles]**
TRIANGLES
90692
In a \(\triangle\text{ABC}, \) perpendicular AD from A on BC meets BC at D. If BD = 8cm, DC = 2cm and AD = 4cm, then:
1 \(\triangle\text{ABC} \) is isosceles.
2 \(\triangle\text{ABC} \) is equilateral.
3 \(\text{AC} = 2\text{AB.} \)
4 \(\triangle\text{ABC} \) is right-angled at A.
Explanation:
D\(\triangle\text{ABC} \) is right-angled at A. in \(\triangle\text{ABC},\ \text{AD}\perp\text{BC} \) BD = 8cm, DC = 2cm, AD = 4cm
In right \(\triangle\text{ACD}, \) AC\(^{2}\) = AD\(^{2}\) + CD\(^{2}\) (Pythagoras Theorem) = (4)\(^{2}\) + (2)\(^{2}\) = 16 + 4 = 20 and in right \(\triangle\text{ABD}, \) AB\(^{2}\) = AD\(^{2}\) + DB\(^{2}\) = (4)\(^{2}\) + (8)\(^{2}\) = 16 + 64 = 80 and BC\(^{2}\) = (BD + DC)\(^{2}\) = (8 + 2 )\(^{2}\) = (10)\(^{2}\) = 100 AB\(^{2}\) + AC\(^{2}\) = 80 + 20 = 100 = BC\(^{2}\) \(\triangle\text{ABC} \) is a right triangle whose \(\angle\text{A}=90^\circ \)
90686
Choose the correct answer from the given four options: Its is given that \(\triangle\text{ABC}\sim\triangle\text{PQR},\text{with}\frac{\text{BC}}{\text{QR}}=\frac{1}{3}. \)\(\text{Then},\frac{\text{ar}(\text{PQR})}{\text{ar}(\text{BCA})} \) is equal to:
1 \(9 \)
2 \(3 \)
3 \(\frac{1}{3} \)
4 \(\frac{1}{9} \)
Explanation:
A\(9 \) Given, \(\triangle\text{ABC}\sim\triangle\text{PQR},\text{and}\frac{\text{BC}}{\text{QR}}=\frac{1}{3}. \) We know that the ratio of the areas of two similar triangles is equal to squal of the ratio of their corresponding sides. \(\therefore\frac{\text{ar}(\text{PQR})}{\text{ar}(\text{BCA})}=\frac{(\text{QR})^2}{(\text{BC})^2} \) \(\Big(\frac{\text{QR}}{\text{BC}}\Big)^2=\Big(\frac{3}{1}\Big)^2=\frac{9}{1}=9 \)
TRIANGLES
90688
In \(\triangle\text{ABC, }\text{AB}=6\sqrt{3 }\text{cm},\text{AC}=12\text{cm} \) and \(\text{BC}=6\text{cm}. \) Then, \(\angle\text{B} \) is: 45º 60º 90º 120º
1 45º
2 60º
3 90º
4 120º
Explanation:
D120º In \(\triangle\text{ABC}, \) \(\text{AB}=6\sqrt{3 }\text{cm},\text{AC}=12\text{cm} \) and \(\text{BC}=6\text{cm} \) \(\text{AB}^2+\text{BC}^2=3\sqrt{3}^2+6^2=108+36=144 \) \(\text{AC}^2=12^2=144 \) \(\Rightarrow\text{AB}^2+\text{BC}^2=\text{AC}^2 \) So, by the Converse of Pythagoras theorem, \(\triangle\text{ABC} \) is a right angled triangle and since AC is the hypotenuse, \(\angle\text{B} \) which is opposite \(\text{AC} = 90^\circ. \)
TRIANGLES
90689
The lenght of the hypotenuse of an isosceles right triangle whose one side is \(4\sqrt{2}\text{cm} \) is:
1 \(12\text{cm}. \)
2 \(8\text{cm}. \)
3 \(8\sqrt{2}\text{ cm}. \)
4 \(12\sqrt{2}\text{ cm}. \)
Explanation:
B\(8\text{cm}. \) Het ABC be an isosceles right triangle. We have,
\(\text{AB}=\text{BC}=4\sqrt{2}\text{ cm} \) \(\Rightarrow\text{AC}^2=\text{AB}^2+\text{BC}^2 \) \(\Rightarrow\text{AC}^2=(4\sqrt{2})^2+(4\sqrt{2})^2 \) \(\Rightarrow\text{AC}^2=32+32 \) \(\Rightarrow\text{AC}^2=64 \) \(\Rightarrow\text{AC}=8\text{cm} \) Thus, the length of hypotenuse is 8cm.
