90680
In an equilateral triangle ABC if \(\text{AD}\perp\text{BC}, \) then:
1 5AB\(^{2}\) = 4AD\(^{2}\)
2 3AB\(^{2}\) = 4AD\(^{2}\)
3 4AB\(^{2}\) = 3AD\(^{2}\)
4 2AB\(^{2}\) = 3AD\(^{2}\)
Explanation:
B3AB\(^{2}\) = 4AD\(^{2}\) \(\triangle\text{ABC} \) is an equilateral triangle and \(\text{AD}\perp\text{BC}. \)
In \(\triangle\text{ABD}, \) applying Pythagoras theorem, we get, \(\text{AB}^2=\text{AD}^2+\text{BD}^2 \) \(\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{BC}\Big)^2\Big(\because\text{BD}=\frac{1}{2}\text{BC}\Big) \) \(\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{AB}\Big)^2\Big(\because\text{AB}=\text{BC}\Big) \) \(\text{AB}^2=\text{AD}^2+\frac{1}{4}\text{AB}^2 \) \(3\text{AB}^2=4\text{AD}^2 \) We got the result as B.
TRIANGLES
90681
If the bisectore of an angle of a triangle bisects the opposite side then the triangle is:
1 Scalene
2 Equilateral
3 isosceles
4 Right-angled
Explanation:
Cisosceles
Let ABC be the triangle and AD be the bisector of \(\angle\text{A}. \) Also, AD bisects the opposite side that is BC. \(\Rightarrow\text{ BD} =\text{ DC} \dots\text{(i)} \) \(\Rightarrow\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}} \) \(\Rightarrow\frac{\text{AB}}{\text{AC}}=1\dots\text{(from(i))} \) \(\Rightarrow\text{AB}=\text{AC} \) So, the triangle is an isosceles triangle.
TRIANGLES
90682
A man goes 24m due west and then 7m due north. How far is he from the starting point?
1 31m.
2 17m.
3 25m.
4 26m.
Explanation:
C25m. Het a man be at O and goes to 24m due west and then 7m due north. Distance of man from starting point be OB So,
In right \(\triangle\text{ABO}, \) OB\(^{2}\) = AB\(^{2}\) + AO\(^{2}\) \(\Rightarrow \)OB\(^{2}\) = (7)\(^{2}\) + (24)\(^{2}\) \(\Rightarrow \)OB\(^{2}\) = 49 + 576 \(\Rightarrow \)OB\(^{2}\) = 625 \(\Rightarrow \)OB = 25m Thus, the distance of man from starting point is 25m.
TRIANGLES
90683
In the given figure, O is the point of intersection of two chords AB and CD such that OB = OD and \(\angle\text{AOC}=45^\circ. \) Then, \(\triangle\text{OAC} \) and \(\triangle\text{ODB} \) are:
1 Equilateral and similar.
2 Equilateral but not similar.
3 Isosceles and similar.
4 Isosceles but not similar.
Explanation:
CIsosceles and similar. In \(\triangle\text{AOC} \) and \(\triangle\text{ODB} \) \(\angle\text{AOC}=\angle\text{DOB} \) ....(Vertically opposite angles) \(\angle\text{OCA}=\angle\text{OBD} \) ....(angels in the same segment) \(\Rightarrow\triangle\text{OAC}\sim\triangle\text{ODB} \) ....(AA criterion for similarity) The two triangles are surely not equil ateral, Since the measure of every angle of an equilateral triangle is 60º. So, the triangles are isosceles and similar.
90680
In an equilateral triangle ABC if \(\text{AD}\perp\text{BC}, \) then:
1 5AB\(^{2}\) = 4AD\(^{2}\)
2 3AB\(^{2}\) = 4AD\(^{2}\)
3 4AB\(^{2}\) = 3AD\(^{2}\)
4 2AB\(^{2}\) = 3AD\(^{2}\)
Explanation:
B3AB\(^{2}\) = 4AD\(^{2}\) \(\triangle\text{ABC} \) is an equilateral triangle and \(\text{AD}\perp\text{BC}. \)
In \(\triangle\text{ABD}, \) applying Pythagoras theorem, we get, \(\text{AB}^2=\text{AD}^2+\text{BD}^2 \) \(\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{BC}\Big)^2\Big(\because\text{BD}=\frac{1}{2}\text{BC}\Big) \) \(\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{AB}\Big)^2\Big(\because\text{AB}=\text{BC}\Big) \) \(\text{AB}^2=\text{AD}^2+\frac{1}{4}\text{AB}^2 \) \(3\text{AB}^2=4\text{AD}^2 \) We got the result as B.
