90702
In trapezium ABCD, if \(\text{AB}\parallel\text{DC},\text{AB}\parallel\text{DC} \), AB = 9cm, DC = 6cm and BD = 12cm, then BO is equal to:
1 7.4cm.
2 7cm..
3 7.5cm.
4 7.2cm
Explanation:
D7.2cm In \(\triangle\text{COD } \)and \(\triangle\text{AOB} \) \(\angle\text{DOC}=\angle\text{AOB} \) [vertically opposite] And \(\angle\text{DCO}=\angle\text{OAB} \) [Alternate angles] \(\Rightarrow\triangle\text{COD}\sim\triangle\text{AOB} \) [similarity] Let OB = xcm \(\therefore\frac{\text{AB}}{\text{CD}}=\frac{\text{OB}}{\text{OD}} \) \(\Rightarrow\frac{9}{6}=\frac{\text{x}}{12-\text{x}} \) \(\Rightarrow108-9\text{x}=6\text{x} \) \(\Rightarrow15\text{x}=108 \) \(\Rightarrow\text{x}=7.2\text{cm} \)
TRIANGLES
90705
In a \(\triangle\text{ABC}, \) AD is the bisector of \(\angle\text{BAC}. \) If AB = 8cm, BD = 6cm and DC = 3cm. Find AC:
1 4cm.
2 6cm.
3 3cm.
4 8cm.
Explanation:
A4cm. Given: In a \(\triangle\text{ABC}, \) AD is the bisector of angle BAC. AB = 8cm, and DC = 3cm and BD = 6cm. To find: AC We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. Hence, \(\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}} \) \(\frac{8}{\text{AC}}=\frac{6}{3} \) \(\text{AC}=\frac{8\times3}{6} \) \(\text{AC}=4\text{cm} \) Hence we got the result A.
TRIANGLES
90706
\(\triangle\text{ABC} \) is an isosceles triangle in which \(\angle\text{C}=90^\circ \) If AC = 6cm, then AB =
1 \(6\sqrt{2}\text{cm}. \)
2 \(6\text{cm}. \)
3 \(2\sqrt{6}\text{cm}. \)
4 \(4\sqrt{2}\text{cm}. \)
Explanation:
A\(6\sqrt{2}\text{cm}. \) \(\triangle\text{ABC} \) is an isosceles with \(\angle\text{C}=90^\circ \)
AC = BC AC = 6cm AB\(^{2}\) = AC\(^{2}\) + BC\(^{2}\) (Pythagoras Theorem) (6)\(^{2}\) + (6)\(^{2}\) = 36 + 36 = 72 (AC = BC) \(\text{AB}=\sqrt{72}=\sqrt{(36\times2}=6\sqrt{2}\text{cm}. \)
TRIANGLES
90707
In the given figure the measure of \(\angle\text{D} \) and \(\angle\text{F} \) are respectively:
1 50°, 40°
2 20°, 30°
3 40°, 50°
4 30°, 20°
Explanation:
B20°, 30°
\(\triangle\text{ABC} \) and \(\triangle\text{DEF,} \) \(\frac{\text{AB}}{\text{AC}}=\frac{\text{EF}}{\text{ED}} \) \(\angle\text{A}=\angle\text{E}=130^\circ \) \(\triangle\text{ABC}\sim\triangle\text{EFD} \) (SAS Similarity) \(\therefore\angle\text{F}=\angle\text{B}=30^\circ \) \(\angle\text{D}=\angle\text{C}=20^\circ \) Hence the correct answer is B.
TRIANGLES
90708
\(\triangle\text{ABC} \) is such that AB = 3cm, BC = 2cm and CA = 2.5cm. If \(\triangle\text{DEF}\sim\triangle\text{ABC} \) and EF = 4cm, then perimeter of \(\triangle\text{DEF} \) is:
1 7.5cm.
2 15cm.
3 22.5cm.
4 30cm.