TRIANGLES
90691
Tick the correct answer and justify: ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is:
1 2 : 1
2 1 : 2
3 4 : 1
4 1 : 4
Explanation:
C4 : 1 \(\triangle\text{ABC}\ \text{and}\ \triangle\text{BDE} \) are both equilateral triangles. \(\therefore\ \triangle\text{ABC}\sim\triangle\text{BDE} \) [Using AAA similar condition]. We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. \(\therefore\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac{\text{AB}^2}{\text{BD}^2} \) \(\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\Big(\frac{\text{BC}}{\text{BD}}\Big)^2 \) [\(\therefore \) AB = BC = CA] \(\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}= \Big(\frac{2\text{BD}}{\text{BD}}\Big)^2 \) \(\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac{4}{1} \) Hard **[Triangles]**
TRIANGLES
90692
In a \(\triangle\text{ABC}, \) perpendicular AD from A on BC meets BC at D. If BD = 8cm, DC = 2cm and AD = 4cm, then:
1 \(\triangle\text{ABC} \) is isosceles.
2 \(\triangle\text{ABC} \) is equilateral.
3 \(\text{AC} = 2\text{AB.} \)
4 \(\triangle\text{ABC} \) is right-angled at A.
Explanation:
D\(\triangle\text{ABC} \) is right-angled at A. in \(\triangle\text{ABC},\ \text{AD}\perp\text{BC} \) BD = 8cm, DC = 2cm, AD = 4cm
In right \(\triangle\text{ACD}, \) AC\(^{2}\) = AD\(^{2}\) + CD\(^{2}\) (Pythagoras Theorem) = (4)\(^{2}\) + (2)\(^{2}\) = 16 + 4 = 20 and in right \(\triangle\text{ABD}, \) AB\(^{2}\) = AD\(^{2}\) + DB\(^{2}\) = (4)\(^{2}\) + (8)\(^{2}\) = 16 + 64 = 80 and BC\(^{2}\) = (BD + DC)\(^{2}\) = (8 + 2 )\(^{2}\) = (10)\(^{2}\) = 100 AB\(^{2}\) + AC\(^{2}\) = 80 + 20 = 100 = BC\(^{2}\) \(\triangle\text{ABC} \) is a right triangle whose \(\angle\text{A}=90^\circ \)
90686
Choose the correct answer from the given four options: Its is given that \(\triangle\text{ABC}\sim\triangle\text{PQR},\text{with}\frac{\text{BC}}{\text{QR}}=\frac{1}{3}. \)\(\text{Then},\frac{\text{ar}(\text{PQR})}{\text{ar}(\text{BCA})} \) is equal to:
1 \(9 \)
2 \(3 \)
3 \(\frac{1}{3} \)
4 \(\frac{1}{9} \)
Explanation:
A\(9 \) Given, \(\triangle\text{ABC}\sim\triangle\text{PQR},\text{and}\frac{\text{BC}}{\text{QR}}=\frac{1}{3}. \) We know that the ratio of the areas of two similar triangles is equal to squal of the ratio of their corresponding sides. \(\therefore\frac{\text{ar}(\text{PQR})}{\text{ar}(\text{BCA})}=\frac{(\text{QR})^2}{(\text{BC})^2} \) \(\Big(\frac{\text{QR}}{\text{BC}}\Big)^2=\Big(\frac{3}{1}\Big)^2=\frac{9}{1}=9 \)
