90655
In \(\triangle\text{ABC}, \) it is given that AB = 9cm, BC = 6cm and CA = 7.5cm. Also, \(\triangle\text{DEF} \) is given such that EF = 8cm and \(\triangle\text{DEF}\sim\triangle\text{ABC}. \) Then, perimeter of \(\triangle\text{DEF} \) 1s:
1 22.5cm
2 25cm
3 27cm
4 30cm
Explanation:
D30cm \(\triangle\text{ABC}\sim\triangle\text{DEF} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{\text{Perimeters of }\triangle\text{ABC}}=\frac{\text{EF}}{\text{BC}} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{\text{AB+BC+AC}}=\frac{8}{6} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{9+6+7.5}=\frac{8}{6} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{22.5}=\frac{4}{3} \) \(\Rightarrow\text{Perimeters of }\triangle\text{DEF}=\frac{4\times22.5}{3} \) \(\Rightarrow\text{Perimeters of }\triangle\text{DEF}=30\text{cm} \)
TRIANGLES
90656
XY is drawn parallel to the base BC of a \(\triangle\text{ABC} \) cutting AB at X and AC at Y. If AB = 4 BX and YC = 2cm, then AY =
1 2cm.
2 4cm.
3 6cm.
4 8cm.
Explanation:
C6cm. In \(\triangle\text{ABC}, \) XY || BC AB = 4BX, YC = 2cm
\(\therefore \) AB = 4BX ? AX + BX = 4BX \(\Rightarrow \)AX = 4BX - BX = 3BX Let AY = x \(\because \) In \(\triangle\text{ABC}, \) XY || BC \(\frac{\text{AX}}{\text{BX}}=\frac{\text{AY}}{\text{CY}}\Rightarrow\frac{\text{3BX}}{\text{BX}}=\frac{\text{x}}{2} \) \(\Rightarrow\frac{3}{1}=\frac{\text{x}}{2}\Rightarrow\text{x}=3\times2=6 \) \(\therefore \) AY = 6cm.
TRIANGLES
90657
A chord of a circle of radius 10cm subtends a right angle at the centre. The length of the chord (in cm) is:
1 \(5\sqrt{2} \)
2 \(10\sqrt{2} \)
3 \(\frac{5}{\sqrt{2}} \)
4 \(10\sqrt{3} \)
Explanation:
B\(10\sqrt{2} \)
In right \(\triangle\text{OAB}, \) AB\(^{2}\) = OA\(^{2}\) + OB\(^{2}\) (Pythagoras Theorem) \(\Rightarrow \)AB\(^{2}\) = (10)\(^{2}\) + (10)\(^{2}\) (OA = OB = 10cm) \(\Rightarrow \)AB\(^{2}\) = 100 + 100 = 200 \(\Rightarrow\text{AB}=\sqrt{200}=10\sqrt{2}\text{cm} \) Thus, the length of the chord is \(10\sqrt{2}\text{cm}. \) Hence, the correct answer is option B.
TRIANGLES
90768
In the given figure, \(\text{DE}\parallel\text{BC} \) AB = 15cm, BD = 6cm, AC = 25cm, then AE is equal to:
1 20cm.
2 18cm.
3 10cm.
4 15cm.
Explanation:
D15cm. Since \(\text{DE} \parallel\text{BC} \),then using Thales theorem, \(\Rightarrow\frac{\text{AB}}{\text{DB}}=\frac{\text{AC}}{\text{EC}} \) \(\Rightarrow\frac{15}{6}=\frac{25}{\text{EC}} \) \(\Rightarrow\text{EC}=10 \text{ cm} \) Now, AE = AC - EC = 25 - 10 = 15cm
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TRIANGLES
90655
In \(\triangle\text{ABC}, \) it is given that AB = 9cm, BC = 6cm and CA = 7.5cm. Also, \(\triangle\text{DEF} \) is given such that EF = 8cm and \(\triangle\text{DEF}\sim\triangle\text{ABC}. \) Then, perimeter of \(\triangle\text{DEF} \) 1s:
1 22.5cm
2 25cm
3 27cm
4 30cm
Explanation:
D30cm \(\triangle\text{ABC}\sim\triangle\text{DEF} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{\text{Perimeters of }\triangle\text{ABC}}=\frac{\text{EF}}{\text{BC}} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{\text{AB+BC+AC}}=\frac{8}{6} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{9+6+7.5}=\frac{8}{6} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{22.5}=\frac{4}{3} \) \(\Rightarrow\text{Perimeters of }\triangle\text{DEF}=\frac{4\times22.5}{3} \) \(\Rightarrow\text{Perimeters of }\triangle\text{DEF}=30\text{cm} \)
TRIANGLES
90656
XY is drawn parallel to the base BC of a \(\triangle\text{ABC} \) cutting AB at X and AC at Y. If AB = 4 BX and YC = 2cm, then AY =
1 2cm.
2 4cm.