TRIANGLES
90681
If the bisectore of an angle of a triangle bisects the opposite side then the triangle is:
1 Scalene
2 Equilateral
3 isosceles
4 Right-angled
Explanation:
Cisosceles
Let ABC be the triangle and AD be the bisector of \(\angle\text{A}. \) Also, AD bisects the opposite side that is BC. \(\Rightarrow\text{ BD} =\text{ DC} \dots\text{(i)} \) \(\Rightarrow\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}} \) \(\Rightarrow\frac{\text{AB}}{\text{AC}}=1\dots\text{(from(i))} \) \(\Rightarrow\text{AB}=\text{AC} \) So, the triangle is an isosceles triangle.
TRIANGLES
90682
A man goes 24m due west and then 7m due north. How far is he from the starting point?
1 31m.
2 17m.
3 25m.
4 26m.
Explanation:
C25m. Het a man be at O and goes to 24m due west and then 7m due north. Distance of man from starting point be OB So,
In right \(\triangle\text{ABO}, \) OB\(^{2}\) = AB\(^{2}\) + AO\(^{2}\) \(\Rightarrow \)OB\(^{2}\) = (7)\(^{2}\) + (24)\(^{2}\) \(\Rightarrow \)OB\(^{2}\) = 49 + 576 \(\Rightarrow \)OB\(^{2}\) = 625 \(\Rightarrow \)OB = 25m Thus, the distance of man from starting point is 25m.
TRIANGLES
90683
In the given figure, O is the point of intersection of two chords AB and CD such that OB = OD and \(\angle\text{AOC}=45^\circ. \) Then, \(\triangle\text{OAC} \) and \(\triangle\text{ODB} \) are:
1 Equilateral and similar.
2 Equilateral but not similar.
3 Isosceles and similar.
4 Isosceles but not similar.
Explanation:
CIsosceles and similar. In \(\triangle\text{AOC} \) and \(\triangle\text{ODB} \) \(\angle\text{AOC}=\angle\text{DOB} \) ....(Vertically opposite angles) \(\angle\text{OCA}=\angle\text{OBD} \) ....(angels in the same segment) \(\Rightarrow\triangle\text{OAC}\sim\triangle\text{ODB} \) ....(AA criterion for similarity) The two triangles are surely not equil ateral, Since the measure of every angle of an equilateral triangle is 60º. So, the triangles are isosceles and similar.
90680
In an equilateral triangle ABC if \(\text{AD}\perp\text{BC}, \) then:
1 5AB\(^{2}\) = 4AD\(^{2}\)
2 3AB\(^{2}\) = 4AD\(^{2}\)
3 4AB\(^{2}\) = 3AD\(^{2}\)
4 2AB\(^{2}\) = 3AD\(^{2}\)
Explanation:
B3AB\(^{2}\) = 4AD\(^{2}\) \(\triangle\text{ABC} \) is an equilateral triangle and \(\text{AD}\perp\text{BC}. \)
In \(\triangle\text{ABD}, \) applying Pythagoras theorem, we get, \(\text{AB}^2=\text{AD}^2+\text{BD}^2 \) \(\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{BC}\Big)^2\Big(\because\text{BD}=\frac{1}{2}\text{BC}\Big) \) \(\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{AB}\Big)^2\Big(\because\text{AB}=\text{BC}\Big) \) \(\text{AB}^2=\text{AD}^2+\frac{1}{4}\text{AB}^2 \) \(3\text{AB}^2=4\text{AD}^2 \) We got the result as B.
TRIANGLES
90681
If the bisectore of an angle of a triangle bisects the opposite side then the triangle is:
1 Scalene
2 Equilateral
3 isosceles
4 Right-angled
Explanation:
Cisosceles
Let ABC be the triangle and AD be the bisector of \(\angle\text{A}. \) Also, AD bisects the opposite side that is BC. \(\Rightarrow\text{ BD} =\text{ DC} \dots\text{(i)} \) \(\Rightarrow\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}} \) \(\Rightarrow\frac{\text{AB}}{\text{AC}}=1\dots\text{(from(i))} \) \(\Rightarrow\text{AB}=\text{AC} \) So, the triangle is an isosceles triangle.
TRIANGLES
90682
A man goes 24m due west and then 7m due north. How far is he from the starting point?
1 31m.
2 17m.
3 25m.
4 26m.
Explanation:
C25m. Het a man be at O and goes to 24m due west and then 7m due north. Distance of man from starting point be OB So,
In right \(\triangle\text{ABO}, \) OB\(^{2}\) = AB\(^{2}\) + AO\(^{2}\) \(\Rightarrow \)OB\(^{2}\) = (7)\(^{2}\) + (24)\(^{2}\) \(\Rightarrow \)OB\(^{2}\) = 49 + 576 \(\Rightarrow \)OB\(^{2}\) = 625 \(\Rightarrow \)OB = 25m Thus, the distance of man from starting point is 25m.