TRIANGLES
90688
In \(\triangle\text{ABC, }\text{AB}=6\sqrt{3 }\text{cm},\text{AC}=12\text{cm} \) and \(\text{BC}=6\text{cm}. \) Then, \(\angle\text{B} \) is: 45º 60º 90º 120º
1 45º
2 60º
3 90º
4 120º
Explanation:
D120º In \(\triangle\text{ABC}, \) \(\text{AB}=6\sqrt{3 }\text{cm},\text{AC}=12\text{cm} \) and \(\text{BC}=6\text{cm} \) \(\text{AB}^2+\text{BC}^2=3\sqrt{3}^2+6^2=108+36=144 \) \(\text{AC}^2=12^2=144 \) \(\Rightarrow\text{AB}^2+\text{BC}^2=\text{AC}^2 \) So, by the Converse of Pythagoras theorem, \(\triangle\text{ABC} \) is a right angled triangle and since AC is the hypotenuse, \(\angle\text{B} \) which is opposite \(\text{AC} = 90^\circ. \)
TRIANGLES
90689
The lenght of the hypotenuse of an isosceles right triangle whose one side is \(4\sqrt{2}\text{cm} \) is:
1 \(12\text{cm}. \)
2 \(8\text{cm}. \)
3 \(8\sqrt{2}\text{ cm}. \)
4 \(12\sqrt{2}\text{ cm}. \)
Explanation:
B\(8\text{cm}. \) Het ABC be an isosceles right triangle. We have,
\(\text{AB}=\text{BC}=4\sqrt{2}\text{ cm} \) \(\Rightarrow\text{AC}^2=\text{AB}^2+\text{BC}^2 \) \(\Rightarrow\text{AC}^2=(4\sqrt{2})^2+(4\sqrt{2})^2 \) \(\Rightarrow\text{AC}^2=32+32 \) \(\Rightarrow\text{AC}^2=64 \) \(\Rightarrow\text{AC}=8\text{cm} \) Thus, the length of hypotenuse is 8cm.
TRIANGLES
90691
Tick the correct answer and justify: ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is:
1 2 : 1
2 1 : 2
3 4 : 1
4 1 : 4
Explanation:
C4 : 1 \(\triangle\text{ABC}\ \text{and}\ \triangle\text{BDE} \) are both equilateral triangles. \(\therefore\ \triangle\text{ABC}\sim\triangle\text{BDE} \) [Using AAA similar condition]. We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. \(\therefore\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac{\text{AB}^2}{\text{BD}^2} \) \(\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\Big(\frac{\text{BC}}{\text{BD}}\Big)^2 \) [\(\therefore \) AB = BC = CA] \(\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}= \Big(\frac{2\text{BD}}{\text{BD}}\Big)^2 \) \(\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac{4}{1} \) Hard **[Triangles]**
TRIANGLES
90692
In a \(\triangle\text{ABC}, \) perpendicular AD from A on BC meets BC at D. If BD = 8cm, DC = 2cm and AD = 4cm, then:
1 \(\triangle\text{ABC} \) is isosceles.
2 \(\triangle\text{ABC} \) is equilateral.
3 \(\text{AC} = 2\text{AB.} \)
4 \(\triangle\text{ABC} \) is right-angled at A.