Explanation:
B15cm. \(\triangle\text{DEF}\sim\triangle\text{ABC} \) AB = 3cm, BC = 2cm, CA = 2.5cm, EF = 4cm. \(\triangle\text{s} \) are similar \(\frac{\text{DE}}{\text{AB}}=\frac{\text{EF}}{\text{BC}}=\frac{\text{FD}}{\text{CA}} \) \(\Rightarrow\frac{\text{DE}}{3}=\frac{4}{2}=\frac{\text{FD}}{2.5} \) Now \(\frac{\text{DE}}{3}=\frac{4}{2} \) \(\Rightarrow\text{DE}=\frac{3\times4}{2}=6\text{cm} \) and \(\text{FD}=\frac{4}{2}\Rightarrow\text{FD}=\frac{4\times2.5}{2}=5\text{cm} \) \(\therefore \) Perimeter of \(\triangle\text{DEF} \) \(=6+4+5=15\text{cm}. \)
90702
In trapezium ABCD, if \(\text{AB}\parallel\text{DC},\text{AB}\parallel\text{DC} \), AB = 9cm, DC = 6cm and BD = 12cm, then BO is equal to:
1 7.4cm.
2 7cm..
3 7.5cm.
4 7.2cm
Explanation:
D7.2cm In \(\triangle\text{COD } \)and \(\triangle\text{AOB} \) \(\angle\text{DOC}=\angle\text{AOB} \) [vertically opposite] And \(\angle\text{DCO}=\angle\text{OAB} \) [Alternate angles] \(\Rightarrow\triangle\text{COD}\sim\triangle\text{AOB} \) [similarity] Let OB = xcm \(\therefore\frac{\text{AB}}{\text{CD}}=\frac{\text{OB}}{\text{OD}} \) \(\Rightarrow\frac{9}{6}=\frac{\text{x}}{12-\text{x}} \) \(\Rightarrow108-9\text{x}=6\text{x} \) \(\Rightarrow15\text{x}=108 \) \(\Rightarrow\text{x}=7.2\text{cm} \)
TRIANGLES
90705
In a \(\triangle\text{ABC}, \) AD is the bisector of \(\angle\text{BAC}. \) If AB = 8cm, BD = 6cm and DC = 3cm. Find AC:
1 4cm.
2 6cm.
3 3cm.
4 8cm.
Explanation:
A4cm. Given: In a \(\triangle\text{ABC}, \) AD is the bisector of angle BAC. AB = 8cm, and DC = 3cm and BD = 6cm. To find: AC We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. Hence, \(\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}} \) \(\frac{8}{\text{AC}}=\frac{6}{3} \) \(\text{AC}=\frac{8\times3}{6} \) \(\text{AC}=4\text{cm} \) Hence we got the result A.
TRIANGLES
90706
\(\triangle\text{ABC} \) is an isosceles triangle in which \(\angle\text{C}=90^\circ \) If AC = 6cm, then AB =
1 \(6\sqrt{2}\text{cm}. \)
2 \(6\text{cm}. \)
3 \(2\sqrt{6}\text{cm}. \)
4 \(4\sqrt{2}\text{cm}. \)
Explanation:
A\(6\sqrt{2}\text{cm}. \) \(\triangle\text{ABC} \) is an isosceles with \(\angle\text{C}=90^\circ \)
AC = BC AC = 6cm AB\(^{2}\) = AC\(^{2}\) + BC\(^{2}\) (Pythagoras Theorem) (6)\(^{2}\) + (6)\(^{2}\) = 36 + 36 = 72 (AC = BC) \(\text{AB}=\sqrt{72}=\sqrt{(36\times2}=6\sqrt{2}\text{cm}. \)
TRIANGLES
90707
In the given figure the measure of \(\angle\text{D} \) and \(\angle\text{F} \) are respectively:
1 50°, 40°
2 20°, 30°
3 40°, 50°
4 30°, 20°
Explanation:
B20°, 30°
\(\triangle\text{ABC} \) and \(\triangle\text{DEF,} \) \(\frac{\text{AB}}{\text{AC}}=\frac{\text{EF}}{\text{ED}} \) \(\angle\text{A}=\angle\text{E}=130^\circ \) \(\triangle\text{ABC}\sim\triangle\text{EFD} \) (SAS Similarity) \(\therefore\angle\text{F}=\angle\text{B}=30^\circ \) \(\angle\text{D}=\angle\text{C}=20^\circ \) Hence the correct answer is B.
