90658
It is given that \(\triangle\text{ABC}\sim\triangle\text{DFE}. \) If \(\angle\text{A}=30^\circ,\angle\text{C}=50^\circ,\text{AB}=5\text{cm},\text{AC}=8\text{cm} \) and \(\text{DF}=7.5\text{cm} \) then which of the following is true?
B\(\text{DE}=12\text{cm},\angle\text{F}=100^\circ \) Given that, \(\angle\text{A}=30^\circ,\angle\text{C}=50^\circ \) \(\triangle\text{ABC}\sim\triangle\text{DFE} \) \(\Rightarrow\angle\text{A}=\angle\text{D}=30^\circ \) \(\angle\text{C}=\angle\text{E}=50^\circ \) Using angle sum property, we can find \(\angle\text{B}=100^\circ \) So, \(\angle\text{B}=\angle\text{F}=100^\circ \) Also, AB = 5cm, AC = 8cm and DF = 7.5cm \(\frac{\text{AB}}{\text{DF}}=\frac{\text{BC}}{\text{FE}}=\frac{\text{AC}}{\text{DE}} \) \(\Rightarrow\frac{5}{7.5}=\frac{\text{BC}}{\text{FE}}=\frac{8}{\text{DE}} \) \(\Rightarrow\frac{5}{7.5}=\frac{8}{\text{DE}}\Rightarrow\frac{8\times7.5}{5}=12\text{cm} \) Hence, DE = 12cm and \(\angle\text{F}=100^\circ \)
TRIANGLES
90660
The shadow of a 5-m-long stick is 2m long. At the same time the langht of the shadow of a 12.5-m-high (in m) is:
1 3.0
2 3.5
3 4.5
4 5.0
Explanation:
D5.0
Let AN be the long stick and AW be its shadow. Let OB be the tree and OW be its shadows. AW = 2cm AN = 5m OW = 12.5m Ratio of actual lengths = ratio of their shadows \(\Rightarrow\frac{\text{OB}}{\text{AN}}=\frac{\text{OW}}{\text{AW}} \) \(\Rightarrow\frac{12.5}{5}=\frac{\text{OW}}{2} \) \(\Rightarrow\text{OW}=\frac{12.5\times2}{5} \) \(\Rightarrow\text{OW}=5.0\text{m} \) So, the height of the tower is 5.m.
TRIANGLES
90663
In \(\triangle\text{ABC},\text{DE }||\text{ BC} \) so that AD = (7x - 4)cm, AE = (5x - 2)cm, DB = (3x + 4)cm and EC = 3x cm. Then, we have:
1 x = 3
2 x = 5
3 x = 4
4 x = 2.5
Explanation:
Cx = 4 In \(\triangle\text{ABC},\text{DE }||\text{ BC} \) By Basic proportionality theorem, \(\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}} \) \(\Rightarrow\frac{7\text{x}-4}{3\text{x}+4}=\frac{5\text{x}-2}{3\text{x}} \) \(\Rightarrow21\text{x}^2-12\text{x}=15\text{x}^2+14\text{x}-8 \) \(\Rightarrow6\text{x}^2-26\text{x}+8=0 \) \(\Rightarrow3\text{x}^2-13\text{x}+4=0 \) \(\Rightarrow(\text{x}-4)(3\text{x}-1)=0 \) \(\Rightarrow\text{x}=4 \) or \(\text{x}=\frac{1}{3} \) If \(\text{x}=\frac{1}{3}, \) then \(\text{AD}=7\text{x}-4=7\Big(\frac{1}{3}\Big)-4=\frac{-5}{3}<0 \) This is not possible since length cannot be negative. \(\Rightarrow \)x = 4
TRIANGLES
90664
If triangles ABC and DEF are similar and AB = 4cm, DE = 6cm, EF = 9cm and FD = 12cm, the perimeter of triangle is:
1 22cm
2 20cm
3 21cm
4 18cm
Explanation:
D18cm ABC ~ DEF AB = 4cm, DE = 6cm, EF = 9cm and FD = 12cm \(\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{AC}}{\text{DF}} \) \(\frac{4}{6}=\frac{\text{BC}}{9}=\frac{\text{AC}}{12} \) \(\text{BC}=(\frac{4.9}{6})=6\text{cm} \) \(\text{AC}=(\frac{12.4}{6})=8\text{cm} \) Perimeter = AB + BC + AC = 4 + 6 + 8 =18cm
90658
It is given that \(\triangle\text{ABC}\sim\triangle\text{DFE}. \) If \(\angle\text{A}=30^\circ,\angle\text{C}=50^\circ,\text{AB}=5\text{cm},\text{AC}=8\text{cm} \) and \(\text{DF}=7.5\text{cm} \) then which of the following is true?