TRIANGLES
90683
In the given figure, O is the point of intersection of two chords AB and CD such that OB = OD and \(\angle\text{AOC}=45^\circ. \) Then, \(\triangle\text{OAC} \) and \(\triangle\text{ODB} \) are:
1 Equilateral and similar.
2 Equilateral but not similar.
3 Isosceles and similar.
4 Isosceles but not similar.
Explanation:
CIsosceles and similar. In \(\triangle\text{AOC} \) and \(\triangle\text{ODB} \) \(\angle\text{AOC}=\angle\text{DOB} \) ....(Vertically opposite angles) \(\angle\text{OCA}=\angle\text{OBD} \) ....(angels in the same segment) \(\Rightarrow\triangle\text{OAC}\sim\triangle\text{ODB} \) ....(AA criterion for similarity) The two triangles are surely not equil ateral, Since the measure of every angle of an equilateral triangle is 60º. So, the triangles are isosceles and similar.
90680
In an equilateral triangle ABC if \(\text{AD}\perp\text{BC}, \) then:
1 5AB\(^{2}\) = 4AD\(^{2}\)
2 3AB\(^{2}\) = 4AD\(^{2}\)
3 4AB\(^{2}\) = 3AD\(^{2}\)
4 2AB\(^{2}\) = 3AD\(^{2}\)
Explanation:
B3AB\(^{2}\) = 4AD\(^{2}\) \(\triangle\text{ABC} \) is an equilateral triangle and \(\text{AD}\perp\text{BC}. \)
In \(\triangle\text{ABD}, \) applying Pythagoras theorem, we get, \(\text{AB}^2=\text{AD}^2+\text{BD}^2 \) \(\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{BC}\Big)^2\Big(\because\text{BD}=\frac{1}{2}\text{BC}\Big) \) \(\text{AB}^2=\text{AD}^2+\Big(\frac{1}{2}\text{AB}\Big)^2\Big(\because\text{AB}=\text{BC}\Big) \) \(\text{AB}^2=\text{AD}^2+\frac{1}{4}\text{AB}^2 \) \(3\text{AB}^2=4\text{AD}^2 \) We got the result as B.
TRIANGLES
90681
If the bisectore of an angle of a triangle bisects the opposite side then the triangle is:
1 Scalene
2 Equilateral
3 isosceles
4 Right-angled
Explanation:
Cisosceles
Let ABC be the triangle and AD be the bisector of \(\angle\text{A}. \) Also, AD bisects the opposite side that is BC. \(\Rightarrow\text{ BD} =\text{ DC} \dots\text{(i)} \) \(\Rightarrow\frac{\text{AB}}{\text{AC}}=\frac{\text{BD}}{\text{DC}} \) \(\Rightarrow\frac{\text{AB}}{\text{AC}}=1\dots\text{(from(i))} \) \(\Rightarrow\text{AB}=\text{AC} \) So, the triangle is an isosceles triangle.
TRIANGLES
90682
A man goes 24m due west and then 7m due north. How far is he from the starting point?
1 31m.
2 17m.
3 25m.
4 26m.
Explanation:
C25m. Het a man be at O and goes to 24m due west and then 7m due north. Distance of man from starting point be OB So,
In right \(\triangle\text{ABO}, \) OB\(^{2}\) = AB\(^{2}\) + AO\(^{2}\) \(\Rightarrow \)OB\(^{2}\) = (7)\(^{2}\) + (24)\(^{2}\) \(\Rightarrow \)OB\(^{2}\) = 49 + 576 \(\Rightarrow \)OB\(^{2}\) = 625 \(\Rightarrow \)OB = 25m Thus, the distance of man from starting point is 25m.
TRIANGLES
90683
In the given figure, O is the point of intersection of two chords AB and CD such that OB = OD and \(\angle\text{AOC}=45^\circ. \) Then, \(\triangle\text{OAC} \) and \(\triangle\text{ODB} \) are:
1 Equilateral and similar.
2 Equilateral but not similar.
3 Isosceles and similar.
4 Isosceles but not similar.
Explanation:
CIsosceles and similar. In \(\triangle\text{AOC} \) and \(\triangle\text{ODB} \) \(\angle\text{AOC}=\angle\text{DOB} \) ....(Vertically opposite angles) \(\angle\text{OCA}=\angle\text{OBD} \) ....(angels in the same segment) \(\Rightarrow\triangle\text{OAC}\sim\triangle\text{ODB} \) ....(AA criterion for similarity) The two triangles are surely not equil ateral, Since the measure of every angle of an equilateral triangle is 60º. So, the triangles are isosceles and similar.