TRIANGLES
90708
\(\triangle\text{ABC} \) is such that AB = 3cm, BC = 2cm and CA = 2.5cm. If \(\triangle\text{DEF}\sim\triangle\text{ABC} \) and EF = 4cm, then perimeter of \(\triangle\text{DEF} \) is:
1 7.5cm.
2 15cm.
3 22.5cm.
4 30cm.
Explanation:
B15cm. \(\triangle\text{DEF}\sim\triangle\text{ABC} \) AB = 3cm, BC = 2cm, CA = 2.5cm, EF = 4cm. \(\triangle\text{s} \) are similar \(\frac{\text{DE}}{\text{AB}}=\frac{\text{EF}}{\text{BC}}=\frac{\text{FD}}{\text{CA}} \) \(\Rightarrow\frac{\text{DE}}{3}=\frac{4}{2}=\frac{\text{FD}}{2.5} \) Now \(\frac{\text{DE}}{3}=\frac{4}{2} \) \(\Rightarrow\text{DE}=\frac{3\times4}{2}=6\text{cm} \) and \(\text{FD}=\frac{4}{2}\Rightarrow\text{FD}=\frac{4\times2.5}{2}=5\text{cm} \) \(\therefore \) Perimeter of \(\triangle\text{DEF} \) \(=6+4+5=15\text{cm}. \)
90702
In trapezium ABCD, if \(\text{AB}\parallel\text{DC},\text{AB}\parallel\text{DC} \), AB = 9cm, DC = 6cm and BD = 12cm, then BO is equal to:
1 7.4cm.
2 7cm..
3 7.5cm.
4 7.2cm
Explanation:
D7.2cm In \(\triangle\text{COD } \)and \(\triangle\text{AOB} \) \(\angle\text{DOC}=\angle\text{AOB} \) [vertically opposite] And \(\angle\text{DCO}=\angle\text{OAB} \) [Alternate angles] \(\Rightarrow\triangle\text{COD}\sim\triangle\text{AOB} \) [similarity] Let OB = xcm \(\therefore\frac{\text{AB}}{\text{CD}}=\frac{\text{OB}}{\text{OD}} \) \(\Rightarrow\frac{9}{6}=\frac{\text{x}}{12-\text{x}} \) \(\Rightarrow108-9\text{x}=6\text{x} \) \(\Rightarrow15\text{x}=108 \) \(\Rightarrow\text{x}=7.2\text{cm} \)
TRIANGLES
90705
In a \(\triangle\text{ABC}, \) AD is the bisector of \(\angle\text{BAC}. \) If AB = 8cm, BD = 6cm and DC = 3cm. Find AC:
1 4cm.
2 6cm.
3 3cm.
4 8cm.
Explanation:
A4cm. Given: In a \(\triangle\text{ABC}, \) AD is the bisector of angle BAC. AB = 8cm, and DC = 3cm and BD = 6cm. To find: AC We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. Hence, \(\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}} \) \(\frac{8}{\text{AC}}=\frac{6}{3} \) \(\text{AC}=\frac{8\times3}{6} \) \(\text{AC}=4\text{cm} \) Hence we got the result A.