Explanation:
D\(\triangle\text{ABC} \) is right-angled at A. in \(\triangle\text{ABC},\ \text{AD}\perp\text{BC} \) BD = 8cm, DC = 2cm, AD = 4cm
In right \(\triangle\text{ACD}, \) AC\(^{2}\) = AD\(^{2}\) + CD\(^{2}\) (Pythagoras Theorem) = (4)\(^{2}\) + (2)\(^{2}\) = 16 + 4 = 20 and in right \(\triangle\text{ABD}, \) AB\(^{2}\) = AD\(^{2}\) + DB\(^{2}\) = (4)\(^{2}\) + (8)\(^{2}\) = 16 + 64 = 80 and BC\(^{2}\) = (BD + DC)\(^{2}\) = (8 + 2 )\(^{2}\) = (10)\(^{2}\) = 100 AB\(^{2}\) + AC\(^{2}\) = 80 + 20 = 100 = BC\(^{2}\) \(\triangle\text{ABC} \) is a right triangle whose \(\angle\text{A}=90^\circ \)
90686
Choose the correct answer from the given four options: Its is given that \(\triangle\text{ABC}\sim\triangle\text{PQR},\text{with}\frac{\text{BC}}{\text{QR}}=\frac{1}{3}. \)\(\text{Then},\frac{\text{ar}(\text{PQR})}{\text{ar}(\text{BCA})} \) is equal to:
1 \(9 \)
2 \(3 \)
3 \(\frac{1}{3} \)
4 \(\frac{1}{9} \)
Explanation:
A\(9 \) Given, \(\triangle\text{ABC}\sim\triangle\text{PQR},\text{and}\frac{\text{BC}}{\text{QR}}=\frac{1}{3}. \) We know that the ratio of the areas of two similar triangles is equal to squal of the ratio of their corresponding sides. \(\therefore\frac{\text{ar}(\text{PQR})}{\text{ar}(\text{BCA})}=\frac{(\text{QR})^2}{(\text{BC})^2} \) \(\Big(\frac{\text{QR}}{\text{BC}}\Big)^2=\Big(\frac{3}{1}\Big)^2=\frac{9}{1}=9 \)
TRIANGLES
90688
In \(\triangle\text{ABC, }\text{AB}=6\sqrt{3 }\text{cm},\text{AC}=12\text{cm} \) and \(\text{BC}=6\text{cm}. \) Then, \(\angle\text{B} \) is: 45º 60º 90º 120º
1 45º
2 60º
3 90º
4 120º
Explanation:
D120º In \(\triangle\text{ABC}, \) \(\text{AB}=6\sqrt{3 }\text{cm},\text{AC}=12\text{cm} \) and \(\text{BC}=6\text{cm} \) \(\text{AB}^2+\text{BC}^2=3\sqrt{3}^2+6^2=108+36=144 \) \(\text{AC}^2=12^2=144 \) \(\Rightarrow\text{AB}^2+\text{BC}^2=\text{AC}^2 \) So, by the Converse of Pythagoras theorem, \(\triangle\text{ABC} \) is a right angled triangle and since AC is the hypotenuse, \(\angle\text{B} \) which is opposite \(\text{AC} = 90^\circ. \)
TRIANGLES
90689
The lenght of the hypotenuse of an isosceles right triangle whose one side is \(4\sqrt{2}\text{cm} \) is:
1 \(12\text{cm}. \)
2 \(8\text{cm}. \)
3 \(8\sqrt{2}\text{ cm}. \)
4 \(12\sqrt{2}\text{ cm}. \)
Explanation:
B\(8\text{cm}. \) Het ABC be an isosceles right triangle. We have,
\(\text{AB}=\text{BC}=4\sqrt{2}\text{ cm} \) \(\Rightarrow\text{AC}^2=\text{AB}^2+\text{BC}^2 \) \(\Rightarrow\text{AC}^2=(4\sqrt{2})^2+(4\sqrt{2})^2 \) \(\Rightarrow\text{AC}^2=32+32 \) \(\Rightarrow\text{AC}^2=64 \) \(\Rightarrow\text{AC}=8\text{cm} \) Thus, the length of hypotenuse is 8cm.