3 6cm.
4 8cm.
Explanation:
C6cm. In \(\triangle\text{ABC}, \) XY || BC AB = 4BX, YC = 2cm
\(\therefore \) AB = 4BX ? AX + BX = 4BX \(\Rightarrow \)AX = 4BX - BX = 3BX Let AY = x \(\because \) In \(\triangle\text{ABC}, \) XY || BC \(\frac{\text{AX}}{\text{BX}}=\frac{\text{AY}}{\text{CY}}\Rightarrow\frac{\text{3BX}}{\text{BX}}=\frac{\text{x}}{2} \) \(\Rightarrow\frac{3}{1}=\frac{\text{x}}{2}\Rightarrow\text{x}=3\times2=6 \) \(\therefore \) AY = 6cm.
TRIANGLES
90657
A chord of a circle of radius 10cm subtends a right angle at the centre. The length of the chord (in cm) is:
1 \(5\sqrt{2} \)
2 \(10\sqrt{2} \)
3 \(\frac{5}{\sqrt{2}} \)
4 \(10\sqrt{3} \)
Explanation:
B\(10\sqrt{2} \)
In right \(\triangle\text{OAB}, \) AB\(^{2}\) = OA\(^{2}\) + OB\(^{2}\) (Pythagoras Theorem) \(\Rightarrow \)AB\(^{2}\) = (10)\(^{2}\) + (10)\(^{2}\) (OA = OB = 10cm) \(\Rightarrow \)AB\(^{2}\) = 100 + 100 = 200 \(\Rightarrow\text{AB}=\sqrt{200}=10\sqrt{2}\text{cm} \) Thus, the length of the chord is \(10\sqrt{2}\text{cm}. \) Hence, the correct answer is option B.
TRIANGLES
90768
In the given figure, \(\text{DE}\parallel\text{BC} \) AB = 15cm, BD = 6cm, AC = 25cm, then AE is equal to:
1 20cm.
2 18cm.
3 10cm.
4 15cm.
Explanation:
D15cm. Since \(\text{DE} \parallel\text{BC} \),then using Thales theorem, \(\Rightarrow\frac{\text{AB}}{\text{DB}}=\frac{\text{AC}}{\text{EC}} \) \(\Rightarrow\frac{15}{6}=\frac{25}{\text{EC}} \) \(\Rightarrow\text{EC}=10 \text{ cm} \) Now, AE = AC - EC = 25 - 10 = 15cm
90655
In \(\triangle\text{ABC}, \) it is given that AB = 9cm, BC = 6cm and CA = 7.5cm. Also, \(\triangle\text{DEF} \) is given such that EF = 8cm and \(\triangle\text{DEF}\sim\triangle\text{ABC}. \) Then, perimeter of \(\triangle\text{DEF} \) 1s:
1 22.5cm
2 25cm
3 27cm
4 30cm
Explanation:
D30cm \(\triangle\text{ABC}\sim\triangle\text{DEF} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{\text{Perimeters of }\triangle\text{ABC}}=\frac{\text{EF}}{\text{BC}} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{\text{AB+BC+AC}}=\frac{8}{6} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{9+6+7.5}=\frac{8}{6} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{22.5}=\frac{4}{3} \) \(\Rightarrow\text{Perimeters of }\triangle\text{DEF}=\frac{4\times22.5}{3} \) \(\Rightarrow\text{Perimeters of }\triangle\text{DEF}=30\text{cm} \)
TRIANGLES
90656
XY is drawn parallel to the base BC of a \(\triangle\text{ABC} \) cutting AB at X and AC at Y. If AB = 4 BX and YC = 2cm, then AY =
1 2cm.
2 4cm.
3 6cm.
4 8cm.
Explanation:
C6cm. In \(\triangle\text{ABC}, \) XY || BC AB = 4BX, YC = 2cm
\(\therefore \) AB = 4BX ? AX + BX = 4BX \(\Rightarrow \)AX = 4BX - BX = 3BX Let AY = x \(\because \) In \(\triangle\text{ABC}, \) XY || BC \(\frac{\text{AX}}{\text{BX}}=\frac{\text{AY}}{\text{CY}}\Rightarrow\frac{\text{3BX}}{\text{BX}}=\frac{\text{x}}{2} \) \(\Rightarrow\frac{3}{1}=\frac{\text{x}}{2}\Rightarrow\text{x}=3\times2=6 \) \(\therefore \) AY = 6cm.
TRIANGLES
90657
A chord of a circle of radius 10cm subtends a right angle at the centre. The length of the chord (in cm) is:
1 \(5\sqrt{2} \)
2 \(10\sqrt{2} \)
3 \(\frac{5}{\sqrt{2}} \)
4 \(10\sqrt{3} \)
Explanation:
B\(10\sqrt{2} \)
In right \(\triangle\text{OAB}, \) AB\(^{2}\) = OA\(^{2}\) + OB\(^{2}\) (Pythagoras Theorem) \(\Rightarrow \)AB\(^{2}\) = (10)\(^{2}\) + (10)\(^{2}\) (OA = OB = 10cm) \(\Rightarrow \)AB\(^{2}\) = 100 + 100 = 200 \(\Rightarrow\text{AB}=\sqrt{200}=10\sqrt{2}\text{cm} \) Thus, the length of the chord is \(10\sqrt{2}\text{cm}. \) Hence, the correct answer is option B.
