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TRIANGLES

90702 In trapezium ABCD, if $AB ∥ DC, AB ∥ DC$ , 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 $△ COD$ and $△ AOB$
$∠ DOC = ∠ AOB$ [vertically opposite]
And $∠ DCO = ∠ OAB$ [Alternate angles]
$\Rightarrow △ COD \sim △ AOB$ [similarity]
Let OB = xcm
$∴ \frac{AB}{CD} = \frac{OB}{OD}$
$\Rightarrow \frac{9}{6} = \frac{x}{12 - x}$
$\Rightarrow 108 - 9 x = 6 x$
$\Rightarrow 15 x = 108$
$\Rightarrow x = 7.2 cm$

TRIANGLES

90705 In a $△ ABC,$ AD is the bisector of $∠ BAC .$ If AB = 8cm, BD = 6cm and DC = 3cm. Find AC:

1 4cm.

2 6cm.

3 3cm.

4 8cm.

Explanation:

A4cm.
Given: In a $△ ABC,$ 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,
$\frac{AB}{AC} = \frac{BD}{DC}$
$\frac{8}{AC} = \frac{6}{3}$
$AC = \frac{8 \times 3}{6}$
$AC = 4 cm$
Hence we got the result A.

TRIANGLES

90706 $△ ABC$ is an isosceles triangle in which $∠ C = 90^{\circ}$ If AC = 6cm, then AB =

1

6 \sqrt{2} cm .

2

6 cm .

3

2 \sqrt{6} cm .

4

4 \sqrt{2} cm .

Explanation:

A $6 \sqrt{2} cm .$
$△ ABC$ is an isosceles with $∠ C = 90^{\circ}$

AC = BC
AC = 6cm
AB $^{2}$ = AC $^{2}$ + BC $^{2}$ (Pythagoras Theorem)
(6) $^{2}$ + (6) $^{2}$ = 36 + 36 = 72 (AC = BC)
$AB = \sqrt{72} = \sqrt{(36 \times 2} = 6 \sqrt{2} cm .$

TRIANGLES

90707 In the given figure the measure of $∠ D$ and $∠ F$ are respectively:

1 50°, 40°

2 20°, 30°

3 40°, 50°

4 30°, 20°

Explanation:

B20°, 30°

$△ ABC$ and $△ DEF,$
$\frac{AB}{AC} = \frac{EF}{ED}$
$∠ A = ∠ E = 130^{\circ}$
$△ ABC \sim △ EFD$ (SAS Similarity)
$∴ ∠ F = ∠ B = 30^{\circ}$
$∠ D = ∠ C = 20^{\circ}$
Hence the correct answer is B.

TRIANGLES

90708 $△ ABC$ is such that AB = 3cm, BC = 2cm and CA = 2.5cm. If $△ DEF \sim △ ABC$ and EF = 4cm, then perimeter of $△ DEF$ is:

1 7.5cm.

2 15cm.

3 22.5cm.

4 30cm.

Explanation:

B15cm.
$△ DEF \sim △ ABC$
AB = 3cm, BC = 2cm, CA = 2.5cm, EF = 4cm.
$△ s$ are similar
$\frac{DE}{AB} = \frac{EF}{BC} = \frac{FD}{CA}$
$\Rightarrow \frac{DE}{3} = \frac{4}{2} = \frac{FD}{2.5}$
Now $\frac{DE}{3} = \frac{4}{2}$
$\Rightarrow DE = \frac{3 \times 4}{2} = 6 cm$
and $FD = \frac{4}{2} \Rightarrow FD = \frac{4 \times 2.5}{2} = 5 cm$
$∴$ Perimeter of $△ DEF$
$= 6 + 4 + 5 = 15 cm .$

TRIANGLES

90702 In trapezium ABCD, if $AB ∥ DC, AB ∥ DC$ , 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 $△ COD$ and $△ AOB$
$∠ DOC = ∠ AOB$ [vertically opposite]
And $∠ DCO = ∠ OAB$ [Alternate angles]
$\Rightarrow △ COD \sim △ AOB$ [similarity]
Let OB = xcm
$∴ \frac{AB}{CD} = \frac{OB}{OD}$
$\Rightarrow \frac{9}{6} = \frac{x}{12 - x}$
$\Rightarrow 108 - 9 x = 6 x$
$\Rightarrow 15 x = 108$
$\Rightarrow x = 7.2 cm$

TRIANGLES

90705 In a $△ ABC,$ AD is the bisector of $∠ BAC .$ If AB = 8cm, BD = 6cm and DC = 3cm. Find AC:

1 4cm.

2 6cm.

3 3cm.