TRIANGLES
90691
Tick the correct answer and justify: ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is:
1 2 : 1
2 1 : 2
3 4 : 1
4 1 : 4
Explanation:
C4 : 1 \(\triangle\text{ABC}\ \text{and}\ \triangle\text{BDE} \) are both equilateral triangles. \(\therefore\ \triangle\text{ABC}\sim\triangle\text{BDE} \) [Using AAA similar condition]. We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. \(\therefore\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac{\text{AB}^2}{\text{BD}^2} \) \(\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\Big(\frac{\text{BC}}{\text{BD}}\Big)^2 \) [\(\therefore \) AB = BC = CA] \(\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}= \Big(\frac{2\text{BD}}{\text{BD}}\Big)^2 \) \(\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac{4}{1} \) Hard **[Triangles]**
TRIANGLES
90692
In a \(\triangle\text{ABC}, \) perpendicular AD from A on BC meets BC at D. If BD = 8cm, DC = 2cm and AD = 4cm, then:
1 \(\triangle\text{ABC} \) is isosceles.
2 \(\triangle\text{ABC} \) is equilateral.
3 \(\text{AC} = 2\text{AB.} \)
4 \(\triangle\text{ABC} \) is right-angled at A.
Explanation:
D\(\triangle\text{ABC} \) is right-angled at A. in \(\triangle\text{ABC},\ \text{AD}\perp\text{BC} \) BD = 8cm, DC = 2cm, AD = 4cm
In right \(\triangle\text{ACD}, \) AC\(^{2}\) = AD\(^{2}\) + CD\(^{2}\) (Pythagoras Theorem) = (4)\(^{2}\) + (2)\(^{2}\) = 16 + 4 = 20 and in right \(\triangle\text{ABD}, \) AB\(^{2}\) = AD\(^{2}\) + DB\(^{2}\) = (4)\(^{2}\) + (8)\(^{2}\) = 16 + 64 = 80 and BC\(^{2}\) = (BD + DC)\(^{2}\) = (8 + 2 )\(^{2}\) = (10)\(^{2}\) = 100 AB\(^{2}\) + AC\(^{2}\) = 80 + 20 = 100 = BC\(^{2}\) \(\triangle\text{ABC} \) is a right triangle whose \(\angle\text{A}=90^\circ \)
90686
Choose the correct answer from the given four options: Its is given that \(\triangle\text{ABC}\sim\triangle\text{PQR},\text{with}\frac{\text{BC}}{\text{QR}}=\frac{1}{3}. \)\(\text{Then},\frac{\text{ar}(\text{PQR})}{\text{ar}(\text{BCA})} \) is equal to:
1 \(9 \)
2 \(3 \)
3 \(\frac{1}{3} \)
4 \(\frac{1}{9} \)
Explanation:
A\(9 \) Given, \(\triangle\text{ABC}\sim\triangle\text{PQR},\text{and}\frac{\text{BC}}{\text{QR}}=\frac{1}{3}. \) We know that the ratio of the areas of two similar triangles is equal to squal of the ratio of their corresponding sides. \(\therefore\frac{\text{ar}(\text{PQR})}{\text{ar}(\text{BCA})}=\frac{(\text{QR})^2}{(\text{BC})^2} \) \(\Big(\frac{\text{QR}}{\text{BC}}\Big)^2=\Big(\frac{3}{1}\Big)^2=\frac{9}{1}=9 \)
TRIANGLES
90688
In \(\triangle\text{ABC, }\text{AB}=6\sqrt{3 }\text{cm},\text{AC}=12\text{cm} \) and \(\text{BC}=6\text{cm}. \) Then, \(\angle\text{B} \) is: 45º 60º 90º 120º
1 45º
2 60º
3 90º
4 120º
Explanation:
D120º In \(\triangle\text{ABC}, \) \(\text{AB}=6\sqrt{3 }\text{cm},\text{AC}=12\text{cm} \) and \(\text{BC}=6\text{cm} \) \(\text{AB}^2+\text{BC}^2=3\sqrt{3}^2+6^2=108+36=144 \) \(\text{AC}^2=12^2=144 \) \(\Rightarrow\text{AB}^2+\text{BC}^2=\text{AC}^2 \) So, by the Converse of Pythagoras theorem, \(\triangle\text{ABC} \) is a right angled triangle and since AC is the hypotenuse, \(\angle\text{B} \) which is opposite \(\text{AC} = 90^\circ. \)
TRIANGLES
90689
The lenght of the hypotenuse of an isosceles right triangle whose one side is \(4\sqrt{2}\text{cm} \) is:
1 \(12\text{cm}. \)
2 \(8\text{cm}. \)
3 \(8\sqrt{2}\text{ cm}. \)
4 \(12\sqrt{2}\text{ cm}. \)
Explanation:
B\(8\text{cm}. \) Het ABC be an isosceles right triangle. We have,
\(\text{AB}=\text{BC}=4\sqrt{2}\text{ cm} \) \(\Rightarrow\text{AC}^2=\text{AB}^2+\text{BC}^2 \) \(\Rightarrow\text{AC}^2=(4\sqrt{2})^2+(4\sqrt{2})^2 \) \(\Rightarrow\text{AC}^2=32+32 \) \(\Rightarrow\text{AC}^2=64 \) \(\Rightarrow\text{AC}=8\text{cm} \) Thus, the length of hypotenuse is 8cm.
TRIANGLES
90691
Tick the correct answer and justify: ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the areas of triangles ABC and BDE is:
1 2 : 1
2 1 : 2
3 4 : 1
4 1 : 4
Explanation:
C4 : 1 \(\triangle\text{ABC}\ \text{and}\ \triangle\text{BDE} \) are both equilateral triangles. \(\therefore\ \triangle\text{ABC}\sim\triangle\text{BDE} \) [Using AAA similar condition]. We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides. \(\therefore\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac{\text{AB}^2}{\text{BD}^2} \) \(\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\Big(\frac{\text{BC}}{\text{BD}}\Big)^2 \) [\(\therefore \) AB = BC = CA] \(\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}= \Big(\frac{2\text{BD}}{\text{BD}}\Big)^2 \) \(\Rightarrow\ \frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{BDE})}=\frac{4}{1} \) Hard **[Triangles]**
TRIANGLES
90692
In a \(\triangle\text{ABC}, \) perpendicular AD from A on BC meets BC at D. If BD = 8cm, DC = 2cm and AD = 4cm, then:
1 \(\triangle\text{ABC} \) is isosceles.
2 \(\triangle\text{ABC} \) is equilateral.
3 \(\text{AC} = 2\text{AB.} \)
4 \(\triangle\text{ABC} \) is right-angled at A.
Explanation:
D\(\triangle\text{ABC} \) is right-angled at A. in \(\triangle\text{ABC},\ \text{AD}\perp\text{BC} \) BD = 8cm, DC = 2cm, AD = 4cm
In right \(\triangle\text{ACD}, \) AC\(^{2}\) = AD\(^{2}\) + CD\(^{2}\) (Pythagoras Theorem) = (4)\(^{2}\) + (2)\(^{2}\) = 16 + 4 = 20 and in right \(\triangle\text{ABD}, \) AB\(^{2}\) = AD\(^{2}\) + DB\(^{2}\) = (4)\(^{2}\) + (8)\(^{2}\) = 16 + 64 = 80 and BC\(^{2}\) = (BD + DC)\(^{2}\) = (8 + 2 )\(^{2}\) = (10)\(^{2}\) = 100 AB\(^{2}\) + AC\(^{2}\) = 80 + 20 = 100 = BC\(^{2}\) \(\triangle\text{ABC} \) is a right triangle whose \(\angle\text{A}=90^\circ \)