B\(\text{DE}=12\text{cm},\angle\text{F}=100^\circ \) Given that, \(\angle\text{A}=30^\circ,\angle\text{C}=50^\circ \) \(\triangle\text{ABC}\sim\triangle\text{DFE} \) \(\Rightarrow\angle\text{A}=\angle\text{D}=30^\circ \) \(\angle\text{C}=\angle\text{E}=50^\circ \) Using angle sum property, we can find \(\angle\text{B}=100^\circ \) So, \(\angle\text{B}=\angle\text{F}=100^\circ \) Also, AB = 5cm, AC = 8cm and DF = 7.5cm \(\frac{\text{AB}}{\text{DF}}=\frac{\text{BC}}{\text{FE}}=\frac{\text{AC}}{\text{DE}} \) \(\Rightarrow\frac{5}{7.5}=\frac{\text{BC}}{\text{FE}}=\frac{8}{\text{DE}} \) \(\Rightarrow\frac{5}{7.5}=\frac{8}{\text{DE}}\Rightarrow\frac{8\times7.5}{5}=12\text{cm} \) Hence, DE = 12cm and \(\angle\text{F}=100^\circ \)
TRIANGLES
90660
The shadow of a 5-m-long stick is 2m long. At the same time the langht of the shadow of a 12.5-m-high (in m) is:
1 3.0
2 3.5
3 4.5
4 5.0
Explanation:
D5.0
Let AN be the long stick and AW be its shadow. Let OB be the tree and OW be its shadows. AW = 2cm AN = 5m OW = 12.5m Ratio of actual lengths = ratio of their shadows \(\Rightarrow\frac{\text{OB}}{\text{AN}}=\frac{\text{OW}}{\text{AW}} \) \(\Rightarrow\frac{12.5}{5}=\frac{\text{OW}}{2} \) \(\Rightarrow\text{OW}=\frac{12.5\times2}{5} \) \(\Rightarrow\text{OW}=5.0\text{m} \) So, the height of the tower is 5.m.
TRIANGLES
90663
In \(\triangle\text{ABC},\text{DE }||\text{ BC} \) so that AD = (7x - 4)cm, AE = (5x - 2)cm, DB = (3x + 4)cm and EC = 3x cm. Then, we have:
1 x = 3
2 x = 5
3 x = 4
4 x = 2.5
Explanation:
Cx = 4 In \(\triangle\text{ABC},\text{DE }||\text{ BC} \) By Basic proportionality theorem, \(\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}} \) \(\Rightarrow\frac{7\text{x}-4}{3\text{x}+4}=\frac{5\text{x}-2}{3\text{x}} \) \(\Rightarrow21\text{x}^2-12\text{x}=15\text{x}^2+14\text{x}-8 \) \(\Rightarrow6\text{x}^2-26\text{x}+8=0 \) \(\Rightarrow3\text{x}^2-13\text{x}+4=0 \) \(\Rightarrow(\text{x}-4)(3\text{x}-1)=0 \) \(\Rightarrow\text{x}=4 \) or \(\text{x}=\frac{1}{3} \) If \(\text{x}=\frac{1}{3}, \) then \(\text{AD}=7\text{x}-4=7\Big(\frac{1}{3}\Big)-4=\frac{-5}{3}<0 \) This is not possible since length cannot be negative. \(\Rightarrow \)x = 4
TRIANGLES
90664
If triangles ABC and DEF are similar and AB = 4cm, DE = 6cm, EF = 9cm and FD = 12cm, the perimeter of triangle is:
1 22cm
2 20cm
3 21cm
4 18cm
Explanation:
D18cm ABC ~ DEF AB = 4cm, DE = 6cm, EF = 9cm and FD = 12cm \(\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{AC}}{\text{DF}} \) \(\frac{4}{6}=\frac{\text{BC}}{9}=\frac{\text{AC}}{12} \) \(\text{BC}=(\frac{4.9}{6})=6\text{cm} \) \(\text{AC}=(\frac{12.4}{6})=8\text{cm} \) Perimeter = AB + BC + AC = 4 + 6 + 8 =18cm
90658
It is given that \(\triangle\text{ABC}\sim\triangle\text{DFE}. \) If \(\angle\text{A}=30^\circ,\angle\text{C}=50^\circ,\text{AB}=5\text{cm},\text{AC}=8\text{cm} \) and \(\text{DF}=7.5\text{cm} \) then which of the following is true?