4 8cm.

Explanation:

A4cm.
Given: In a $△ ABC,$ 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,
$\frac{AB}{AC} = \frac{BD}{DC}$
$\frac{8}{AC} = \frac{6}{3}$
$AC = \frac{8 \times 3}{6}$
$AC = 4 cm$
Hence we got the result A.

TRIANGLES

90706 $△ ABC$ is an isosceles triangle in which $∠ C = 90^{\circ}$ If AC = 6cm, then AB =

1

6 \sqrt{2} cm .

2

6 cm .

3

2 \sqrt{6} cm .

4

4 \sqrt{2} cm .

Explanation:

A $6 \sqrt{2} cm .$
$△ ABC$ is an isosceles with $∠ C = 90^{\circ}$

AC = BC
AC = 6cm
AB $^{2}$ = AC $^{2}$ + BC $^{2}$ (Pythagoras Theorem)
(6) $^{2}$ + (6) $^{2}$ = 36 + 36 = 72 (AC = BC)
$AB = \sqrt{72} = \sqrt{(36 \times 2} = 6 \sqrt{2} cm .$

TRIANGLES

90707 In the given figure the measure of $∠ D$ and $∠ F$ are respectively:

1 50°, 40°

2 20°, 30°

3 40°, 50°

4 30°, 20°

Explanation:

B20°, 30°

$△ ABC$ and $△ DEF,$
$\frac{AB}{AC} = \frac{EF}{ED}$
$∠ A = ∠ E = 130^{\circ}$
$△ ABC \sim △ EFD$ (SAS Similarity)
$∴ ∠ F = ∠ B = 30^{\circ}$
$∠ D = ∠ C = 20^{\circ}$
Hence the correct answer is B.

TRIANGLES

90708 $△ ABC$ is such that AB = 3cm, BC = 2cm and CA = 2.5cm. If $△ DEF \sim △ ABC$ and EF = 4cm, then perimeter of $△ DEF$ is:

1 7.5cm.

2 15cm.

3 22.5cm.

4 30cm.

Explanation:

B15cm.
$△ DEF \sim △ ABC$
AB = 3cm, BC = 2cm, CA = 2.5cm, EF = 4cm.
$△ s$ are similar
$\frac{DE}{AB} = \frac{EF}{BC} = \frac{FD}{CA}$
$\Rightarrow \frac{DE}{3} = \frac{4}{2} = \frac{FD}{2.5}$
Now $\frac{DE}{3} = \frac{4}{2}$
$\Rightarrow DE = \frac{3 \times 4}{2} = 6 cm$
and $FD = \frac{4}{2} \Rightarrow FD = \frac{4 \times 2.5}{2} = 5 cm$
$∴$ Perimeter of $△ DEF$
$= 6 + 4 + 5 = 15 cm .$

TRIANGLES

90702 In trapezium ABCD, if $AB ∥ DC, AB ∥ DC$ , 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 $△ COD$ and $△ AOB$
$∠ DOC = ∠ AOB$ [vertically opposite]
And $∠ DCO = ∠ OAB$ [Alternate angles]
$\Rightarrow △ COD \sim △ AOB$ [similarity]
Let OB = xcm
$∴ \frac{AB}{CD} = \frac{OB}{OD}$
$\Rightarrow \frac{9}{6} = \frac{x}{12 - x}$
$\Rightarrow 108 - 9 x = 6 x$
$\Rightarrow 15 x = 108$
$\Rightarrow x = 7.2 cm$

TRIANGLES

90705 In a $△ ABC,$ AD is the bisector of $∠ BAC .$ If AB = 8cm, BD = 6cm and DC = 3cm. Find AC:

1 4cm.

2 6cm.

3 3cm.

4 8cm.

Explanation:

A4cm.
Given: In a $△ ABC,$ 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,
$\frac{AB}{AC} = \frac{BD}{DC}$
$\frac{8}{AC} = \frac{6}{3}$
$AC = \frac{8 \times 3}{6}$
$AC = 4 cm$
Hence we got the result A.

TRIANGLES

90706 $△ ABC$ is an isosceles triangle in which $∠ C = 90^{\circ}$ If AC = 6cm, then AB =

1

6 \sqrt{2} cm .

2

6 cm .

3

2 \sqrt{6} cm .

4

4 \sqrt{2} cm .