TRIANGLES
90706
\(\triangle\text{ABC} \) is an isosceles triangle in which \(\angle\text{C}=90^\circ \) If AC = 6cm, then AB =
1 \(6\sqrt{2}\text{cm}. \)
2 \(6\text{cm}. \)
3 \(2\sqrt{6}\text{cm}. \)
4 \(4\sqrt{2}\text{cm}. \)
Explanation:
A\(6\sqrt{2}\text{cm}. \) \(\triangle\text{ABC} \) is an isosceles with \(\angle\text{C}=90^\circ \)
AC = BC AC = 6cm AB\(^{2}\) = AC\(^{2}\) + BC\(^{2}\) (Pythagoras Theorem) (6)\(^{2}\) + (6)\(^{2}\) = 36 + 36 = 72 (AC = BC) \(\text{AB}=\sqrt{72}=\sqrt{(36\times2}=6\sqrt{2}\text{cm}. \)
TRIANGLES
90707
In the given figure the measure of \(\angle\text{D} \) and \(\angle\text{F} \) are respectively:
1 50°, 40°
2 20°, 30°
3 40°, 50°
4 30°, 20°
Explanation:
B20°, 30°
\(\triangle\text{ABC} \) and \(\triangle\text{DEF,} \) \(\frac{\text{AB}}{\text{AC}}=\frac{\text{EF}}{\text{ED}} \) \(\angle\text{A}=\angle\text{E}=130^\circ \) \(\triangle\text{ABC}\sim\triangle\text{EFD} \) (SAS Similarity) \(\therefore\angle\text{F}=\angle\text{B}=30^\circ \) \(\angle\text{D}=\angle\text{C}=20^\circ \) Hence the correct answer is B.
TRIANGLES
90708
\(\triangle\text{ABC} \) is such that AB = 3cm, BC = 2cm and CA = 2.5cm. If \(\triangle\text{DEF}\sim\triangle\text{ABC} \) and EF = 4cm, then perimeter of \(\triangle\text{DEF} \) is:
1 7.5cm.
2 15cm.
3 22.5cm.
4 30cm.
Explanation:
B15cm. \(\triangle\text{DEF}\sim\triangle\text{ABC} \) AB = 3cm, BC = 2cm, CA = 2.5cm, EF = 4cm. \(\triangle\text{s} \) are similar \(\frac{\text{DE}}{\text{AB}}=\frac{\text{EF}}{\text{BC}}=\frac{\text{FD}}{\text{CA}} \) \(\Rightarrow\frac{\text{DE}}{3}=\frac{4}{2}=\frac{\text{FD}}{2.5} \) Now \(\frac{\text{DE}}{3}=\frac{4}{2} \) \(\Rightarrow\text{DE}=\frac{3\times4}{2}=6\text{cm} \) and \(\text{FD}=\frac{4}{2}\Rightarrow\text{FD}=\frac{4\times2.5}{2}=5\text{cm} \) \(\therefore \) Perimeter of \(\triangle\text{DEF} \) \(=6+4+5=15\text{cm}. \)
90702
In trapezium ABCD, if \(\text{AB}\parallel\text{DC},\text{AB}\parallel\text{DC} \), AB = 9cm, DC = 6cm and BD = 12cm, then BO is equal to:
1 7.4cm.
2 7cm..
3 7.5cm.
4 7.2cm
Explanation:
D7.2cm In \(\triangle\text{COD } \)and \(\triangle\text{AOB} \) \(\angle\text{DOC}=\angle\text{AOB} \) [vertically opposite] And \(\angle\text{DCO}=\angle\text{OAB} \) [Alternate angles] \(\Rightarrow\triangle\text{COD}\sim\triangle\text{AOB} \) [similarity] Let OB = xcm \(\therefore\frac{\text{AB}}{\text{CD}}=\frac{\text{OB}}{\text{OD}} \) \(\Rightarrow\frac{9}{6}=\frac{\text{x}}{12-\text{x}} \) \(\Rightarrow108-9\text{x}=6\text{x} \) \(\Rightarrow15\text{x}=108 \) \(\Rightarrow\text{x}=7.2\text{cm} \)
TRIANGLES
90705
In a \(\triangle\text{ABC}, \) AD is the bisector of \(\angle\text{BAC}. \) If AB = 8cm, BD = 6cm and DC = 3cm. Find AC:
1 4cm.
2 6cm.
3 3cm.
4 8cm.
Explanation:
A4cm. Given: In a \(\triangle\text{ABC}, \) AD is the bisector of angle BAC. AB = 8cm, and DC = 3cm and BD = 6cm. To find: AC We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. Hence, \(\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}} \) \(\frac{8}{\text{AC}}=\frac{6}{3} \) \(\text{AC}=\frac{8\times3}{6} \) \(\text{AC}=4\text{cm} \) Hence we got the result A.