TRIANGLES
90768
In the given figure, \(\text{DE}\parallel\text{BC} \) AB = 15cm, BD = 6cm, AC = 25cm, then AE is equal to:
1 20cm.
2 18cm.
3 10cm.
4 15cm.
Explanation:
D15cm. Since \(\text{DE} \parallel\text{BC} \),then using Thales theorem, \(\Rightarrow\frac{\text{AB}}{\text{DB}}=\frac{\text{AC}}{\text{EC}} \) \(\Rightarrow\frac{15}{6}=\frac{25}{\text{EC}} \) \(\Rightarrow\text{EC}=10 \text{ cm} \) Now, AE = AC - EC = 25 - 10 = 15cm
90655
In \(\triangle\text{ABC}, \) it is given that AB = 9cm, BC = 6cm and CA = 7.5cm. Also, \(\triangle\text{DEF} \) is given such that EF = 8cm and \(\triangle\text{DEF}\sim\triangle\text{ABC}. \) Then, perimeter of \(\triangle\text{DEF} \) 1s:
1 22.5cm
2 25cm
3 27cm
4 30cm
Explanation:
D30cm \(\triangle\text{ABC}\sim\triangle\text{DEF} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{\text{Perimeters of }\triangle\text{ABC}}=\frac{\text{EF}}{\text{BC}} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{\text{AB+BC+AC}}=\frac{8}{6} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{9+6+7.5}=\frac{8}{6} \) \(\Rightarrow\frac{\text{Perimeters of }\triangle\text{DEF}}{22.5}=\frac{4}{3} \) \(\Rightarrow\text{Perimeters of }\triangle\text{DEF}=\frac{4\times22.5}{3} \) \(\Rightarrow\text{Perimeters of }\triangle\text{DEF}=30\text{cm} \)
TRIANGLES
90656
XY is drawn parallel to the base BC of a \(\triangle\text{ABC} \) cutting AB at X and AC at Y. If AB = 4 BX and YC = 2cm, then AY =
1 2cm.
2 4cm.
3 6cm.
4 8cm.
Explanation:
C6cm. In \(\triangle\text{ABC}, \) XY || BC AB = 4BX, YC = 2cm
\(\therefore \) AB = 4BX ? AX + BX = 4BX \(\Rightarrow \)AX = 4BX - BX = 3BX Let AY = x \(\because \) In \(\triangle\text{ABC}, \) XY || BC \(\frac{\text{AX}}{\text{BX}}=\frac{\text{AY}}{\text{CY}}\Rightarrow\frac{\text{3BX}}{\text{BX}}=\frac{\text{x}}{2} \) \(\Rightarrow\frac{3}{1}=\frac{\text{x}}{2}\Rightarrow\text{x}=3\times2=6 \) \(\therefore \) AY = 6cm.
TRIANGLES
90657
A chord of a circle of radius 10cm subtends a right angle at the centre. The length of the chord (in cm) is:
1 \(5\sqrt{2} \)
2 \(10\sqrt{2} \)
3 \(\frac{5}{\sqrt{2}} \)
4 \(10\sqrt{3} \)
Explanation:
B\(10\sqrt{2} \)
In right \(\triangle\text{OAB}, \) AB\(^{2}\) = OA\(^{2}\) + OB\(^{2}\) (Pythagoras Theorem) \(\Rightarrow \)AB\(^{2}\) = (10)\(^{2}\) + (10)\(^{2}\) (OA = OB = 10cm) \(\Rightarrow \)AB\(^{2}\) = 100 + 100 = 200 \(\Rightarrow\text{AB}=\sqrt{200}=10\sqrt{2}\text{cm} \) Thus, the length of the chord is \(10\sqrt{2}\text{cm}. \) Hence, the correct answer is option B.
TRIANGLES
90768
In the given figure, \(\text{DE}\parallel\text{BC} \) AB = 15cm, BD = 6cm, AC = 25cm, then AE is equal to:
1 20cm.
2 18cm.
3 10cm.
4 15cm.
Explanation:
D15cm. Since \(\text{DE} \parallel\text{BC} \),then using Thales theorem, \(\Rightarrow\frac{\text{AB}}{\text{DB}}=\frac{\text{AC}}{\text{EC}} \) \(\Rightarrow\frac{15}{6}=\frac{25}{\text{EC}} \) \(\Rightarrow\text{EC}=10 \text{ cm} \) Now, AE = AC - EC = 25 - 10 = 15cm