Explanation:

A $6 \sqrt{2} cm .$
$△ ABC$ is an isosceles with $∠ C = 90^{\circ}$

AC = BC
AC = 6cm
AB $^{2}$ = AC $^{2}$ + BC $^{2}$ (Pythagoras Theorem)
(6) $^{2}$ + (6) $^{2}$ = 36 + 36 = 72 (AC = BC)
$AB = \sqrt{72} = \sqrt{(36 \times 2} = 6 \sqrt{2} cm .$

TRIANGLES

90707 In the given figure the measure of $∠ D$ and $∠ F$ are respectively:

1 50°, 40°

2 20°, 30°

3 40°, 50°

4 30°, 20°

Explanation:

B20°, 30°

$△ ABC$ and $△ DEF,$
$\frac{AB}{AC} = \frac{EF}{ED}$
$∠ A = ∠ E = 130^{\circ}$
$△ ABC \sim △ EFD$ (SAS Similarity)
$∴ ∠ F = ∠ B = 30^{\circ}$
$∠ D = ∠ C = 20^{\circ}$
Hence the correct answer is B.

TRIANGLES

90708 $△ ABC$ is such that AB = 3cm, BC = 2cm and CA = 2.5cm. If $△ DEF \sim △ ABC$ and EF = 4cm, then perimeter of $△ DEF$ is:

1 7.5cm.

2 15cm.

3 22.5cm.

4 30cm.

Explanation:

B15cm.
$△ DEF \sim △ ABC$
AB = 3cm, BC = 2cm, CA = 2.5cm, EF = 4cm.
$△ s$ are similar
$\frac{DE}{AB} = \frac{EF}{BC} = \frac{FD}{CA}$
$\Rightarrow \frac{DE}{3} = \frac{4}{2} = \frac{FD}{2.5}$
Now $\frac{DE}{3} = \frac{4}{2}$
$\Rightarrow DE = \frac{3 \times 4}{2} = 6 cm$
and $FD = \frac{4}{2} \Rightarrow FD = \frac{4 \times 2.5}{2} = 5 cm$
$∴$ Perimeter of $△ DEF$
$= 6 + 4 + 5 = 15 cm .$

TRIANGLES

90702 In trapezium ABCD, if $AB ∥ DC, AB ∥ DC$ , 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 $△ COD$ and $△ AOB$
$∠ DOC = ∠ AOB$ [vertically opposite]
And $∠ DCO = ∠ OAB$ [Alternate angles]
$\Rightarrow △ COD \sim △ AOB$ [similarity]
Let OB = xcm
$∴ \frac{AB}{CD} = \frac{OB}{OD}$
$\Rightarrow \frac{9}{6} = \frac{x}{12 - x}$
$\Rightarrow 108 - 9 x = 6 x$
$\Rightarrow 15 x = 108$
$\Rightarrow x = 7.2 cm$

TRIANGLES

90705 In a $△ ABC,$ AD is the bisector of $∠ BAC .$ If AB = 8cm, BD = 6cm and DC = 3cm. Find AC:

1 4cm.

2 6cm.

3 3cm.

4 8cm.

Explanation:

A4cm.
Given: In a $△ ABC,$ 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,
$\frac{AB}{AC} = \frac{BD}{DC}$
$\frac{8}{AC} = \frac{6}{3}$
$AC = \frac{8 \times 3}{6}$
$AC = 4 cm$
Hence we got the result A.

TRIANGLES

90706 $△ ABC$ is an isosceles triangle in which $∠ C = 90^{\circ}$ If AC = 6cm, then AB =

1

6 \sqrt{2} cm .

2

6 cm .

3

2 \sqrt{6} cm .

4

4 \sqrt{2} cm .

Explanation:

A $6 \sqrt{2} cm .$
$△ ABC$ is an isosceles with $∠ C = 90^{\circ}$

AC = BC
AC = 6cm
AB $^{2}$ = AC $^{2}$ + BC $^{2}$ (Pythagoras Theorem)
(6) $^{2}$ + (6) $^{2}$ = 36 + 36 = 72 (AC = BC)
$AB = \sqrt{72} = \sqrt{(36 \times 2} = 6 \sqrt{2} cm .$

TRIANGLES

90707 In the given figure the measure of $∠ D$ and $∠ F$ are respectively:

1 50°, 40°

2 20°, 30°

3 40°, 50°

4 30°, 20°

Explanation:

B20°, 30°

$△ ABC$ and $△ DEF,$
$\frac{AB}{AC} = \frac{EF}{ED}$
$∠ A = ∠ E = 130^{\circ}$
$△ ABC \sim △ EFD$ (SAS Similarity)
$∴ ∠ F = ∠ B = 30^{\circ}$
$∠ D = ∠ C = 20^{\circ}$
Hence the correct answer is B.

TRIANGLES

90708 $△ ABC$ is such that AB = 3cm, BC = 2cm and CA = 2.5cm. If $△ DEF \sim △ ABC$ and EF = 4cm, then perimeter of $△ DEF$ is:

1 7.5cm.