TRIANGLES
90706
\(\triangle\text{ABC} \) is an isosceles triangle in which \(\angle\text{C}=90^\circ \) If AC = 6cm, then AB =
1 \(6\sqrt{2}\text{cm}. \)
2 \(6\text{cm}. \)
3 \(2\sqrt{6}\text{cm}. \)
4 \(4\sqrt{2}\text{cm}. \)
Explanation:
A\(6\sqrt{2}\text{cm}. \) \(\triangle\text{ABC} \) is an isosceles with \(\angle\text{C}=90^\circ \)
AC = BC AC = 6cm AB\(^{2}\) = AC\(^{2}\) + BC\(^{2}\) (Pythagoras Theorem) (6)\(^{2}\) + (6)\(^{2}\) = 36 + 36 = 72 (AC = BC) \(\text{AB}=\sqrt{72}=\sqrt{(36\times2}=6\sqrt{2}\text{cm}. \)
TRIANGLES
90707
In the given figure the measure of \(\angle\text{D} \) and \(\angle\text{F} \) are respectively:
1 50°, 40°
2 20°, 30°
3 40°, 50°
4 30°, 20°
Explanation:
B20°, 30°
\(\triangle\text{ABC} \) and \(\triangle\text{DEF,} \) \(\frac{\text{AB}}{\text{AC}}=\frac{\text{EF}}{\text{ED}} \) \(\angle\text{A}=\angle\text{E}=130^\circ \) \(\triangle\text{ABC}\sim\triangle\text{EFD} \) (SAS Similarity) \(\therefore\angle\text{F}=\angle\text{B}=30^\circ \) \(\angle\text{D}=\angle\text{C}=20^\circ \) Hence the correct answer is B.
TRIANGLES
90708
\(\triangle\text{ABC} \) is such that AB = 3cm, BC = 2cm and CA = 2.5cm. If \(\triangle\text{DEF}\sim\triangle\text{ABC} \) and EF = 4cm, then perimeter of \(\triangle\text{DEF} \) is:
1 7.5cm.
2 15cm.
3 22.5cm.
4 30cm.
Explanation:
B15cm. \(\triangle\text{DEF}\sim\triangle\text{ABC} \) AB = 3cm, BC = 2cm, CA = 2.5cm, EF = 4cm. \(\triangle\text{s} \) are similar \(\frac{\text{DE}}{\text{AB}}=\frac{\text{EF}}{\text{BC}}=\frac{\text{FD}}{\text{CA}} \) \(\Rightarrow\frac{\text{DE}}{3}=\frac{4}{2}=\frac{\text{FD}}{2.5} \) Now \(\frac{\text{DE}}{3}=\frac{4}{2} \) \(\Rightarrow\text{DE}=\frac{3\times4}{2}=6\text{cm} \) and \(\text{FD}=\frac{4}{2}\Rightarrow\text{FD}=\frac{4\times2.5}{2}=5\text{cm} \) \(\therefore \) Perimeter of \(\triangle\text{DEF} \) \(=6+4+5=15\text{cm}. \)
90702
In trapezium ABCD, if \(\text{AB}\parallel\text{DC},\text{AB}\parallel\text{DC} \), AB = 9cm, DC = 6cm and BD = 12cm, then BO is equal to:
1 7.4cm.
2 7cm..
3 7.5cm.
4 7.2cm
Explanation:
D7.2cm In \(\triangle\text{COD } \)and \(\triangle\text{AOB} \) \(\angle\text{DOC}=\angle\text{AOB} \) [vertically opposite] And \(\angle\text{DCO}=\angle\text{OAB} \) [Alternate angles] \(\Rightarrow\triangle\text{COD}\sim\triangle\text{AOB} \) [similarity] Let OB = xcm \(\therefore\frac{\text{AB}}{\text{CD}}=\frac{\text{OB}}{\text{OD}} \) \(\Rightarrow\frac{9}{6}=\frac{\text{x}}{12-\text{x}} \) \(\Rightarrow108-9\text{x}=6\text{x} \) \(\Rightarrow15\text{x}=108 \) \(\Rightarrow\text{x}=7.2\text{cm} \)
TRIANGLES
90705
In a \(\triangle\text{ABC}, \) AD is the bisector of \(\angle\text{BAC}. \) If AB = 8cm, BD = 6cm and DC = 3cm. Find AC:
1 4cm.