B\(\text{DE}=12\text{cm},\angle\text{F}=100^\circ \) Given that, \(\angle\text{A}=30^\circ,\angle\text{C}=50^\circ \) \(\triangle\text{ABC}\sim\triangle\text{DFE} \) \(\Rightarrow\angle\text{A}=\angle\text{D}=30^\circ \) \(\angle\text{C}=\angle\text{E}=50^\circ \) Using angle sum property, we can find \(\angle\text{B}=100^\circ \) So, \(\angle\text{B}=\angle\text{F}=100^\circ \) Also, AB = 5cm, AC = 8cm and DF = 7.5cm \(\frac{\text{AB}}{\text{DF}}=\frac{\text{BC}}{\text{FE}}=\frac{\text{AC}}{\text{DE}} \) \(\Rightarrow\frac{5}{7.5}=\frac{\text{BC}}{\text{FE}}=\frac{8}{\text{DE}} \) \(\Rightarrow\frac{5}{7.5}=\frac{8}{\text{DE}}\Rightarrow\frac{8\times7.5}{5}=12\text{cm} \) Hence, DE = 12cm and \(\angle\text{F}=100^\circ \)
TRIANGLES
90660
The shadow of a 5-m-long stick is 2m long. At the same time the langht of the shadow of a 12.5-m-high (in m) is:
1 3.0
2 3.5
3 4.5
4 5.0
Explanation:
D5.0
Let AN be the long stick and AW be its shadow. Let OB be the tree and OW be its shadows. AW = 2cm AN = 5m OW = 12.5m Ratio of actual lengths = ratio of their shadows \(\Rightarrow\frac{\text{OB}}{\text{AN}}=\frac{\text{OW}}{\text{AW}} \) \(\Rightarrow\frac{12.5}{5}=\frac{\text{OW}}{2} \) \(\Rightarrow\text{OW}=\frac{12.5\times2}{5} \) \(\Rightarrow\text{OW}=5.0\text{m} \) So, the height of the tower is 5.m.
TRIANGLES
90663
In \(\triangle\text{ABC},\text{DE }||\text{ BC} \) so that AD = (7x - 4)cm, AE = (5x - 2)cm, DB = (3x + 4)cm and EC = 3x cm. Then, we have:
1 x = 3
2 x = 5
3 x = 4
4 x = 2.5
Explanation:
Cx = 4 In \(\triangle\text{ABC},\text{DE }||\text{ BC} \) By Basic proportionality theorem, \(\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}} \) \(\Rightarrow\frac{7\text{x}-4}{3\text{x}+4}=\frac{5\text{x}-2}{3\text{x}} \) \(\Rightarrow21\text{x}^2-12\text{x}=15\text{x}^2+14\text{x}-8 \) \(\Rightarrow6\text{x}^2-26\text{x}+8=0 \) \(\Rightarrow3\text{x}^2-13\text{x}+4=0 \) \(\Rightarrow(\text{x}-4)(3\text{x}-1)=0 \) \(\Rightarrow\text{x}=4 \) or \(\text{x}=\frac{1}{3} \) If \(\text{x}=\frac{1}{3}, \) then \(\text{AD}=7\text{x}-4=7\Big(\frac{1}{3}\Big)-4=\frac{-5}{3}<0 \) This is not possible since length cannot be negative. \(\Rightarrow \)x = 4
TRIANGLES
90664
If triangles ABC and DEF are similar and AB = 4cm, DE = 6cm, EF = 9cm and FD = 12cm, the perimeter of triangle is:
1 22cm
2 20cm
3 21cm
4 18cm
Explanation:
D18cm ABC ~ DEF AB = 4cm, DE = 6cm, EF = 9cm and FD = 12cm \(\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{AC}}{\text{DF}} \) \(\frac{4}{6}=\frac{\text{BC}}{9}=\frac{\text{AC}}{12} \) \(\text{BC}=(\frac{4.9}{6})=6\text{cm} \) \(\text{AC}=(\frac{12.4}{6})=8\text{cm} \) Perimeter = AB + BC + AC = 4 + 6 + 8 =18cm
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
TRIANGLES
90658
It is given that \(\triangle\text{ABC}\sim\triangle\text{DFE}. \) If \(\angle\text{A}=30^\circ,\angle\text{C}=50^\circ,\text{AB}=5\text{cm},\text{AC}=8\text{cm} \) and \(\text{DF}=7.5\text{cm} \) then which of the following is true?