2 15cm.

3 22.5cm.

4 30cm.

Explanation:

B15cm.
$△ DEF \sim △ ABC$
AB = 3cm, BC = 2cm, CA = 2.5cm, EF = 4cm.
$△ s$ are similar
$\frac{DE}{AB} = \frac{EF}{BC} = \frac{FD}{CA}$
$\Rightarrow \frac{DE}{3} = \frac{4}{2} = \frac{FD}{2.5}$
Now $\frac{DE}{3} = \frac{4}{2}$
$\Rightarrow DE = \frac{3 \times 4}{2} = 6 cm$
and $FD = \frac{4}{2} \Rightarrow FD = \frac{4 \times 2.5}{2} = 5 cm$
$∴$ Perimeter of $△ DEF$
$= 6 + 4 + 5 = 15 cm .$

TRIANGLES

90702 In trapezium ABCD, if $AB ∥ DC, AB ∥ DC$ , 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 $△ COD$ and $△ AOB$
$∠ DOC = ∠ AOB$ [vertically opposite]
And $∠ DCO = ∠ OAB$ [Alternate angles]
$\Rightarrow △ COD \sim △ AOB$ [similarity]
Let OB = xcm
$∴ \frac{AB}{CD} = \frac{OB}{OD}$
$\Rightarrow \frac{9}{6} = \frac{x}{12 - x}$
$\Rightarrow 108 - 9 x = 6 x$
$\Rightarrow 15 x = 108$
$\Rightarrow x = 7.2 cm$

TRIANGLES

90705 In a $△ ABC,$ AD is the bisector of $∠ BAC .$ If AB = 8cm, BD = 6cm and DC = 3cm. Find AC:

1 4cm.

2 6cm.

3 3cm.

4 8cm.

Explanation:

A4cm.
Given: In a $△ ABC,$ 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,
$\frac{AB}{AC} = \frac{BD}{DC}$
$\frac{8}{AC} = \frac{6}{3}$
$AC = \frac{8 \times 3}{6}$
$AC = 4 cm$
Hence we got the result A.

TRIANGLES

90706 $△ ABC$ is an isosceles triangle in which $∠ C = 90^{\circ}$ If AC = 6cm, then AB =

1

6 \sqrt{2} cm .

2

6 cm .

3

2 \sqrt{6} cm .

4

4 \sqrt{2} cm .

Explanation:

A $6 \sqrt{2} cm .$
$△ ABC$ is an isosceles with $∠ C = 90^{\circ}$

AC = BC
AC = 6cm
AB $^{2}$ = AC $^{2}$ + BC $^{2}$ (Pythagoras Theorem)
(6) $^{2}$ + (6) $^{2}$ = 36 + 36 = 72 (AC = BC)
$AB = \sqrt{72} = \sqrt{(36 \times 2} = 6 \sqrt{2} cm .$

TRIANGLES

90707 In the given figure the measure of $∠ D$ and $∠ F$ are respectively:

1 50°, 40°

2 20°, 30°

3 40°, 50°

4 30°, 20°

Explanation:

B20°, 30°

$△ ABC$ and $△ DEF,$
$\frac{AB}{AC} = \frac{EF}{ED}$
$∠ A = ∠ E = 130^{\circ}$
$△ ABC \sim △ EFD$ (SAS Similarity)
$∴ ∠ F = ∠ B = 30^{\circ}$
$∠ D = ∠ C = 20^{\circ}$
Hence the correct answer is B.

TRIANGLES

90708 $△ ABC$ is such that AB = 3cm, BC = 2cm and CA = 2.5cm. If $△ DEF \sim △ ABC$ and EF = 4cm, then perimeter of $△ DEF$ is:

1 7.5cm.

2 15cm.

3 22.5cm.

4 30cm.

Explanation:

B15cm.
$△ DEF \sim △ ABC$
AB = 3cm, BC = 2cm, CA = 2.5cm, EF = 4cm.
$△ s$ are similar
$\frac{DE}{AB} = \frac{EF}{BC} = \frac{FD}{CA}$
$\Rightarrow \frac{DE}{3} = \frac{4}{2} = \frac{FD}{2.5}$
Now $\frac{DE}{3} = \frac{4}{2}$
$\Rightarrow DE = \frac{3 \times 4}{2} = 6 cm$
and $FD = \frac{4}{2} \Rightarrow FD = \frac{4 \times 2.5}{2} = 5 cm$
$∴$ Perimeter of $△ DEF$
$= 6 + 4 + 5 = 15 cm .$