2 6cm.
3 3cm.
4 8cm.
Explanation:
A4cm. Given: In a \(\triangle\text{ABC}, \) AD is the bisector of angle BAC. AB = 8cm, and DC = 3cm and BD = 6cm. To find: AC We know that the internal bisector of angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle. Hence, \(\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}} \) \(\frac{8}{\text{AC}}=\frac{6}{3} \) \(\text{AC}=\frac{8\times3}{6} \) \(\text{AC}=4\text{cm} \) Hence we got the result A.
TRIANGLES
90706
\(\triangle\text{ABC} \) is an isosceles triangle in which \(\angle\text{C}=90^\circ \) If AC = 6cm, then AB =
1 \(6\sqrt{2}\text{cm}. \)
2 \(6\text{cm}. \)
3 \(2\sqrt{6}\text{cm}. \)
4 \(4\sqrt{2}\text{cm}. \)
Explanation:
A\(6\sqrt{2}\text{cm}. \) \(\triangle\text{ABC} \) is an isosceles with \(\angle\text{C}=90^\circ \)
AC = BC AC = 6cm AB\(^{2}\) = AC\(^{2}\) + BC\(^{2}\) (Pythagoras Theorem) (6)\(^{2}\) + (6)\(^{2}\) = 36 + 36 = 72 (AC = BC) \(\text{AB}=\sqrt{72}=\sqrt{(36\times2}=6\sqrt{2}\text{cm}. \)
TRIANGLES
90707
In the given figure the measure of \(\angle\text{D} \) and \(\angle\text{F} \) are respectively:
1 50°, 40°
2 20°, 30°
3 40°, 50°
4 30°, 20°
Explanation:
B20°, 30°
\(\triangle\text{ABC} \) and \(\triangle\text{DEF,} \) \(\frac{\text{AB}}{\text{AC}}=\frac{\text{EF}}{\text{ED}} \) \(\angle\text{A}=\angle\text{E}=130^\circ \) \(\triangle\text{ABC}\sim\triangle\text{EFD} \) (SAS Similarity) \(\therefore\angle\text{F}=\angle\text{B}=30^\circ \) \(\angle\text{D}=\angle\text{C}=20^\circ \) Hence the correct answer is B.
TRIANGLES
90708
\(\triangle\text{ABC} \) is such that AB = 3cm, BC = 2cm and CA = 2.5cm. If \(\triangle\text{DEF}\sim\triangle\text{ABC} \) and EF = 4cm, then perimeter of \(\triangle\text{DEF} \) is:
1 7.5cm.
2 15cm.
3 22.5cm.
4 30cm.
Explanation:
B15cm. \(\triangle\text{DEF}\sim\triangle\text{ABC} \) AB = 3cm, BC = 2cm, CA = 2.5cm, EF = 4cm. \(\triangle\text{s} \) are similar \(\frac{\text{DE}}{\text{AB}}=\frac{\text{EF}}{\text{BC}}=\frac{\text{FD}}{\text{CA}} \) \(\Rightarrow\frac{\text{DE}}{3}=\frac{4}{2}=\frac{\text{FD}}{2.5} \) Now \(\frac{\text{DE}}{3}=\frac{4}{2} \) \(\Rightarrow\text{DE}=\frac{3\times4}{2}=6\text{cm} \) and \(\text{FD}=\frac{4}{2}\Rightarrow\text{FD}=\frac{4\times2.5}{2}=5\text{cm} \) \(\therefore \) Perimeter of \(\triangle\text{DEF} \) \(=6+4+5=15\text{cm}. \)