B\(\text{DE}=12\text{cm},\angle\text{F}=100^\circ \) Given that, \(\angle\text{A}=30^\circ,\angle\text{C}=50^\circ \) \(\triangle\text{ABC}\sim\triangle\text{DFE} \) \(\Rightarrow\angle\text{A}=\angle\text{D}=30^\circ \) \(\angle\text{C}=\angle\text{E}=50^\circ \) Using angle sum property, we can find \(\angle\text{B}=100^\circ \) So, \(\angle\text{B}=\angle\text{F}=100^\circ \) Also, AB = 5cm, AC = 8cm and DF = 7.5cm \(\frac{\text{AB}}{\text{DF}}=\frac{\text{BC}}{\text{FE}}=\frac{\text{AC}}{\text{DE}} \) \(\Rightarrow\frac{5}{7.5}=\frac{\text{BC}}{\text{FE}}=\frac{8}{\text{DE}} \) \(\Rightarrow\frac{5}{7.5}=\frac{8}{\text{DE}}\Rightarrow\frac{8\times7.5}{5}=12\text{cm} \) Hence, DE = 12cm and \(\angle\text{F}=100^\circ \)
TRIANGLES
90660
The shadow of a 5-m-long stick is 2m long. At the same time the langht of the shadow of a 12.5-m-high (in m) is:
1 3.0
2 3.5
3 4.5
4 5.0
Explanation:
D5.0
Let AN be the long stick and AW be its shadow. Let OB be the tree and OW be its shadows. AW = 2cm AN = 5m OW = 12.5m Ratio of actual lengths = ratio of their shadows \(\Rightarrow\frac{\text{OB}}{\text{AN}}=\frac{\text{OW}}{\text{AW}} \) \(\Rightarrow\frac{12.5}{5}=\frac{\text{OW}}{2} \) \(\Rightarrow\text{OW}=\frac{12.5\times2}{5} \) \(\Rightarrow\text{OW}=5.0\text{m} \) So, the height of the tower is 5.m.
TRIANGLES
90663
In \(\triangle\text{ABC},\text{DE }||\text{ BC} \) so that AD = (7x - 4)cm, AE = (5x - 2)cm, DB = (3x + 4)cm and EC = 3x cm. Then, we have:
1 x = 3
2 x = 5
3 x = 4
4 x = 2.5
Explanation:
Cx = 4 In \(\triangle\text{ABC},\text{DE }||\text{ BC} \) By Basic proportionality theorem, \(\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}} \) \(\Rightarrow\frac{7\text{x}-4}{3\text{x}+4}=\frac{5\text{x}-2}{3\text{x}} \) \(\Rightarrow21\text{x}^2-12\text{x}=15\text{x}^2+14\text{x}-8 \) \(\Rightarrow6\text{x}^2-26\text{x}+8=0 \) \(\Rightarrow3\text{x}^2-13\text{x}+4=0 \) \(\Rightarrow(\text{x}-4)(3\text{x}-1)=0 \) \(\Rightarrow\text{x}=4 \) or \(\text{x}=\frac{1}{3} \) If \(\text{x}=\frac{1}{3}, \) then \(\text{AD}=7\text{x}-4=7\Big(\frac{1}{3}\Big)-4=\frac{-5}{3}<0 \) This is not possible since length cannot be negative. \(\Rightarrow \)x = 4
TRIANGLES
90664
If triangles ABC and DEF are similar and AB = 4cm, DE = 6cm, EF = 9cm and FD = 12cm, the perimeter of triangle is:
1 22cm
2 20cm
3 21cm
4 18cm
Explanation:
D18cm ABC ~ DEF AB = 4cm, DE = 6cm, EF = 9cm and FD = 12cm \(\frac{\text{AB}}{\text{DE}}=\frac{\text{BC}}{\text{EF}}=\frac{\text{AC}}{\text{DF}} \) \(\frac{4}{6}=\frac{\text{BC}}{9}=\frac{\text{AC}}{12} \) \(\text{BC}=(\frac{4.9}{6})=6\text{cm} \) \(\text{AC}=(\frac{12.4}{6})=8\text{cm} \) Perimeter = AB + BC + AC = 4 + 6 + 8 =18cm
